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