A ball of radius 14 has a round hole of radius 9 drilled through its center. Find the volume of the resulting solid.
Please Help!
You must use integration?. Then let's use shells to find the volume of the portion removed, then subtract from the volume of the sphere.
$\displaystyle V_{s}=\frac{4}{3}{\pi}(14)^{3}=\frac{10976\pi}{3}$
Now, the volume of the 'hole':
$\displaystyle V_{h}=2{\pi}\int_{0}^{9}x(2\sqrt{196-x^{2}})dx$
=$\displaystyle 4\pi\int_{0}^{9}x\sqrt{196-x^{2}}dx$
Integrate and subtract that result from the total volume of the sphere.
You can also use washers to solve this.
This can be found by revolving the region bound between the circle of radius 14 and the line x = 9 (we would have to use the shell method here). do you see why this is true? did you draw a sketch?
if you want to use the disk method, we can get the same result by revolving the region bound between the circle of radius 14 and the line y = 9.
can you continue?
EDIT: Ah, ok. galactus did it another way.