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The integral to solve is as follows:

$$\iiint(x^2-y^2)^2 dV$$

Bounded in the first octan by $z=\frac{1}{x^2+y^2},x+y=2,x+y=1$.

In order to solve it and practice transition to cylindrical bounds, I tried to turn it into a cylindrical triple-integral. The bounds, as I saw it, turn into:

$z=\frac{1}{r^2}, r=\frac{1}{\cos\theta+\sin\theta},r=\frac{2}{\cos\theta+\sin\theta}$ while $0\le \theta \le \frac{\pi}{2}$ as we are in the first octant. So I get the following integral:

$$\int_0^\frac{\pi}{2}\int_\frac{1}{cos\theta+sin\theta}^\frac{2}{{cos\theta+sin\theta}}\int_0^\frac{1}{r^2}r^5 (cos^2(2\theta))dzdrd\theta$$

Solving it until the last integration leads to:

$$\frac{15}{4}\int_0^\frac{\pi}{2}\frac{\cos^2(2\theta)}{(\cos\theta+\sin\theta)^4}d\theta$$

Which leads me to believe the boundaries I have for $r$ are wrong, yet I can't figure out why - it seems logical to convert the boundaries to that. Any help in spotting the mistake will be extremely appreciated!

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  • $\begingroup$ Draw the picture in the $xy$-plane. There isn't a unique region determined in the first quadrant. Either way, your $\theta$ limits are wrong, and one of your equations for $r$ is wrong. $\endgroup$
    – Ted Shifrin
    May 31, 2014 at 1:24
  • $\begingroup$ My post had the second $x+y$ wrong, corrected now. $\endgroup$
    – Studentmath
    May 31, 2014 at 1:26

1 Answer 1

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HINT: Write the denominator as $(1+\sin2\theta)^2$. You should get an integral of something like $(\sec 2\theta-\tan 2\theta)^2$, which is quite doable.

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