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# The sphere.
Given a **sphere** of radius $R$, what is **the volume it encloses**, and what is **its surface area**?
![[1 teaching/summer program 2023/puzzles-and-problems/---files/sphere 2023-08-22 15.09.46.excalidraw.svg]]
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**Archimedes** (c. ~250 BC) made the following discovery:
![[1 teaching/summer program 2023/puzzles-and-problems/---files/sphere 2023-08-23 10.13.49.excalidraw.svg]]
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That the volume of a cone of radius $R$ and height $2R$ plus the volume of a sphere of radius $R$ gives the volume of a cylinder of radius $R$ and height $2R$ !
Archimedes considered this to be his most profound and favorite theorem, despite having made many other mathematical and scientific discoveries. He loved it so much that it is said he wanted a diagram of a cone and cylinder inscribed in a cylinder on his tombstone.
But why is this true? We shall see it in this worksheet.
Recall, a cone of any base area $B$ and height $H$ has volume $\frac{1}{3}BH$.
## Volume of a sphere.
![[---images/---assets/---icons/question-icon.svg]] First notice the following, a cone of radius $R$ and height $2R$ has the same volume as **two cones** of radius $R$ and height $R$ each. Show indeed they have the same volume.
![[1 teaching/summer program 2023/puzzles-and-problems/---files/sphere 2023-08-23 10.45.46.excalidraw.svg]]
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So we need to show that the sphere of radius $R$ has the same volume as a
We will use the following principle to find the volume of a sphere:
**If two solids of the same height have the same cross-sectional area at each height, then they have the same volume.**
![[1 teaching/summer program 2023/puzzles-and-problems/---files/sphere 2023-08-22 08.28.33.excalidraw.svg]]
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The idea is that if you slice up the object horizontally, and shift them around horizontally, or even redistribute them horizontally, the volume should remain the same. So as long as two solids at each height has the same cross-sectional area, then they ought to have the same volume.
So consider two solids:
(1) A sphere of radius $R$
(2) A circular cylinder of radius $R$ and height $2R$, but with two cones of radius $R$ and height $R$ **drilled out** from each side, so that the vertices of the two cones meet at the center of the cylinder:
![[1 teaching/summer program 2023/puzzles-and-problems/---files/sphere 2023-08-22 11.08.15.excalidraw.svg]]
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We claim that these two solids, which have the same height of $2R$, have the same cross-sectional area at any height $h$ measured from the center of these objects, they are $A_1$ and $A_2$ as shown below:
![[1 teaching/summer program 2023/puzzles-and-problems/---files/sphere 2023-08-22 08.40.39.excalidraw.svg]]
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![[---images/---assets/---icons/question-icon.svg]] Show that this is indeed the case:
(1) Find the **cross-sectional area** $A_1$ of the sphere at a distance $h$ from the center, and write it in terms of $h$ and $R$. This cross-section has a shape of a circle.
(2) Find the **cross-sectional area** $A_{2}$ of the second solid at a distance $h$ from its center, and write it in terms of $h$ and $R$. This cross-section has a shape that is an **annulus** (a circle removed from a circle).
Are the two areas $A_1$ and $A_2$ above the same?
![[---images/---assets/---icons/question-icon.svg]] Now, use the principle stated in the beginning (that two solids of the same height with the same cross-sectional areas at each height have the same volume), find the volume of the sphere of radius $R$ by finding the volume of the second solid.
The volume of a sphere of radius $R$ is.......?
Great job! You just derived the formula for the volume of a sphere!
### Surface area of a sphere.
How can we reason what is the surface area of a sphere?
Consider first partition the surface of the sphere into smaller shapes, and then cut them into wedges towards the center, as follows:
![[1 teaching/summer program 2023/puzzles-and-problems/---files/sphere 2023-08-22 11.43.41.excalidraw.svg]]
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