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I'm trying to calculate the volume between the surfaces $z=x^2+y^2$ and $z=2ax+2by$ where $a>0,b>0$. Here's what I've tried:

First I noticed the projection of the volume to the xy plane is a circle: $(x-a)^2+(y-b)^2\leq a^2+b^2$. Using this I simplified the calculation of the integral for the volume a little. Marking $B$ as the circle we get that the volume is:

$$\iint_{}^{B} (2ax+2ay-x^2-y^2) = \iint_{}^{B} (a^2+b^2)-\iint_{}^{B} ((x-a)^2+(y-b)^2) $$

Using the symmetry of the circle we get:

$$\iint_{}^{B} ((x-a)^2+(y-b)^2) = 2\iint_{}^{B} ((x-a)^2)$$

And we can also use the formula for the area of a circle to get:

$$\iint_{}^{B} (a^2+b^2) = \pi (a^2+b^2)^2$$

So all I have left to do is calculate $\iint_{}^{B} ((x-a)^2)$, but this is where I get stuck. Trying to do it using iterated integrals becomes too complex (we have only covered Cartesian coordinates, so I can't use something like polar coordinates here).

I know the result is supposed to be $(1/2)\pi (a^2+b^2)^2$.

Assistance would be appreciated. Thanks!

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I think a classmate of yours asked this yesterday. – Antonio Vargas May 27 '12 at 6:32
@AntonioVargas Looks like he did! Small world. Thanks, I'll have a look and see if there's anything useful there. – ro44 May 27 '12 at 6:39
@AntonioVargas No luck. The only answer provided does all the steps I already mentioned in my post, except the one I'm actually stuck on: evaluating the integral. Anyway, I'm not sure what to do with these two questions. Do we merge them in some way? – ro44 May 27 '12 at 6:42
Come on guys and girls, I'm sure some of you could give me some insights into this problem :). – ro44 May 27 '12 at 10:53
up vote 0 down vote accepted

Integrate in $y$ first; this gives the additional factor of $2\sqrt{(a^2+b^2)-(x-a)^2}$, so that the question reduces to $$\int_{a-\sqrt{a^2+b^2}}^{a+\sqrt{a^2+b^2}} (x-a)^2 \sqrt{(a^2+b^2)-(x-a)^2}\,dx$$ which may look scary, but is in fact a typical trigonometric substitution problem. Namely, $x=a+\sqrt{a^2+b^2}\sin \theta$ turns the integral into a multiple of $$\int_{-\pi/2}^{\pi/2} \sin^2\theta\, \cos^2\theta \,d\theta$$ Of course, with polar integrals this would have been much easier.

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Yeeeeeeeeeees. You are a wizard of integrals. Thank you! – ro44 May 28 '12 at 1:04

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