A ball of radius 14 has a round hole of radius 9 drilled through its center. Find the volume of the resulting solid.

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- Dec 9th 2007, 10:13 AMDelVolume by Integration
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! - Dec 9th 2007, 11:07 AMgalactus
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. - Dec 9th 2007, 11:09 AMJhevon
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.