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Convert the integral $\int_{-2}^{2} \int_0^{\sqrt{4-x^2}} e^{-(x^2+y^2)} dy\,dx$ into polar coordinates don't need to solve.

my solution is currently $\int_0^2 \int_0^\pi re^{-r^2} d\theta\, dr$

not sure if this is correct, I have subbed in $r\cos\theta$ and $r\sin\theta$ and multiplied by the jacobian, not sure if limits are correct, keep getting confused as to which way around they go, thanks in advance.

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Yes it is absolutely correct indeed we are integrating $e^{-(x^2+y^2)}$ on half circle with radius $2$ centered at the origin with $y\ge 0$ and therefore

  • $\theta \in [0,\pi]$
  • $r \in [0,2]$

and you have correctly transformed $e^{-(x^2+y^2)}$ in $e^{-r^2}$ taking into account the $r$ factor from the Jacobian.

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Your answer is correct. I would have changed the order of integration but there is nothing wrong with your integral.

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