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I'm having trouble converting this double integral from Cartesian to polar:

$$\int_0^1\int_0^{\sqrt{(1-x^2)}}e^{x^2+y^2}dydx$$

What I know: $$e^{x^2+y^2}dydx = re^{r^2}drd\theta$$ $$\sqrt{1-x^2} = \sqrt{r^2\sin^2{\theta}} = r\sin{\theta}$$

My issue:

I can't figure out how to convert the limits. I graphed $y=\sqrt{1-x^2}$ using wolfram alpha, and found that it forms half of an ellipse on the positive side of the x-axis with the origin being the center. Since it forms an ellipse, I don't know the limits of r because the radius is not uniform around the shape. I assume that the limits of $\theta$ are $0\le\theta\le{\pi/2}$ since the limits of X in Cartesian form are $0\le{x}\le1$ and the shape is an ellipse centered around the origin. Could I get some guidance in the conversion of the limits?

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If $y=\sqrt{1-x^2}$ then $x^2+y^2=1$.

The region is just the first quadrant unit circle centered at origin.

Hence $r$ should take value from $0$ to $1$.

and $\theta$ should take value from $0$ to $\frac{\pi}{2}$.

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  • $\begingroup$ Wow I feel so dumb for not trying to put both variables on the same side of the equation. Thanks so much! $\endgroup$ – Mdomin45 Dec 2 '16 at 8:19
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If $y=\sqrt{1-x^2}$ then $x^2+y^2=1$.

This is a circle ! Thus $r \in [0,1]$ and $ \theta \in [0, \frac{\pi}{2}]$

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