# Volume by Integration

• Dec 9th 2007, 10:13 AM
Del
Volume 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.

• Dec 9th 2007, 11:07 AM
galactus
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.

$V_{s}=\frac{4}{3}{\pi}(14)^{3}=\frac{10976\pi}{3}$

Now, the volume of the 'hole':

$V_{h}=2{\pi}\int_{0}^{9}x(2\sqrt{196-x^{2}})dx$

= $4\pi\int_{0}^{9}x\sqrt{196-x^{2}}dx$

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 AM
Jhevon
Quote:

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