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I've tried my best in solving the following double integral. I just wanted to be sure I've done the right thing here (b is the vertical axis whilst a is the horizontal axis).

$\int\int(ab^2)\sqrt(a^2+b^2)dadb$ for $1\leq a^2+b^2\leq9, b \geq 0$

I ended up solving for the following:

$2*\int_{0}^{\pi/2}\int_{1}^{3}r^5Cos\theta Sin^2\theta drd\theta$

Am I thinking along the right lines here?

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Your integral is correct. Because of the $ab^2$ factor I'm not sure how much you have gained by going to polar coordinates. But if you now substitute $u = \sin^2 \theta$ the angle integral becomes easy, so I guess the approach is useful.

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  • $\begingroup$ many thanks. I'll also try to integrate it without the polar transformation. $\endgroup$
    – John_dydx
    Nov 3, 2014 at 23:12

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