Answer:
- r
- a circle
- an annulus
- 2πr³
- (4/3)πr³
Note: the last sentence in the problem statement text should read, "Therefore, the volume of the sphere is 4/3πr³ by Cavalieri's principle.
Explanation:
I believe it can help a lot if you have seen and understand this derivation of the volume of a sphere. Here is the basic idea.
Shown in the attachment is a cross section of half the volume under consideration. Basically, it is showing one cone and the (red) top hemisphere in a (green) cylinder of radius R and height R. (The problem text refers to a sphere and two cones in a cylinder of height 2R. This is the top half of that geometry.) Actually, only the left edge of the cone is represented here, in order to avoid cluttering the diagram.
We can use this figure to think about a horizontal cross section (cut plane) of this geometry at height h from the center of the sphere. We want to consider the annulus of inner radius C between the cylinder of radius R and the cone, and we want to consider the circle of radius S where the cut plane intersects the hemisphere.
Because the cone has a height of R and a radius of R, the radius C of the cross section will be the same as the height h. That is, in our figure, h = C. We know from the Pythagorean theorem that ...
h² + S² = R²
S² = R² - C² . . . . . . subtract h² and substitute C for h
The area of the circular cross section of the hemisphere is πS², and the area of the annulus between the cylinder and cone is π(R² - C²). The above equation tells us these areas are the same.
By Cavalieri's principle, since the cross sections have the same area at every height, the volume of the space between the cylinder and cone is the same as the volume of the hemisphere. Using the formulas for the volumes of cylinder and cone, we find the difference to be ...
difference volume = hemisphere volume = πr²·r - 1/3πr²·r = 2/3πr³
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Once this approach to the sphere volume formula derivation is understood, filling in the blanks in your problem statement may become much simpler.