# Why is volume of a sphere 4/3× pi×r³?

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I got curious since I learned formula can be derived but since my level in maths is at gcse level so could I get a relative simpler ans as I been thinking about this?

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I got curious since I learned formula can be derived but since my level in maths is at gcse level so could I get a relative simpler ans as I been thinking about this?

**Laotsu**)I got curious since I learned formula can be derived but since my level in maths is at gcse level so could I get a relative simpler ans as I been thinking about this?

https://www.youtube.com/watch?v=YokKp3pwVFc

In the same way, the surface area of a sphere is 4*pi*r^2. You can "prove" this by equating the sphere to an open ended cylinder. Then the sphere is cut up into cones, each of volume r*basearea/3, just like the circle is cut up into wedges. The surface area of the sphere must equate to all the base areas of the cones, so the overall volume of the cylinder is

r*4*pi*r^2/3

or

4/3*pi*r^3

I'll have a quick dig round to see if I can find a pic of this.

Edit ...

https://www.researchgate.net/figure/...ig27_314913282

Isn't perfect, but its ok. Imagine for each of the grey patches on the surface of the sphere, the cone which is generated by connecting the edges up to the center. This splits the sphere up into cones of height r where the surface area of the sphere is equal to the bases areas of the cones. So volume is

r/3*4*pi*r^2 as mentioned above

Much like the circle is cut up into pizza slices, I imagine the sphere being unrolled into a spiky sheet, where each spike is of height ~r and is one of the cones and the (base) sheet is area 4*pi*r^2.

Last edited by mqb2766; 1 month ago

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Do you understand why the area of a circle is pi*r^2. Basically, cut it up into thin wedges where the circumference is 2*pi*r, then it can be re-arranged into length of thin wedges (triangles) of height r and area r/2*baselength. The sum of them must be (r/2)*2*pi*r as their bases correspond to the circumference

https://www.youtube.com/watch?v=YokKp3pwVFc

In the same way, the surface area of a sphere is 4*pi*r^2. You can "prove" this by equating the sphere to an open ended cylinder. Then the sphere is cut up into cones, each of volume r*basearea/3, just like the circle is cut up into wedges. The surface area of the sphere must equate to all the base areas of the cones, so the overall volume of the cylinder is

r*4*pi*r^2/3

or

4/3*pi*r^3

I'll have a quick dig round to see if I can find a pic of this.

Edit ...

https://www.researchgate.net/figure/...ig27_314913282

Isn't perfect, but its ok. Imagine for each of the grey patches on the surface of the sphere, the cone which is generated by connecting the edges up to the center. This splits the sphere up into cones of height r where the surface area of the sphere is equal to the bases areas of the cones. So volume is

r/3*4*pi*r^2 as mentioned above

Much like the circle is cut up into pizza slices, I imagine the sphere being unrolled into a spiky sheet, where each spike is of height ~r and is one of the cones and the (base) sheet is area 4*pi*r^2.

**mqb2766**)Do you understand why the area of a circle is pi*r^2. Basically, cut it up into thin wedges where the circumference is 2*pi*r, then it can be re-arranged into length of thin wedges (triangles) of height r and area r/2*baselength. The sum of them must be (r/2)*2*pi*r as their bases correspond to the circumference

https://www.youtube.com/watch?v=YokKp3pwVFc

In the same way, the surface area of a sphere is 4*pi*r^2. You can "prove" this by equating the sphere to an open ended cylinder. Then the sphere is cut up into cones, each of volume r*basearea/3, just like the circle is cut up into wedges. The surface area of the sphere must equate to all the base areas of the cones, so the overall volume of the cylinder is

r*4*pi*r^2/3

or

4/3*pi*r^3

I'll have a quick dig round to see if I can find a pic of this.

Edit ...

https://www.researchgate.net/figure/...ig27_314913282

Isn't perfect, but its ok. Imagine for each of the grey patches on the surface of the sphere, the cone which is generated by connecting the edges up to the center. This splits the sphere up into cones of height r where the surface area of the sphere is equal to the bases areas of the cones. So volume is

r/3*4*pi*r^2 as mentioned above

Much like the circle is cut up into pizza slices, I imagine the sphere being unrolled into a spiky sheet, where each spike is of height ~r and is one of the cones and the (base) sheet is area 4*pi*r^2.

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That's an interesting way of looking at it, thanks. However, some or relies on the vol of a cone being 1/3 of the vol of a cylinder. Which begs the question....where does the 1/3 come from? It's never been intuitively obvious. Can you explain it??

**dextrous63**)That's an interesting way of looking at it, thanks. However, some or relies on the vol of a cone being 1/3 of the vol of a cylinder. Which begs the question....where does the 1/3 come from? It's never been intuitively obvious. Can you explain it??

Is decent ... pic about 1/3 of the way down.

http://www.math.brown.edu/~banchoff/...section02.html

Edit - have to admit I was a bit sloppy in talking about pyramids, cones, ... but really it doesn't matter what the base shape is.

Last edited by mqb2766; 1 month ago

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If you have a cube of side length h and have three square based cones (pyramids), then the bases cover 3 faces and remainder of the cones cover the rest of the cube so each cone is 1/3*h^3. Again, give me a sec and I'll see if I can dig out a pic.

**mqb2766**)If you have a cube of side length h and have three square based cones (pyramids), then the bases cover 3 faces and remainder of the cones cover the rest of the cube so each cone is 1/3*h^3. Again, give me a sec and I'll see if I can dig out a pic.

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I know what you mean. Just doesn't seem to "feel" right when trying to mentally picture things. I can sort of imagine "cone-ising" such shapes and using the volume of revolution integration to identify the 1/3 bit.

**dextrous63**)I know what you mean. Just doesn't seem to "feel" right when trying to mentally picture things. I can sort of imagine "cone-ising" such shapes and using the volume of revolution integration to identify the 1/3 bit.

TBH, its probably more to do with what is normally taught. Splitting a cube into 3 pyramids is a simple generalization of splitting a rectangle into 2 to get the area of a triangle, but its not necessarily normally covered.

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#7

Thanks for the link. Have only skim read it, but I like the casual "throwaway" comment about how the vol of a 4d shape at the end of paragraph 3, from which point on the author casually sidesteps the somewhat elephant in the room of "proof". Lol

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Thanks for the link. Have only skim read it, but I like the casual "throwaway" comment about how the vol of a 4d shape at the end of paragraph 3, from which point on the author casually sidesteps the somewhat elephant in the room of "proof". Lol

**dextrous63**)Thanks for the link. Have only skim read it, but I like the casual "throwaway" comment about how the vol of a 4d shape at the end of paragraph 3, from which point on the author casually sidesteps the somewhat elephant in the room of "proof". Lol

Too late on a Sunday to read anything ...

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#9

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My kids were happy enough with the explanation, so tetrahedrons are 1/6 because they're 1/2 a square based pyramid. I think they'd rebel if I mentioned volumes of revolution.

TBH, its probably more to do with what is normally taught. Splitting a cube into 3 pyramids is a simple generalization of splitting a rectangle into 2 to get the area of a triangle, but its not necessarily normally covered.

**mqb2766**)My kids were happy enough with the explanation, so tetrahedrons are 1/6 because they're 1/2 a square based pyramid. I think they'd rebel if I mentioned volumes of revolution.

TBH, its probably more to do with what is normally taught. Splitting a cube into 3 pyramids is a simple generalization of splitting a rectangle into 2 to get the area of a triangle, but its not necessarily normally covered.

Last edited by dextrous63; 1 month ago

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