Using polar coordinates to find the volume of the given solid

serg_yegi
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1. Use polar coordinates to find the volume of the given solid.
2. Inside the sphere x^2 + y^2 + z^2 = 16 and outside the cylinder x^2 + y^2 = 4.
 
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serg_yegi said:
1. Use polar coordinates to find the volume of the given solid.



2. Inside the sphere x^2 + y^2 + z^2 = 16 and outside the cylinder x^2 + y^2 = 4.
What have you tried? You have to show an effort first.
 
I've tried doing a double integral from theta=0 to theta=2pi and r=0 to r=4 of ((16-r^2)^(1/2))r (dr)(dtheta)
 
The limits of integration for r should be from r = 2 to r = 4. If you go from r = 0, you're getting the volume inside the cylinder, which you don't want.
 
Ok. Is the actual integral correct?
 
Yes, I don't see anything wrong with it.
 
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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