Calculate Volume of Solid with Hole: Ball of Radius 12 and Hole of Radius 7

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Homework Statement



A ball of radius 12 has a round hole of radius 7 drilled through its center. Find the volume of the resulting solid.

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The Attempt at a Solution



I tried to do it as though it is a ball with a radius of 12 missing a ball with a radius of 7 from its center and got an answer around 2900. But it is actually a ball with a radius of 12 with a hole of radius of center, if that makes sense. Like you could see through the ball. I have no idea how to approach this..
 
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Do you need an exact answer? You could approximate it very easily by using the volume of a sphere with 12 as it's radius and subtract the volume of a cylinder with radius 7 and height 12?

edit: The exact answer is obtained (one way) by using the equation for a circle (x*x + y*y = r*r, solve for the necessary variable) and the indefinite integral formula for finding volume (remember to choose the right limits).

First consider the case where the semi-circle is touching the Y axis. What is the radius of the hole generated by the volume created by revolution about the y axis?

picture:

http://img132.imageshack.us/img132/1650/helpbn7.jpg
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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