So, I was practicing some problems and found this:

$$\int_0^2 \int_0^{\sqrt{4-x^2}} e^{x^2+y^2} dydx.$$

While converting from Cartesian to polar coordinates, $\theta$ limits will be from $0$ to $\pi /2$ but what about the $r$ limits? I am stuck on this. Is it $0$ to $2cosec θ$?

  • $\begingroup$ Thanks @Robert Z $\endgroup$ – Korra Feb 19 at 11:04

The integration region corresponds to the part of a circle with radius $2$ centered at $(0,0)$ that lies in the first quadrant. So the distance to the origin, $r$, is between $0$ and $2$. the integral becomes $$ \int_0^{\pi/2} \int_0^2 r e^{r^2} dr d \theta = \frac{\pi}{2} [\frac 12 e^{r^2}]_0^2 = \frac{\pi}{4}(e^4-1) $$

  • $\begingroup$ Thanks I arrived at the same result $\endgroup$ – Korra Feb 19 at 11:09

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