1
$\begingroup$

I want to evaluate the integral $\int_{0}^{1}\int_{0}^{1}\sqrt{x^{2}+y^{2}}dxdy$ using the substitution $x=r\cos\theta$ and $y=r\sin\theta$ where the Jacobian is given as $|J(r, \theta)|=r$ and thus $dxdy=rdrd\theta$.

I need to set the range for $r$ and $\theta$. For $\theta$, it seems obvious that it ranges from $0$ to $\frac{\pi}{2}$ to cover all parts of the rectangle. But I am confused with the range for $r$. Would it be OK to set minimum $r$ as $0$ and maximum $r$ as $\sqrt{2}$, which is the diagonal of the square domain? I think I am confused with the fundamentals of integration.

$\endgroup$
4
  • $\begingroup$ I tried setting $r$ between $0$ and $\sqrt{2}$ and WolframAlpha tells me that the calculated value is different from the answer. $\endgroup$ – Senna Dec 14 '19 at 7:52
  • $\begingroup$ No this is two integrals in polar. $\endgroup$ – Ninad Munshi Dec 14 '19 at 7:53
  • $\begingroup$ @NinadMunshi Then, should I divide the domain into two triangles? Why does the simple integration not work properly? $\endgroup$ – Senna Dec 14 '19 at 7:59
  • $\begingroup$ Because the inner most bounds of a double integral are curves, not points. They would go from one function of $r$ to another (yes in this case the origin counts as a "curve" because multiple $\theta$'s end up there). As you can see for a square, the outermost $r$ bounds would be a piecewise function of $\theta$. $\endgroup$ – Ninad Munshi Dec 14 '19 at 11:12
2
$\begingroup$

In the square, $r$ will go from $0$ to $\sqrt2$, as the furthest point from the origin within the square is $(1,1)$. But the range of allowable $\theta$ will depend on $r$. If $r<1$ we get all $\theta$ between $0$ and $\pi/2$. But for $1<r<\sqrt2$ the smallest $\theta$ we get corresponds to the point $(1,\sqrt{r^2-1})$ and so is $\theta=\tan^{-1}\sqrt{r^2-1}$. The largest theta we get is its complement. Therefore \begin{align} \int_0^1\int_0^1\sqrt{x^2+y^2}\,dx\,dy &=\int_0^1\int_0^{\pi/2}r^2\,d\theta\,dr +\int_1^{\sqrt2}\int_{\tan^{-1}\sqrt{r^2-1}}^{\pi/2-\tan^{-1}\sqrt{r^2-1}} r^2\,d\theta\,dr\\ &=\frac\pi2\int_0^1r^2\,dr+\int_1^{\sqrt2}\left(\frac\pi2 -2\tan^{-1}\sqrt{r^2-1}\right)r^2\,dr. \end{align} Good luck with that second integral!

$\endgroup$
3
$\begingroup$

With symmetry we cut the line in half at $y=x$ and claim that the integral is equal to twice its value on the bottom triangle. The $r$ bounds go from the origin to the line $x=1$, which translates to $r =\sec\theta$. Then the integral becomes

$$2 \int_0^{\frac{\pi}{4}} \int_0^{\sec\theta} r^2 dr d\theta = \frac{2}{3} \int_0^{\frac{\pi}{4}} \sec^3\theta d\theta$$

Which you can take from here with trig identities, or use the substitution $\tan\theta = \sinh(t)$:

$$\frac{2}{3} \int_0^{\sinh^{-1}(1)} \cosh^2(t) \: dt = \frac{1}{3} \int_0^{\sinh^{-1}(1)} 1 + \cosh(2t)\:dt $$ $$ = \frac{t}{3} + \frac{1}{3}\sinh(t)\cosh(t)\Biggr|_0^{\sinh^{-1}(1)} = \frac{1}{3}\sinh^{-1}(1) + \frac{\sqrt{2}}{3}$$

$\endgroup$
1
$\begingroup$

The integrand $\sqrt{x^2+y^2}$ is symmetric about the line $y=x$ yielding $$\iint_{[0,1]\times [0,1]}\sqrt{x^2+y^2}\;dydx=2\int_0^1\int_0^x\sqrt{x^2+y^2}\;dydx=2\int_0^\frac{\pi}{4}\int_0^{\sec(\theta)}r^2\;drd\theta=\frac{2}{3}\int_0^\frac{\pi}{4}\sec^3(\theta)\;d\theta$$ $$=\frac{1}{3}(\sec({\pi\over 4})\tan({\pi\over4})+\ln(\sec({\pi\over4})+\tan({\pi\over4})))=\frac{1}{3}(\sqrt2+\ln(1+\sqrt2))$$

$\endgroup$

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.