Integrating in polar coordinates (volume)

rhyno89
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Homework Statement



The solid bounded by the parabolids z = 3x^2 + 3y^2 -7 and z = -x^2 -y^2 + 9

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The Attempt at a Solution



Ok so i set the two z equations into polar form and came up with 3r^2 = 7 and r^2 = 9
I thought that r went from (7/3) ^(1/2) to 3 and that theta went from 0 to 2pi.
I figured the equation to integrate was -4r^2 + 16. (Outer bounds -inner bounds) Anyway, the answer the book gives is different than the two different ways I tried. Just need help setting up the integral...thanks
 
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This is a 3-D problem; you were given z coordinate values in each case. Work with those.
 
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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