Volumes of solids of revolution

Koey
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Homework Statement


A circle with a radius of (a/2) is bored through the centre of a sphere of radius a. Find the volume of the remaining solid.


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The Attempt at a Solution


I've been trying this for an hour now and I've been trying to find the remaining volume by finding the area of the cylinder bored + the little side bits on each side. I've just been getting really messy answers which aren't close to the actual answer.

Urgently need a hand :)
 
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The easiest way to do this is in my opinion:
Set the sphere in the center of an xyz-axis system with the axis of the cilinder the y-axis. (for example)
Now consider just the first quadrant in the xy-axis system.
You should know that you can make a function of the curve you see ( f(x)= sqrt( (a^2)/4 - x^2 ) )

To find the volume of the cylinder inside the sphere you only have to rotate the function around the y-axis while integrating from 0 to a/4. Multiply by two and you have the total volume of the cylinder within the sphere.

I'm sure you will be able to solve it now ;).
Goodluck
 
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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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