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A framework for solving parabolic partial differential equations

A new algorithm solves complicated partial differential equations by breaking them down into simpler problems, potentially guiding computer graphics and geometry processing.


Computer graphics and geometry processing research provide the tools needed to simulate physical phenomena like fire and flames, aiding the creation of visual effects in video games and movies as well as the fabrication of complex geometric shapes using tools like 3D printing.

Under the hood, mathematical problems called partial differential equations (PDEs) model these natural processes. Among the many PDEs used in physics and computer graphics, a class called second-order parabolic PDEs explain how phenomena can become smooth over time. The most famous example in this class is the heat equation, which predicts how heat diffuses along a surface or in a volume over time.

Researchers in geometry processing have designed numerous algorithms to solve these problems on curved surfaces, but their methods often apply only to linear problems or to a single PDE. A more general approach by researchers from MIT’s Computer Science and Artificial Intelligence Laboratory (CSAIL) tackles a general class of these potentially nonlinear problems.

In a paper recently published in the Transactions on Graphics journal and presented at the SIGGRAPH conference, they describe an algorithm that solves different nonlinear parabolic PDEs on triangle meshes by splitting them into three simpler equations that can be solved with techniques graphics researchers already have in their software toolkit. This framework can help better analyze shapes and model complex dynamical processes.

“We provide a recipe: If you want to numerically solve a second-order parabolic PDE, you can follow a set of three steps,” says lead author Leticia Mattos Da Silva SM ’23, an MIT PhD student in electrical engineering and computer science (EECS) and CSAIL affiliate. “For each of the steps in this approach, you’re solving a simpler problem using simpler tools from geometry processing, but at the end, you get a solution to the more challenging second-order parabolic PDE.”

To accomplish this, Da Silva and her coauthors used Strang splitting, a technique that allows geometry processing researchers to break the PDE down into problems they know how to solve efficiently.

First, their algorithm advances a solution forward in time by solving the heat equation (also called the “diffusion equation”), which models how heat from a source spreads over a shape. Picture using a blow torch to warm up a metal plate — this equation describes how heat from that spot would diffuse over it. 
This step can be completed easily with linear algebra.

Now, imagine that the parabolic PDE has additional nonlinear behaviors that are not described by the spread of heat. This is where the second step of the algorithm comes in: it accounts for the nonlinear piece by solving a Hamilton-Jacobi (HJ) equation, a first-order nonlinear PDE.

While generic HJ equations can be hard to solve, Mattos Da Silva and coauthors prove that their splitting method applied to many important PDEs yields an HJ equation that can be solved via convex optimization algorithms. Convex optimization is a standard tool for which researchers in geometry processing already have efficient and reliable software. In the final step, the algorithm advances a solution forward in time using the heat equation again to advance the more complex second-order parabolic PDE forward in time.

Among other applications, the framework could help simulate fire and flames more efficiently. “There’s a huge pipeline that creates a video with flames being simulated, but at the heart of it is a PDE solver,” says Mattos Da Silva. For these pipelines, an essential step is solving the G-equation, a nonlinear parabolic PDE that models the front propagation of the flame and can be solved using the researchers’ framework.

The team’s algorithm can also solve the diffusion equation in the logarithmic domain, where it becomes nonlinear. Senior author Justin Solomon, associate professor of EECS and leader of the CSAIL Geometric Data Processing Group, previously developed a state-of-the-art technique for optimal transport that requires taking the logarithm of the result of heat diffusion. Mattos Da Silva’s framework provided more reliable computations by doing diffusion directly in the logarithmic domain. This enabled a more stable way to, for example, find a geometric notion of average among distributions on surface meshes like a model of a koala.

Even though their framework focuses on general, nonlinear problems, it can also be used to solve linear PDE. For instance, the method solves the Fokker-Planck equation, where heat diffuses in a linear way, but there are additional terms that drift in the same direction heat is spreading. In a straightforward application, the approach modeled how swirls would evolve over the surface of a triangulated sphere. The result resembles purple-and-brown latte art.

The researchers note that this project is a starting point for tackling the nonlinearity in other PDEs that appear in graphics and geometry processing head-on. For example, they focused on static surfaces but would like to apply their work to moving ones, too. Moreover, their framework solves problems involving a single parabolic PDE, but the team would also like to tackle problems involving coupled parabolic PDE. These types of problems arise in biology and chemistry, where the equation describing the evolution of each agent in a mixture, for example, is linked to the others’ equations.

Mattos Da Silva and Solomon wrote the paper with Oded Stein, assistant professor at the University of Southern California’s Viterbi School of Engineering. Their work was supported, in part, by an MIT Schwarzman College of Computing Fellowship funded by Google, a MathWorks Fellowship, the Swiss National Science Foundation, the U.S. Army Research Office, the U.S. Air Force Office of Scientific Research, the U.S. National Science Foundation, MIT-IBM Watson AI Lab, the Toyota-CSAIL Joint Research Center, Adobe Systems, and Google Research.

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August 28, 2024

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Framework for solving parabolic partial differential equations could guide computer graphics and geometry processing

by Alex Shipps, Massachusetts Institute of Technology

Computer graphics and geometry processing research provide the tools needed to simulate physical phenomena like fire and flames, aiding the creation of visual effects in video games and movies as well as the fabrication of complex geometric shapes using tools like 3D printing.

Under the hood, mathematical problems called partial differential equations (PDEs) model these natural processes. Among the many PDEs used in physics and computer graphics , a class called second-order parabolic PDEs explain how phenomena can become smooth over time. The most famous example in this class is the heat equation, which predicts how heat diffuses along a surface or in a volume over time.

Researchers in geometry processing have designed numerous algorithms to solve these problems on curved surfaces, but their methods often apply only to linear problems or to a single PDE. A more general approach by researchers from MIT's Computer Science and Artificial Intelligence Laboratory (CSAIL) tackles a general class of these potentially nonlinear problems.

In a paper recently published in the ACM Transactions on Graphics journal and presented at the SIGGRAPH conference, they describe an algorithm that solves different nonlinear parabolic PDEs on triangle meshes by splitting them into three simpler equations that can be solved with techniques graphics researchers already have in their software toolkit. This framework can help better analyze shapes and model complex dynamical processes.

"We provide a recipe: If you want to numerically solve a second-order parabolic PDE, you can follow a set of three steps," says lead author Leticia Mattos Da Silva, an MIT Ph.D. student in electrical engineering and computer science (EECS) and CSAIL affiliate. "For each of the steps in this approach, you're solving a simpler problem using simpler tools from geometry processing, but at the end, you get a solution to the more challenging second-order parabolic PDE."

To accomplish this, Mattos Da Silva and her co-authors used Strang splitting, a technique that allows geometry processing researchers to break the PDE down into problems they know how to solve efficiently.

First, their algorithm advances a solution forward in time by solving the heat equation (also called the "diffusion equation"), which models how heat from a source spreads over a shape. Picture using a blow torch to warm up a metal plate—this equation describes how heat from that spot would diffuse over it. This step can be completed easily with linear algebra.

Now, imagine that the parabolic PDE has additional nonlinear behaviors that are not described by the spread of heat. This is where the second step of the algorithm comes in: it accounts for the nonlinear piece by solving a Hamilton-Jacobi (HJ) equation, a first-order nonlinear PDE.

While generic HJ equations can be hard to solve, Mattos Da Silva and co-authors prove that their splitting method applied to many important PDEs yields an HJ equation that can be solved via convex optimization algorithms. Convex optimization is a standard tool for which researchers in geometry processing already have efficient and reliable software. In the final step, the algorithm advances a solution forward in time using the heat equation again to advance the more complex second-order parabolic PDE forward in time.

Among other applications, the framework could help simulate fire and flames more efficiently. "There's a huge pipeline that creates a video with flames being simulated, but at the heart of it is a PDE solver," says Mattos Da Silva. For these pipelines, an essential step is solving the G-equation, a nonlinear parabolic PDE that models the front propagation of the flame and can be solved using the researchers' framework.

The team's algorithm can also solve the diffusion equation in the logarithmic domain, where it becomes nonlinear. Senior author Justin Solomon, associate professor of EECS and leader of the CSAIL Geometric Data Processing Group, had previously developed a state-of-the-art technique for optimal transport that requires taking the logarithm of the result of heat diffusion.

Mattos Da Silva's framework provided more reliable computations by doing diffusion directly in the logarithmic domain. This enabled a more stable way, for example, to find a geometric notion of average among distributions on surface meshes like a model of a koala.

Even though their framework focuses on general, nonlinear problems, it can also be used to solve linear PDE. For instance, the method solves the Fokker-Planck equation, where heat diffuses in a linear way, but there are additional terms that drift in the same direction heat is spreading. In a straightforward application, the approach modeled how swirls would evolve over the surface of a triangulated sphere. The result resembles purple-and-brown latte art.

The researchers note that this project is a starting point for tackling the nonlinearity in other PDEs that appear in graphics and geometry processing head-on. For example, they focused on static surfaces but would like to apply their work to moving ones, too. Moreover, their framework solves problems involving a single parabolic PDE, but the team would also like to tackle problems involving coupled parabolic PDE. These types of problems arise in biology and chemistry, where the equation describing the evolution of each agent in a mixture, for example, is linked to the others' equations.

Mattos Da Silva and Solomon wrote the paper with Oded Stein, assistant professor at the University of Southern California's Viterbi School of Engineering.

Journal information: ACM Transactions on Graphics

Provided by Massachusetts Institute of Technology

This story is republished courtesy of MIT News ( web.mit.edu/newsoffice/ ), a popular site that covers news about MIT research, innovation and teaching.

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A framework for solving parabolic partial differential equations

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Computer graphics and geometry processing research provide the tools needed to simulate physical phenomena like fire and flames, aiding the creation of visual effects in video games and movies as well as the fabrication of complex geometric shapes using tools like 3D printing.

Under the hood, mathematical problems called partial differential equations (PDEs) model these natural processes. Among the many PDEs used in physics and computer graphics, a class called second-order parabolic PDEs explain how phenomena can become smooth over time. The most famous example in this class is the heat equation, which predicts how heat diffuses along a surface or in a volume over time.

Researchers in geometry processing have designed numerous algorithms to solve these problems on curved surfaces, but their methods often apply only to linear problems or to a single PDE. A more general approach by researchers from MIT’s Computer Science and Artificial Intelligence Laboratory (CSAIL) tackles a general class of these potentially nonlinear problems.  In a paper recently published in the Transactions on Graphics journal and presented at the SIGGRAPH conference, they describe an algorithm that solves different nonlinear parabolic PDEs on triangle meshes by splitting them into three simpler equations that can be solved with techniques graphics researchers already have in their software toolkit. This framework can help better analyze shapes and model complex dynamical processes.

“We provide a recipe: If you want to numerically solve a second-order parabolic PDE, you can follow a set of three steps,” says lead author Leticia Mattos Da Silva SM ’23, an MIT PhD student in electrical engineering and computer science (EECS) and CSAIL affiliate. “For each of the steps in this approach, you’re solving a simpler problem using simpler tools from geometry processing, but at the end, you get a solution to the more challenging second-order parabolic PDE.” To accomplish this, Da Silva and her coauthors used Strang splitting, a technique that allows geometry processing researchers to break the PDE down into problems they know how to solve efficiently.

First, their algorithm advances a solution forward in time by solving the heat equation (also called the “diffusion equation”), which models how heat from a source spreads over a shape. Picture using a blow torch to warm up a metal plate — this equation describes how heat from that spot would diffuse over it. 
This step can be completed easily with linear algebra.

Now, imagine that the parabolic PDE has additional nonlinear behaviors that are not described by the spread of heat. This is where the second step of the algorithm comes in: it accounts for the nonlinear piece by solving a Hamilton-Jacobi (HJ) equation, a first-order nonlinear PDE.  While generic HJ equations can be hard to solve, Mattos Da Silva and coauthors prove that their splitting method applied to many important PDEs yields an HJ equation that can be solved via convex optimization algorithms. Convex optimization is a standard tool for which researchers in geometry processing already have efficient and reliable software. In the final step, the algorithm advances a solution forward in time using the heat equation again to advance the more complex second-order parabolic PDE forward in time.


Among other applications, the framework could help simulate fire and flames more efficiently. “There’s a huge pipeline that creates a video with flames being simulated, but at the heart of it is a PDE solver,” says Mattos Da Silva. For these pipelines, an essential step is solving the G-equation, a nonlinear parabolic PDE that models the front propagation of the flame and can be solved using the researchers’ framework.

The team’s algorithm can also solve the diffusion equation in the logarithmic domain, where it becomes nonlinear. Senior author Justin Solomon, associate professor of EECS and leader of the CSAIL Geometric Data Processing Group, previously developed a state-of-the-art technique for optimal transport that requires taking the logarithm of the result of heat diffusion. Mattos Da Silva’s framework provided more reliable computations by doing diffusion directly in the logarithmic domain. This enabled a more stable way to, for example, find a geometric notion of average among distributions on surface meshes like a model of a koala. Even though their framework focuses on general, nonlinear problems, it can also be used to solve linear PDE. For instance, the method solves the Fokker-Planck equation, where heat diffuses in a linear way, but there are additional terms that drift in the same direction heat is spreading. In a straightforward application, the approach modeled how swirls would evolve over the surface of a triangulated sphere. The result resembles purple-and-brown latte art.

The researchers note that this project is a starting point for tackling the nonlinearity in other PDEs that appear in graphics and geometry processing head-on. For example, they focused on static surfaces but would like to apply their work to moving ones, too. Moreover, their framework solves problems involving a single parabolic PDE, but the team would also like to tackle problems involving coupled parabolic PDE. These types of problems arise in biology and chemistry, where the equation describing the evolution of each agent in a mixture, for example, is linked to the others’ equations.

Mattos Da Silva and Solomon wrote the paper with Oded Stein, assistant professor at the University of Southern California’s Viterbi School of Engineering. Their work was supported, in part, by an MIT Schwarzman College of Computing Fellowship funded by Google, a MathWorks Fellowship, the Swiss National Science Foundation, the U.S. Army Research Office, the U.S. Air Force Office of Scientific Research, the U.S. National Science Foundation, MIT-IBM Watson AI Lab, the Toyota-CSAIL Joint Research Center, Adobe Systems, and Google Research.

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Lesson Solved problems on volume of spheres

Volume of Sphere Exercises

Volume of sphere practice problems with answers.

You will find ten (10) practice problems below regarding the sphere’s volume. This set of problems should provide you enough practice on how to use the formula to find volume of the sphere.

Use the formula below as a guide. Good luck!

Problem 1: Find the volume of a sphere that has a radius of [latex]5[/latex] yards. Use [latex]\pi=3.14[/latex]. Round your answer to the nearest tenth.

Therefore, the volume of the sphere is approximately [latex]523.3[/latex] [latex]\text{yd}^3[/latex].

Problem 2: The radius of a sphere is [latex]1.5[/latex] centimeters. What is its volume? Use [latex]\pi=3.1416[/latex]. Round your answer to the nearest hundredth.

Therefore, the volume of the sphere is approximately [latex]14.14[/latex] [latex]\text{cm}^3[/latex].

Problem 3: Calculate the volume of a sphere with a radius of [latex]8[/latex] inches. Use [latex]\pi=22/7[/latex]. Round your answer to the nearest whole number.

Therefore, the volume of the sphere is approximately [latex]2,146[/latex] [latex]\text{in}^3[/latex].

Problem 4: Determine the volume of a sphere that has a diameter of [latex]14[/latex] meters. Use [latex]\pi=3.14[/latex]. Round your answer to the nearest hundredth.

Therefore, the volume of the sphere is approximately [latex]1,436.03[/latex] [latex]\text{m}^3[/latex].

Problem 5: If the diameter of a sphere is [latex]7[/latex] feet, what is its volume? Use [latex]\pi=3.1416[/latex]. Round your answer to the nearest thousandth.

Therefore, the volume of the sphere is approximately [latex]179.595[/latex] [latex]\text{ft}^3[/latex].

Problem 6: Find the volume of a sphere whose diameter is [latex]10.5[/latex] centimeters. Use [latex]\pi=22/7[/latex]. Round your answer to the nearest whole number.

Therefore, the volume of the sphere is approximately [latex]606[/latex] [latex]\text{cm}^3[/latex].

Problem 7: The volume of the sphere is [latex]3,719[/latex] cubic kilometers. What is the radius of the sphere? Round your answer to the nearest hundredth. Use [latex]\pi=3.14[/latex].

Therefore, the radius of the sphere is [latex]9.61[/latex] [latex]\text{km}[/latex].

Problem 8: Given that the volume of the sphere is [latex]1,000[/latex] cubic inches. Calculate its radius. Round your answer to the nearest tenth. Use [latex]\pi=22/7[/latex].

Therefore, the radius of the sphere is [latex]6.2[/latex] [latex]\text{in.}[/latex].

Problem 9: The surface area of a sphere is [latex]5,544[/latex] [latex]\text{ft}^2[/latex]. Find the volume of the sphere. Use [latex]\pi=22/7[/latex] as an approximation.

Find the radius of the sphere using the surface area formula.

That means, the radius of the sphere is [latex]21[/latex] [latex]\text{ft}[/latex].

Use the radius from previous step to determine the volume of the sphere.

Therefore, the volume of the sphere is [latex]38,808[/latex] [latex]\text{ft}^3[/latex].

Problem 10: If the surface area of a sphere is [latex]9,856[/latex] [latex]\text{m}^2[/latex], what is its volume? Round you answer to one decimal place. Use [latex]\pi=22/7[/latex] as an approximation.

That means the radius of the sphere is [latex]28[/latex] [latex]\text{m}[/latex].

Now use the value of the radius to calculate the volume of the sphere.

Therefore, the volume of the sphere is [latex]91,989.3[/latex] [latex]\text{m}^3[/latex].

You might also like these tutorials:

• Volume of Sphere Formula

FINDING THE VOLUME OF A SPHERE IN A REAL WORLD CONTEXT

We can find the volume of a sphere using the volume of a cylinder.

Cylinder is a solid which has a circular base.

We know the fact that the volume of any solid is equal to the product of base area and height of the solid.

So, the volume of a right circular cylinder of base radius ‘r’ and height ‘h’ is given by

V = (Base Area) x (Height)

The base of a cylinder is a circle, so for a cylinder,

Base Area =  π r 2

Volume of a cylinder =  π r 2 h cubic units

Consider a right circular cylinder and three right circular cones of same base radius and height as that of the cylinder.

The contents of three cones will exactly occupy the cylinder.

When we model the volume of a sphere, we will be getting the following result.

3 x (Volume of a cone)  =  Volume of cylinder

3 x (Volume of a cone)  =  π r 2 h

Volume of the cone = 1 /3  ·  π r 2 h cubic units

Consider a sphere and two right circular cones of same base radius and height such that twice the radius of the sphere is equal to the height of the cones.

Then we can observe that the contents of two cones will exactly occupy the sphere.

Volume of sphere  = 2 x (Volume of a cone)

Volume of a sphere = 2 x (1 /3  ·  π r 2 h)

Volume of a sphere = 2 /3  ·  π r 2 h

A sphere always has a height which is  equal to twice the radius.

So, substitute 2r for h.

Volume of sphere = 2/3 ·  π r 2 (2r)

Volume of sphere =  4 /3  ·  π r 3  cubic units

Example 1 :

Soccer balls come in several different sizes. One of the soccer balls has a diameter of 24 centimeters. What is the volume of this soccer ball ? Round your answer to the nearest tenth. Use the approximate of value of    ∏ , that is 3.14.

Because  soccer ball is in the shape of sphere, we can use the formula of volume of a sphere to find volume of the soccer ball.

Write the formula to find volume of a sphere.

V = 4 /3  ·  π r 3  ----(1)

To find the volume, we need the radius of the sphere. But, the diameter is given, that is  24 cm. So, find the radius.

r = diameter/2

Substitute  π  ≈ 3.14 and  r = 12 in (1).

V ≈ 4/3  ·  3.14  · 12 3

V ≈ 4/3  ·  3.14  · 1728

V ≈ 7,234.6

So, the volume of the soccer ball is about 7,234.6 cubic cm.

Example 2 :

Jose measures the diameter of a ball as 14 inches. How many cubic  inches of air can the ball hold, to the nearest tenth ? Use the approximate of value of    ∏ , that is 3.14.

To know how much air the ball can hold, we have to find the volume of the ball.

Because a  ball is in the shape of sphere, we can use the formula of volume of a sphere to find volume of a ball.

To find the volume, we need the radius of the sphere. But, the diameter is given, that is  14 cm. So, find the radius.

Substitute  π   ≈  3.14 and  r = 7 in (1).

V ≈ 4/3  ·  3.14  · 7 3

V ≈ 4/3  ·  3.14  · 343

V ≈ 1,436

So, the ball can hold about 1,436 cubic inches.

Example 3 :

Air is leaking from a spherical-shaped advertising balloon at the rate of 26 cubic feet per minute. If the radius of the ball is 7  feet, how long would it take for the balloon to empty fully ? Round your answer to the nearest minute. Use the approximate of value of  π , that is 3.14.

To know how long it would take for the balloon to empty fully, first we have to find the volume of air in the balloon.

Because the balloon is in the shape of sphere, we can use the formula of volume of a sphere to find volume of air in the balloon.

Substitute  π   ≈ 3.14 and  r = 7 in (1).

V ≈ 4/3 · 3.14 · 7 3

So, the volume of air in the balloon is about 1,436 feet.

Air is leaking from the balloon at the rate of 26 cubic feet per minute.

To know how long it would take for the balloon  to empty 1,426 cubic feet of air, divide 1,426 by 26.

= 1,426 / 26

   ≈ 55

So, it would take about 55 minutes for the balloon to empty fully.

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We always appreciate your feedback.

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How to Calculate the Volume of a Sphere

Last Updated: May 8, 2024 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 7 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 2,275,865 times.

In geometric terms, a sphere is defined as a set of points that are a given distance from a given point. The given point is called the center of the sphere, and the given distance from the center of the sphere to the set of points that form the sphere is called the radius of the sphere. [1] X Research source

Many commonly-used objects such as balls or globes are spheres. If you want to calculate the volume of a sphere, you just have to find its radius and plug it into a simple formula, V = ⁴⁄₃πr³. [2] X Research source

• If you're only given the surface area of the sphere, then you can find the radius by finding the square root of the surface area divided by 4π. In that case, r = root (surface area/4π) [5] X Research source

• If the radius was 2 inches (5.1 cm), for example, then to cube it, you would find 2 3 , which is 2 x 2 x 2, or 8.

Practice Problems and Answers

Community Q&A

• Make sure your measurements are all in the same unit. If they aren't, you will need to convert them. Thanks Helpful 51 Not Helpful 27
• Note that the "*" symbol is used as a multiplication sign to avoid confusion with the variable "x". Thanks Helpful 51 Not Helpful 32
• Don't forget to use cubed units (e.g. 31 ft³ ). Thanks Helpful 6 Not Helpful 1

Things You'll Need

• Calculator (reason: to calculate problems that would be annoying to do without it)
• Pencil and paper (not needed if you have an advanced calculator)

You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/sphere.html
• ↑ https://www.mathsisfun.com/geometry/sphere.html
• ↑ https://www.omnicalculator.com/math/sphere-volume#volume-of-a-sphere-formula
• ↑ https://www.mathsisfun.com/definitions/diameter.html
• ↑ https://www.web-formulas.com/Math_Formulas/Geometry_Surface_of_Sphere.aspx
• ↑ https://www.mathsisfun.com/geometry/sphere-volume-area.html
• ↑ https://www.mathopenref.com/pi.html

About This Article

To calculate the volume of a sphere, use the formula v = ⁴⁄₃πr³, where r is the radius of the sphere. If you don't have the radius, you can find it by dividing the diameter by 2. Once you have the radius, plug it into the formula and solve to find the volume. For more tips, including examples you can use for practice, read on! Did this summary help you? Yes No

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Volume Problem Solving Questions - Sphere, Cone, Cylinder, Cuboid - Higher Tier A/A*

Subject: Mathematics

Age range: 14-16

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Last updated

22 February 2018

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Great resource for stretching my year 10 set and thinking about volume in a slightly different way. They really enjoyed it and got quite competitive. thank you!

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Love this challenge sheet as a stretch and challenge task, only query is if you have the solutions for the other tasks? I find it odd that you have only included solutions to task 1.

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Good level of challenge, perfect for a problem solving lesson on volume with year 10. Thanks a lot

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Volume of a Sphere

Volume of a sphere ( using 3.14 ) worksheets.

• Volume of a Sphere (Radius Given)
• Volume of a Sphere (Diameter Given)
• Volume of a Sphere (Radius/Diameter Given)

Volume of a Sphere ( In Terms of Pi ) Worksheets

• Volume of a Sphere(Radius/Diameter Given)

Volume of a Sphere ( Radius Given)  Worksheets

Volume of a Sphere (Radius Given) Worksheet 1 – This worksheet features images of 12 spheres. The radius of each sphere is provided, and you must round the volume to the nearest tenth. Volume of a Sphere (Radius Given) Worksheet 1 RTF Volume of a Sphere (Radius Given) Worksheet 1 PDF Preview Volume of a Sphere (Radius Given) Worksheet 1 In Your Web Browser View Answers

Volume of a Sphere (Radius Given) Worksheet 3 – This worksheet features images of 12 spheres. The radius of each sphere is provided, and you must round the volume to the nearest tenth. Volume of a Sphere (Radius Given) Worksheet 3 RTF Volume of a Sphere (Radius Given) Worksheet 3 PDF Preview Volume of a Sphere (Radius Given) Worksheet 3 In Your Web Browser View Answers

Volume of a Sphere ( Diameter  Given)  Worksheets

Volume of a Sphere (Diameter Given) Worksheet 1 – This worksheet features images of 12 spheres. The diameter of each sphere is provided, and you must round the volume to the nearest tenth. Volume of a Sphere (Diameter Given) Worksheet 1 RTF Volume of a Sphere (Diameter Given) Worksheet 1 PDF Preview Volume of a Sphere (Diameter Given) Worksheet 1 In Your Web Browser View Answers

Volume of a Sphere (Diameter Given) Worksheet 2 – This worksheet features images of 12 spheres. The diameter of each sphere is provided, and you must round the volume to the nearest tenth. Volume of a Sphere (Diameter Given) Worksheet 2 RTF Volume of a Sphere (Diameter Given) Worksheet 2 PDF Preview Volume of a Sphere (Diameter Given) Worksheet 2 In Your Web Browser View Answers

Volume of a Sphere (Diameter Given) Worksheet 3 – This worksheet features images of 12 spheres. The diameter of each sphere is provided, and you must round the volume to the nearest tenth. Volume of a Sphere (Diameter Given) Worksheet 3 RTF Volume of a Sphere (Diameter Given) Worksheet 3 PDF Preview Volume of a Sphere (Diameter Given) Worksheet 3 In Your Web Browser View Answers

Volume of a Sphere ( Radius / Diameter  Given)  Worksheets

Volume of a Sphere (Radius or Diameter Given) Worksheet 1 – This worksheet features images of 12 spheres. The radius or diameter of each sphere is provided, and you must round the volume to the nearest tenth. Volume of a Sphere (Radius or Diameter Given) Worksheet 1 RTF Volume of a Sphere (Radius or Diameter Given) Worksheet 1 PDF Preview Volume of a Sphere (Radius or Diameter Given) Worksheet 1 In Your Web Browser View Answers

Volume of a Sphere (Radius or Diameter Given) Worksheet 2 – This worksheet features images of 12 spheres. The radius or diameter of each sphere is provided, and you must round the volume to the nearest tenth. Volume of a Sphere (Radius or Diameter Given) Worksheet 2 RTF Volume of a Sphere (Radius or Diameter Given) Worksheet 2 PDF Preview Volume of a Sphere (Radius or Diameter Given) Worksheet 2 In Your Web Browser View Answers

Volume of a Sphere (Radius or Diameter Given) Worksheet 3 – This worksheet features images of 12 spheres. The radius or diameter of each sphere is provided, and you must round the volume to the nearest tenth. Volume of a Sphere (Radius or Diameter Given) Worksheet 3 RTF Volume of a Sphere (Radius or Diameter Given) Worksheet 3 PDF Preview Volume of a Sphere (Radius or Diameter Given) Worksheet 3 In Your Web Browser View Answers

Volume of a Sphere in  Terms of Pi  (Radius Given )  Worksheets

Volume of a Sphere (In Terms of Pi) Worksheet 1 – This worksheet features images of 12 spheres. The radius of each sphere is provided, and you must calculate the volume in terms of Pi. Volume of a Sphere (In Terms of Pi) Worksheet 1 RTF Volume of a Sphere (In Terms of Pi) Worksheet 1 PDF Preview Volume of a Sphere (In Terms of Pi) Worksheet 1 In Your Web Browser View Answers

Volume of a Sphere (In Terms of Pi) Worksheet 2 – This worksheet features images of 12 spheres. The radius of each sphere is provided, and you must calculate the volume in terms of Pi. Volume of a Sphere (In Terms of Pi) Worksheet 2 RTF Volume of a Sphere (In Terms of Pi) Worksheet 2 PDF Preview Volume of a Sphere (In Terms of Pi) Worksheet 2 In Your Web Browser View Answers

Volume of a Sphere (In Terms of Pi) Worksheet 3 – This worksheet features images of 12 spheres. The radius of each sphere is provided, and you must calculate the volume in terms of Pi. Volume of a Sphere (In Terms of Pi) Worksheet 3 RTF Volume of a Sphere (In Terms of Pi) Worksheet 3 PDF Preview Volume of a Sphere (In Terms of Pi) Worksheet 3 In Your Web Browser View Answers

Volume of a Sphere in  Terms of Pi  (Diameter Given )  Worksheets

Volume of a Sphere (In Terms of Pi) Worksheet 1  – This worksheet features images of 12 spheres. The radius of each sphere is provided, and you must calculate the volume in terms of Pi. Volume of a Sphere (In Terms of Pi) Worksheet 1 RTF Volume of a Sphere (In Terms of Pi) Worksheet 1 PDF Preview Volume of a Sphere (In Terms of Pi) Worksheet 1 In Your Web Browser View Answers

Volume of a Sphere (In Terms of Pi) Worksheet 2 – This worksheet features images of 12 spheres. The diameter of each sphere is provided, and you must calculate the volume in terms of Pi. Volume of a Sphere (In Terms of Pi) Worksheet 2 RTF Volume of a Sphere (In Terms of Pi) Worksheet 2 PDF Preview Volume of a Sphere (In Terms of Pi) Worksheet 2 In Your Web Browser View Answers

Volume of a Sphere (In Terms of Pi) Worksheet 3 – This worksheet features images of 12 spheres. The diameter of each sphere is provided, and you must calculate the volume in terms of Pi. Volume of a Sphere (In Terms of Pi) Worksheet 3 RTF Volume of a Sphere (In Terms of Pi) Worksheet 3 PDF Preview Volume of a Sphere (In Terms of Pi) Worksheet 3 In Your Web Browser View Answers

Volume of a Sphere in  Terms of Pi  (Radius/Diameter Given )  Worksheets

Volume of a Sphere (In Terms of Pi) Worksheet 1 – This worksheet features images of 12 spheres. The radius or diameter of each sphere is provided, and you must calculate the volume in terms of Pi. Volume of a Sphere (In Terms of Pi) Worksheet 1 RTF Volume of a Sphere (In Terms of Pi) Worksheet 1 PDF Preview Volume of a Sphere (In Terms of Pi) Worksheet 1 In Your Web Browser View Answers

Volume of a Sphere (In Terms of Pi) Worksheet 2 – This worksheet features images of 12 spheres. The radius or diameter of each sphere is provided, and you must calculate the volume in terms of Pi. Volume of a Sphere (In Terms of Pi) Worksheet 2 RTF Volume of a Sphere (In Terms of Pi) Worksheet 2 PDF Preview Volume of a Sphere (In Terms of Pi) Worksheet 2 In Your Web Browser View Answers

Volume of a Sphere (In Terms of Pi) Worksheet 3 – This worksheet features images of 12 spheres. The radius or diameter of each sphere is provided, and you must calculate the volume in terms of Pi. Volume of a Sphere (In Terms of Pi) Worksheet 3 RTF Volume of a Sphere (In Terms of Pi) Worksheet 3 PDF Preview Volume of a Sphere (In Terms of Pi) Worksheet 3 In Your Web Browser View Answers

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Volume of a Sphere

Introduction, what is the volume of a sphere, volume of the sphere formula, derivation of volume of sphere, the volume of a solid sphere, the volume of a hollow sphere.

• Calculating the Volume of a Sphere

Solved Examples

Practice problems, frequently asked questions.

The volume of a sphere is fundamental in mathematics and various scientific disciplines. The volume of a sphere refers to the amount of space enclosed by the spherical shape. It can be calculated using a simple formula using the sphere's radius.

The volume of a sphere is essentially the amount of space that can be occupied by any spherical shape. Imagine drawing a circle on a flat surface, then taking a circular disc, attaching a string along its diameter, and rotating it around that string. The resulting shape is a sphere. The unit used to measure the volume of a sphere is typically given as `(\text{unit})^3`, where the unit could be meters, centimeters, inches, or feet, depending on the system of measurement being used. 

The formula to calculate the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( V \) represents the volume and \( r \) represents the radius of the sphere. This formula allows us to determine the amount of space enclosed within a sphere based on its radius.

According to Archimedes, if we have a cylinder, a cone, and a sphere, all with the same radius \( r \) and sharing the same cross-sectional area, their volumes are related in a specific ratio: `1:2:3`. From this, we can derive a relationship between the volumes of the cylinder, cone, and sphere.

We start with the formula for the volume of a cylinder, which is \( V_{\text{cylinder}} = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Now, the volume of a cone is one-third of the volume of the cylinder, so \( V_{\text{cone}} = \frac{1}{3} \pi r^2 h \).

Using these formulas, we can express the volume of the sphere in terms of the cylinder and cone volumes. By subtracting the volume of the cone from the volume of the cylinder, we get the volume of the sphere: \( V_{\text{sphere}} = V_{\text{cylinder}} - V_{\text{cone}} \).

Now, substituting the formulas for cylinder and cone volumes, we have \( V_{\text{sphere}} = \pi r^2 h - \frac{1}{3} \pi r^2 h \).

Simplifying this, we find that \( V_{\text{sphere}} = \frac{2}{3} \pi r^2 h \).

In this case, the height of the cylinder is the same as the diameter of the sphere, which is \( 2r \). Therefore, we can rewrite the equation as \( V_{\text{sphere}} = \frac{2}{3} \pi r^2 (2r) = \frac{4}{3} \pi r^3 \).

So, the volume of a solid sphere is \( \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere. This formula is widely used in various fields of science and engineering, as it allows us to determine the amount of space occupied by spherical objects and to make calculations related to their properties and interactions.

The volume of a hollow sphere is the amount of space enclosed between the outer and inner surfaces of a hollow sphere. To calculate the volume of a hollow sphere, we subtract the volume of the inner sphere from the outer volume sphere.

If the outer radius of the hollow sphere is \( R \) and the inner radius is \( r \) (where \( R > r \)), the formula for the volume of the hollow sphere is:

\( V = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi r^3 \)

We can simplify this to:

\( V = \frac{4}{3} \pi (R^3 - r^3) \)

Calculating the Volume of a Sphere

`1`. Understand the formula:  The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( V \) represents the volume and \( r \) represents the radius of the sphere.

`2`. Measure the radius:  Begin by measuring the radius of the sphere. The radius is the distance from the center of the sphere to any point on its surface.

`3`. Plug the radius into the formula:  Once you have the radius value, plug it into the formula \( V = \frac{4}{3} \pi r^3 \). 

`4`. Calculate:  Perform the necessary calculations. First, cube the radius (multiply it by itself three times), then multiply by \( \frac{4}{3} \pi \).

`5`. Get the Volume:  After completing the calculations, you will have the volume of the sphere in cubic units. 

Example: Calculate the volume of a sphere with a radius of \( r = 3 \) units. Express the volume in terms of `pi`.

Using the formula for the volume of a sphere \( V = \frac{4}{3} \pi r^3 \), we can plug in the given radius:

\( V = \frac{4}{3} \pi (3)^3 \)

Now, let's perform the calculations step by step:

`1`. Cube the radius: \( 3^3 = 27 \).

`2`. Multiply by \( \frac{4}{3} \pi \): \( \frac{4}{3} \pi \times 27 \).

`3`. Calculate: \( \frac{4}{3} \times 27 \times \pi = 36 \pi \) cubic units.

Therefore, the volume of the sphere with a radius of `3` units is \( 36 \pi \) cubic units.

Example `1`. Find the volume of a solid sphere with a radius of `6` centimeters. Consider \( \pi \ = 3.14 \).

Given: Radius (\( r \)) `= 6` `cm`

Using the formula for the volume of a solid sphere: \( V = \frac{4}{3} \pi r^3 \)

Substitute the radius: \( V = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = 288 \pi\ = 904.32\) cubic centimeters.

The volume of the solid sphere is \( 288 \pi \) cubic centimeters.

Example `2`. A hollow sphere has an outer radius of `10` meters and an inner radius of `8` meters. Calculate the volume of the hollow sphere. Write your answer in terms of \( \pi \).

Given: Outer Radius (\( R \)) `= 10` `m`, Inner Radius (\( r \)) `= 8` `m`

Using the formula for the volume of a hollow sphere: \( V = \frac{4}{3} \pi (R^3 - r^3) \)

Substitute the radii: \( V = \frac{4}{3} \pi (10^3 - 8^3) \)

\( V = \frac{4}{3} \pi (1000) - \frac{4}{3} \pi (512) \)

\( V = \frac{4}{3} \pi (1000 - 512) \)

\( V = \frac{1952}{3} \pi \) cubic meters.

The volume of the hollow sphere is \( \frac{1952}{3} \pi \) cubic meters.

Example `3`. A spherical balloon has a volume of \( 500 \pi \) cubic inches. Calculate its radius.

Given: Volume (\( V \)) `=` \( 500 \pi \) cubic inches

Rearrange the formula to solve for \( r \): \( r = \sqrt[3]{\frac{3V}{4\pi}} \)

Substitute the volume: \( r = \sqrt[3]{\frac{3 \times 500 \pi}{4\pi}} \)

\( r = \sqrt[3]{\frac{1500}{4}} \)

\( r = \sqrt[3]{375} \)

\( r \approx 7.28 \) inches (rounded to two decimal places).

The radius of the spherical balloon is approximately `7.28` inches.

Example `4`. The total surface area of a sphere is \( 144\pi \) square units. Find the total volume of the sphere in terms of `pi`.

Given: Total surface area (\( A \)) `=` \( 144\pi \) square units

We know that the formula for the total surface area (\( A \)) of a sphere is given by:

\( A = 4\pi r^2 \)

Given the total surface area, we can rearrange the formula to solve for the radius (\( r \)):

\( r = \sqrt{\frac{A}{4\pi}} \)

Substitute the given surface area:

\( r = \sqrt{\frac{144\pi}{4\pi}} = \sqrt{36} = 6 \)

Now that we have the radius (\( r \)), we can use it to find the volume (\( V \)) of the sphere using the formula:

\( V = \frac{4}{3}\pi r^3 \)

Substitute the radius:

\( V = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 288\pi \) cubic units.

Therefore, the total volume of the sphere is \( 288\pi \) cubic units.

Example `5`. Three spheres of radius `4` `cm` fit inside a tube. Calculate the volume of the tube that is not filled. Consider \( \pi \ = 3.14 \).

In order to find the unfilled volume of the tube, we need to find the difference between the volume of the cylindrical tube and the combined volume of the `3` spheres.

We know that the formula for the volume of a cylinder (\( V_1 \)) of a sphere is given by:

\( V = \pi r^2 h \), 

where \(h \) is the height of the cylinder

\(r \) is the radius of the base

Note: 

• The height of the cylinder is `3` times the diameter of each sphere `=` \(3 \times 8\) Hence `h = 24` `cm`.
• The radius of the base of the cylinder is equal to the radius of each sphere  `= 4 cm`. Hence `r = 4` `cm`

Substituting `h = 24` and `r = 4` into the formula \( V_1= \pi r^2 h \), we get the volume of the cylinder as 

\( V_1 = \pi \times4^2 \times 24 \)

\( V_1 = \pi \times16 \times 24 \)

\( V_1 = 384 \pi\ cm^3 \)

Substituting `r = 4` into the formula \( V_2 = \frac{4}{3}\pi r^3 \), we get the volume of each sphere as 

\( V_2 = \frac{256}{3}\pi\)

Hence the combined volume of the `3` spheres \(=3 \times \frac{256}{3}\pi = 256\pi\ cm^3\) 

Unfilled volume of the tube `=` Volume of the cylindrical tube `-` Combined volume of the `3` spheres

Unfilled volume of the tube `=` \( 384 \pi\ cm^3 -  256 \pi\ cm^3\) 

Unfilled volume of the tube `=` \( 128 \pi\ cm^3\)  `=`  \( 401.92\ cm^3\)

Q`1`. What is the volume of a sphere with a radius of `5` meters? Give the answer in terms of `pi`.

• \( \frac{256}{3}\pi \) cubic meters
• \( \frac{500}{3}\pi \) cubic meters
• \( \frac{625}{3}\pi \) cubic meters
• \( \frac{1000}{3}\pi \) cubic meters

Q`2`. A hollow sphere has an outer radius of `12` centimeters and an inner radius of `8` centimeters. What is the volume of the hollow space? Give the answer in terms of `pi`.

• \( \frac{128}{3}\pi \) cubic centimeters
• \( \frac{1912}{3}\pi \) cubic centimeters
• \( \frac{4864}{3}\pi \) cubic centimeters
• \( \frac{384}{3}\pi \) cubic centimeters

Q`3`. The volume of a sphere is \( 288 \pi \) cubic inches. What is its radius?

Q`4`. The total surface area of a sphere is \( 100\pi \) square units. Find the total volume of the sphere in terms of `pi`.

• \( \frac{384}{3}\pi \) cubic meters
• \( \frac{725}{3}\pi \) cubic meters
• \( \frac{800}{3}\pi \) cubic meters

Q`5`. A spherical tank has a volume of \( 8000 \) cubic feet. What is its diameter, rounded to the nearest whole number .

Q`1`. What is the formula for the volume of a sphere?

Answer:  The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( V \) represents the volume and \( r \) represents the radius of the sphere.

Q`2`. How do I calculate the volume of a sphere if I only know its diameter?

Answer:  If you only have the diameter (\( d \)) of the sphere, you can calculate the radius (\( r \)) by dividing the diameter by `2` (\( r = \frac{d}{2} \)), and then use the formula \( V = \frac{4}{3} \pi r^3 \) to find the volume.

Q`3`. Can I find the volume of a hollow sphere?

Answer:  Yes, the volume of a hollow sphere can be calculated by subtracting the volume of the inner sphere from the volume of the outer sphere. The formula for the volume of a hollow sphere is \( V = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi r^3 \), where \( R \) is the outer radius and \( r \) is the inner radius.

Q`4`. What are the units for the volume of a sphere?

Answer:  The volume of a sphere is typically measured in cubic units, such as cubic meters `(\text{m}^3)`, cubic centimeters `(\text{cm}^3)`, cubic inches `("in"^3)`, or cubic feet `(\text{ft}^3)`, depending on the system of measurement being used.

Q`5`. How can I use the volume of a sphere in real-life applications?

Answer:  The volume of a sphere is used in various fields, including mathematics, physics, engineering, and astronomy. It is employed in calculating the capacities of spherical containers, understanding buoyancy in fluids, modeling planetary bodies, and designing objects with spherical components, among other applications.

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Practice Problems on Surface Area and Volume of a Sphere

In this article, we are going to study about an important chapter of school mathematics. This article will explain concepts related to the sphere and solve questions and unsolved questions.

Important Formulas Related to Sphere

A sphere is a three-dimensional figure that resembles a ball and various formulas related to a sphere are:

• r is Radius of the Hemisphere

Q1. A spherical water tank has a radius of 8 meters. Calculate the volume of water it can hold.

Q2. Volume of a sphere is 512π m³. Calculate the diameter of the sphere.

Q3. Given a sphere of diameter of 30 meters. Find the volume of the sphere.

Q4. Given a sphere of radius 12 meters. Find the surface area of the sphere.

Q5. Given a sphere of diameter 18 meters. Find the surface area of the sphere.

Q6. Given a hemisphere of radius 6 meters. Find the surface area of the hemisphere.

Q7. Given a solid hemisphere of radius 10 meters. Find the surface area of the solid hemisphere.

Q8. Given a sphere whose surface area is 7392πm². Find the diameter of the sphere.

Q9. Given a hemisphere of radius 7 meters. Find the volume of the hemisphere.

Q10. Given a hemisphere of radius 12 meters. Find the volume of the hemisphere.

Practice Problems on Surface Area and Volume of a Sphere with Solutions

Problem 1: a spherical water tank has a radius of 5 meters. calculate the volume of water it can hold..

Volume of sphere = 4/3πr 3 So, radius = 5m Volume = 4/3 × π × r× r × r = 4/3 × 3.14 × 5 × 5 × 5 = 523.33 m 3

Problem 2: Volume of a sphere is 288πm 3 . Calculate the diameter of the sphere.

Volume of a sphere is 288π m 3 According to formula, ⇒ 4/3 × π × r × r × r = 288π ⇒ r = 6m So, diameter of the sphere = 2r = 12 m.

Problem 3: Given a sphere of diameter of 20m. Find the volume of the sphere.

Given, Diameter(D) = 20 m Radius(r) = D/2 = 10 m Volume = 4/3πr 3 = 4/3 × π × 10 × 10 × 10 = 4186 m 3 So, volume of the sphere of diameter 20 m is 4186 m 3

Problem 4: Given a sphere of radius 10m. Find the surface area of the sphere.

Given, Radius = 10 m Volume = 4πr 2 = 4 × π × 10 × 10 = 1256m 2 So, surface area of the sphere of radius 10m is 1256 m 2 .

Problem 5: Given a sphere of diameter 14m. Find the surface area of the sphere.

Given, Diameter = 14 m Radius = 7 m Volume = 4πr 2 = 4 × π × 7 × 7 = 615.44 m 2 So, surface area of the sphere of radius 5m is 1256 m 2 .

Problem 6: Given a hemisphere of radius 5m. Find the surface area of the hemisphere.

Given, Radius = 5 m Volume = 2πr 2 = 2 × π × 5 × 5 = 157 m 2 So, surface area of the hemisphere of radius 5m is 157 m 2 .

Problem 7: Given a solid hemisphere of radius 8m. Find the surface area of the solid hemisphere.

Given, Radius = 7 m Volume = 3πr 2 = 3 × π × 7 × 7 = 461.58 m 2 So, surface area of the solid hemisphere of radius 7m is 461.58 m 2 .

Problem 8: Given a sphere whose surface area is 5544m 2 . Find the diameter sphere.

Given, Surface Area = 5544 cm 2 ⇒ 4π×r×r = 5544 ⇒ 4× 3.14 × r×r = 5544 ⇒ r×r = 441 ⇒ r = 21m So, diameter of the sphere is 42 m.

Problem 9: Given a hemisphere of radius 5m. Find the volume of the hemisphere.

Given, Radius = 5m Volume = 2/3πr 3 = 2/3 × π × 5 × 5 × 5 = 261.66 m 3 So, volume of the hemisphere of radius 5 m is 261.66 m 3 .

Problem 10: Given a hemisphere of radius 8m. Find the volume of the hemisphere.

Given, Radius = 8 m Volume = 2/3πr 3 = 2/3 × π × 8 × 8 × 8 = 1071.78 m 3 So, volume of the hemisphere of radius 8m is 1071.78 m 3

FAQs on Surface Area and Volume of a Sphere

Does sphere comes under two-dimensional geometry or three-dimensional geometry.

Sphere comes under the category of three-dimensional geometry.

If we are given the radius of a sphere, then find the diameter of the sphere.

To find the diameter of a sphere, we use the formula, Diameter = 2 × Radius of Sphere

Can we find radius of sphere if we are given the surface area value of the sphere?

Yes, we can calculate the radius of sphere if the surface area of the sphere is given.

Find surface area of solid hemissphere which have radius = 4 cm.

To calculate the surface area of a solid hemisphere we use the formula, Surface Area = 3πr 2 Surface Area = 3 × 3.14 × 4 × 4 = 48 × 3.14 = 150.72 cm 2

Find radius of a sphere whose surface area is 154 cm 2 .

To calculate radius of a sphere we use the formula, Surface Area = 4πr 2 ⇒ 154 = 4 × 3.14 × r × r ⇒ r×r = 12.26 ⇒ r = 3.5 cm

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Volume of a Sphere Worksheets

The volume of a sphere and hemisphere worksheets meticulously created for 8th grade and high school students help them learn the know-how of calculating the volume of spheres with a set of practice exercises offering integer and decimal dimensions with two levels of difficulty. Find the volume of spheres and hemispheres and find the missing parameters as well. Grab some of these worksheets for free!

Volume of a Sphere - Integers | Easy

Navigate through this batch of printable volume of sphere worksheets presenting the radius measure as integers. Plug in the value in the formula V = 4/3 π r 3 and compute the volume of the spheres.

• Download the set

Volume of a Sphere - Integers | Moderate

Figure out the radius measure from the diameter and apply it to determine the volume of the spheres presented as 3D shapes and as word problems in these pdf worksheets for grade 8.

Volume of a Sphere - Decimals | Easy

The radius measure is expressed as a decimal. Compute the volume by substituting the radius and the value of pi. Round off your answers to two decimal places as well.

Volume of a Sphere - Decimals | Moderate

Know how to calculate the volume of spheres by dividing the radius by 2. Assign the values in the formula and compute the volume of the spheres presented as geometric figures and as real-life word problems.

Volume of a Sphere and Hemisphere - Integers

Hemisphere is the exact half of the sphere and the formula for the volume of a hemisphere is V = 2/3 π r 3 . Find the volume of the spheres and hemispheres using appropriate formulas.

Volume of a Sphere and Hemisphere - Decimals

Find a mix of spheres and hemispheres in these printable 3D shapes worksheets. Compute the volume of the spheres and hemispheres using the given dimensions expressed as decimals.

Volume of a Sphere and Hemisphere | Missing parameter - Level 1

The volume the spheres and hemispheres are provided in terms of pi in this compilation of high school worksheets. Rearrange the formula, making the missing dimension the subject and solve for radius.

Volume of a Sphere and Hemisphere | Missing parameter - Level 2

Reiterate the concept of changing the subject, substitute the volume and the value of pi in the formula. Determine the missing dimensions in these pdf volume of spheres and hemispheres worksheets.

Related Worksheets

» Volume of Cones

» Volume of Cylinders

» Volume of Prisms

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