Doing some old exam questions, this popped up: Calculate $$\int_Kx^2(x^2+y^2)\mathop{d(x,y)}$$ where $$K = \{(x,y)\in \mathbb{R}^2\vert x^2+y^2\le 1\}$$ I'm sure polar coordinates are the way to go, so the integral is equal to:

$$\int_{0}^{2\pi} \int_{0}^{1}r^4\cos^2(\varphi)r\mathop{drd\varphi}=\frac{\pi}{6}$$ (checked with calculator)

I'm unsure because WA says that $$\int_{0}^{1} \int_{0}^{\sqrt{1-x^2}}x^2(x^2+y^2)\mathop{dydx} = \frac{\pi}{24}$$

Am I using polar coordinates correctly? Did I make some other mistake?


Note that the domain of second integral is only a quarter circle, so both of the answers are correct.

  • 2
    $\begingroup$ For OP: always draw the regions that the limits correspond to. It will help you decide if you made a mistake, someone else did, or if there is only a mistake at a symmetry level (like in this case.. only 1/4 of the region was being considered). $\endgroup$ – Cameron Williams Aug 13 at 16:37

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