1
$\begingroup$

Compute $$\iint_D(x^2-y^2)e^{2xy}dxdy,$$ where $D=\{(x,y):x^2+y^2\leq 1, \ -x\leq y\leq x, \ x\geq 0\}.$

The area is a circlesector disk with radius $1$ in the first and fourth quadrant. Going over to polar coordinates I get

$$\left\{ \begin{array}{rcr} x & = & r\cos{\theta} \\ y & = & r\sin{\theta} \\ \end{array}, \ \ \implies E:\left\{ \begin{array}{rcr} 0 \leq r\leq 1 \\ -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \\ \end{array} \right. \right.$$

and $$J(r,\theta)=\frac{d(x,y)}{d(r,\theta)}=r.$$

So $$\iint_D(x^2-y^2)e^{2xy}r \ dxdy=\iint_Er^3(\cos^2{\theta}-\sin{2\theta})e^{r^22\cos{\theta}\sin{\theta}}drd\theta= \\ =\iint_Er^3\cos{2\theta} e^{r^2\sin{2\theta}}drd\theta = 2\int_0^{4/\pi}\cos{2\theta}\cdot\left(\int_0^1 r^3e^{r^2\sin{2\theta}}dr\right)d\theta.$$

I have no idea how to compute the inner integral. I seem to get quite complicated integrals everytime I do this.

$\endgroup$
  • $\begingroup$ You might be able to do it by parts -- but you'd probably be even better off switching the order of integration and doing the $\theta$ integral first. $\endgroup$ – Y. Forman Feb 11 '18 at 17:36
  • $\begingroup$ Isn't that equally difficult? I've tried it. $\endgroup$ – Parseval Feb 11 '18 at 17:50
  • $\begingroup$ The $\theta$ integral is equally difficult, or the resulting $r$ integral is equally difficult? $\endgroup$ – Y. Forman Feb 11 '18 at 17:58
  • $\begingroup$ The $\theta$ integral. I still get the integrand to $\cos{2\theta}e^{r^2\sin{2\theta}}$. $\endgroup$ – Parseval Feb 11 '18 at 18:08
  • 1
    $\begingroup$ You can now substitute $u = \sin 2 \theta$. Remember $r$ is constant with respect to $\theta$. $\endgroup$ – Y. Forman Feb 11 '18 at 18:15
0
$\begingroup$

Call $\alpha = \sin(2\theta)$ for simplicity.

$$\int_0^1 r^3 e^{\alpha r^2}\ dr = \int_0^1 \frac{d}{d\alpha} r e^{\alpha r^2} = \frac{d}{d\alpha} \int_0^1 r e^{\alpha r^2}\ dr$$

The latter can easily be done by parts once to get

$$\int r e^{\alpha r}\ dr = \frac{e^{a r} (a r-1)}{a^2}$$

Hence with the extrema it becomes

$$\frac{e^a (a-1)+1}{a^2}$$

Hence

$$\frac{d}{d\alpha} \left(\frac{e^a (a-1)+1}{a^2}\right) = \frac{\left(a^2-2 a+2\right) e^a-2}{a^3}$$

Getting back $\alpha = \sin(2\theta)$ and thou hast

$$\frac{\left(\sin^2(2\theta)-2 \sin(2\theta)+2\right) e^{\sin(2\theta)}-2}{\sin^3(2\theta)}$$

$\endgroup$
  • $\begingroup$ And integrating $\cos{2\theta}\cdot\text{your answer}$ is simple? $\endgroup$ – Parseval Feb 11 '18 at 18:10
  • $\begingroup$ @Parseval Not really, but this is what comes out by HIS final integral. I didn't check if the whole procedure up to that is correct. $\endgroup$ – Von Neumann Feb 11 '18 at 18:14
0
$\begingroup$

By letting $x=\frac{u+v}{\sqrt{2}},y=\frac{u-v}{\sqrt{2}}$ the given integral becomes

$$\begin{eqnarray*} \iint_{\substack{u^2+v^2\leq 1 \\ u,v\geq 0}} 2uv\,e^{u^2-v^2}\,du\,dv &=& \int_{0}^{1}\int_{0}^{\pi/2}\rho^3\sin(2\theta)e^{\rho^2\cos(2\theta)}\,d\theta\,d\rho\\&=&\frac{1}{2}\int_{0}^{1}\int_{0}^{\pi}\rho^3\sin(\theta)e^{\rho^2\cos(\theta)}\,d\theta\,d\rho\\&=&\frac{1}{2}\int_{0}^{1}2\rho \sinh(\rho^2)\,d\rho\\&=&\frac{1}{2}\left[\cosh(\rho^2)\right]_{0}^{1}=\frac{\cosh(1)-1}{2}=\color{red}{\sinh^2\left(\tfrac{1}{2}\right)}.\end{eqnarray*}$$

$\endgroup$
0
$\begingroup$

Switch the order of integration to do the $\theta$ integral first. Substitute $w = r^2\sin2\theta$, $dw = 2r^2\cos2\theta$.

$$\begin{align} \int_0^1 \int_0^\frac\pi4 2r^3 \cos2\theta e^{r^2\sin2\theta}d\theta dr &= \int_0^1 \int_0^{r^2} re^w dwdr \\ &= \int_0^1 r(e^{r^2}-1)dr \\ &= \frac12\int_0^12re^{r^2}dr - \int_0^1rdr \\&= \frac12(e-1)-\frac12 \\&=\frac e2 -1 \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.