Calculate the double integral $ \iint x e^{-\sqrt{x^2+y^2}}$ using polar coordinates.

Question

Calculate the integral $R \iint f(x,y) \ dx dy$ of $f(x,y)= x e^{-\sqrt{x^2+y^2}}$ using polar coordinates if $R$ is the first quadrant ($x \ge 0, y \ge 0$). I figured out the integral bounds for the 2nd integral ($0 \le\Theta\le \frac{\pi}{2}$) but I'm not sure how to find the bounds for the first integral ($r$). Help?

Answer 1

In polar coordinates, $f(r,\theta) = r\cos\theta \cdot e^{-r}$.

\begin{align} \iint_R f(x,y) \,\mathrm{d}x\,\mathrm{d}y &= \iint_R f(r,\theta) \cdot r\,\mathrm{d}r\,\mathrm{d}\theta \\ &= \int_0^{+\infty} \int_0^{\pi/2} r^2e^{-r}\cos\theta\,\mathrm{d}r\,\mathrm{d}\theta \\ &= \left(\int_0^{+\infty} r^2e^{-r}\,\mathrm{d}r \right) \left(\int_0^{\pi/2} \cos\theta\,\mathrm{d}\theta \right) \end{align}

For the first integral, we can apply integration by parts twice, and note that $r^n e^{-r}|^{+\infty}_0 = 0 \forall n \in \Bbb N^*$. (due to l'Hôpital's Rule)

\begin{align} \int_0^{+\infty} r^2e^{-r}\,\mathrm{d}r &= -\int_0^{+\infty} r^2 \,\mathrm{d}(e^{-r}) \\ &= \left. -r^2 e^{-r}\right|^{+\infty}_0 + \int_0^{+\infty} 2r e^{-r} \,\mathrm{d}r \\ &= 0 + 2\int_0^{+\infty} r e^{-r} \,\mathrm{d}r \\ &= 0 - 2\int_0^{+\infty} r \,\mathrm{d}(e^{-r}) \\ &= \left. -r e^{-r}\right|^{+\infty}_0 + 2\int_0^{+\infty} e^{-r} \,\mathrm{d}r \\ &= \left. -2e^{-r} \right|^{+\infty}_0 \\ &= 2 \end{align}

For the second integral, it's equal to $\sin\theta|^{\pi/2}_0 = 1$.

Hence $$\iint_R f(x,y) \,\mathrm{d}x\,\mathrm{d}y = 2\cdot1 = 2.$$