Question

Having some trouble with a double integral in polar coordinates, constrained by a disk of r=2. I've made a few attempts, but I'll post a screenshot of my latest try. Any help is appreciated. :)

Answer 1

Actually, this image is my latest attempt.

Answer 2

I'm also going to go run a few miles. Hopefully that'll clear my head. :P

Answer 3

In your work, one thing in your latest attempt is that:
\( \displaystyle \int_{0}^{2\pi} \int_{-2/\cos \theta}^{2/\cos \theta} \left( r^2 \right)^{5/2} \ r \ dr \ d\theta = \int_{0}^{2\pi} \int_{-2/\cos \theta}^{2/\cos \theta} \color{red}{r^7} \ dr \ d\theta\)

\(r^5 \cdot r = r^6 \), not \(r^7 \)

Once you change that, these later problems might adjust. But I'll note them also. Later on in the integration, you write this result:
\( \displaystyle \int_{0}^{2\pi} \left[ r^7 \right]_{-2/\cos \theta}^{2/\cos \theta} \ d\theta = \int_{0}^{2\pi} \left[ \left( \dfrac{2}{\cos \theta}\right)^8 \color{red}+ \ \left( \dfrac{2}{\cos \theta}\right)^8 \right] \ d\theta \)
It should be a subtraction instead of addition: F(b) - F(a).

Finally, at this last point:
\( \displaystyle \int_0^{2\pi} \left( \cos \theta \right)^{-8} \ d \theta \)
You can't use integral power rule here, you'd probably have to keep it as \( \sec \theta \) and try to find a trig identity to simplify. For example, you might pull out \(\sec ^2 \theta\) and rewrite the remaining \( \sec^6 \theta = (\tan^2 \theta + 1 )^3 \). Then apply a u-substitution.

I can't quite comment on whether the conversion to polar coordinates was correct since I haven't done enough polar coordinates double integrals to know. :P

Answer 4

Ah, thanks for the leads. That might be just what I needed. I'm reasonably confident in the conversion to polar. I'll post back once I've worked through it.

Answer 5

Ok, just getting back to this (I'd started another problem when I responded). This integral:\[\left\{\left[\frac{2}{\cos (\theta )}\right]^7+\left[\frac{2}{\cos (\theta )}\right]^7\right\}\] should be added, because the limits are from a positive value minus a negative value. Otherwise, you'd end up with zero. More in a minute. :)

Answer 6

Hmm... Judging by the limits of your Cartesian integral, it looks like you're integrating within a square region in the x-y plane rather than a disk.