I have the following problem:
Sketch the region of integration for the double integral
$$\int_{0}^{2} \int_{0}^{ \pi} y dy dx$$
Rewrite the rectangular double integral as a polar double integral, and evaluate the polar integral.
Now if I didn't have to convert the integral limits I would know what to do but I'm confused as how I do that.
I know polar coordinates have the form
$$ f(r\cos\theta,r\sin\theta) rdr d\theta$$
and I know how to convert the function, which would give me
$$\iint_R r\sin\theta rdrd\theta$$
but I do not know how to convert the limits of integration and sketch them. Any help?
2 Answers
$\int_{0}^{2} \int_{0}^{ \pi} y dy dx$
The region is bound by $x = 0, x = 2, y = 0, y = pi$
Convert to polar $x = r \cos \theta\\ y = r \sin \theta$
$x=0\\ r\cos \theta = 0\\ \theta = \frac {\pi}{2}$
$x=2\\ r\cos \theta = 2\\ r = 2 \sec \theta$
$y=0\\ r\sin \theta = 0\\ \theta = 0$
$y=\pi\\ r\sin \theta = \pi\\ r = \pi \csc \theta$
What is the angle $\theta$ at the point $(2,\pi)$?
$\theta = \tan^{-1} \frac {\pi}2$
put it together
$\int_{0}^{\tan^{-1}\frac{\pi}2} \int_{0}^{2\sec \theta} (r \sin\theta) r\;dr\;d\theta + \int_{\tan^{-1}\frac{\pi}2}^{\frac {\pi}{2}}\int_{0}^{\pi\csc \theta} (r \sin\theta) r\;dr\;d\theta$
You may find it helpful to take one of the integrals and substitute $\phi = \frac {\pi}{2} - \theta$
$\int_{0}^{\tan^{-1}\frac2{\pi}}\int_{0}^{\pi\sec \phi} (r \cos\phi) r\;dr\;d\phi$
You want $0\leq x\leqslant 2$ and $0\leqslant y\leqslant \pi$, which is a rectangle in the first quadrant with one apex at the origin, and two sides on the axii.
So then we want to integrate $\theta$ over that right angle: $0\leqslant \theta\leqslant \pi/2$.
But where shall we integrate $r$?
Polarising the limits gives: $0\leqslant r\cos\theta\leqslant 2$ and $0\leqslant r\sin\theta\leqslant \pi$
So $0\leqslant r\leq \min\{ \frac 2{\cos\theta}, \frac \pi{\sin\theta}\}$
Combining and separating into a disjoint union:
$${\left\{(r,\theta): 0\leqslant \theta < \arctan\frac \pi 2, 0\leqslant r\leqslant \frac 2{\cos\theta}\right\}}\\\cup{\left\{(r,\theta):\arctan\frac \pi2\leqslant \theta\leqslant \frac \pi 2, 0\leqslant r\leqslant \frac \pi{\sin\theta}\right\}} $$
PS: The answer should obviously be: $\pi^2$ , which is most easily obtained from the original expression, contrary to the normal intent of employing a change of variables.
