# Converting limits of integration

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?

• The limits are currently in terms of $x,y$ and you just need to resolve them in terms of $r,\theta$. Think about how to parametrize a rectangular region in terms of polar coordinates. Commented Nov 28, 2016 at 23:56
• Can I do this using these two equations : $r = √ ( x^2 + y^2 )$ $θ = tan^-{1} ( y / x )$ Commented Nov 28, 2016 at 23:57
• Why would anyone want to use polar coordinates to compute that integral, other than to make students miserable and confused ... ? Commented Nov 28, 2016 at 23:58
• @CatalinZara I asked myself the same question, seems awfully unnecessary to me. Commented Nov 28, 2016 at 23:59
• @B.Goddard I agree that the instructor's job is to help students learn, not to make students feel good. But that doesn't mean forcing them through a path that pretty much guarantees unnecessary hard work and confusion. I teach my students to look for a change of coordinates if the integral looks easier in the new coordinates, not unnecessarily more complicated. Commented Nov 29, 2016 at 0:07

$\int_{0}^{2} \int_{0}^{ \pi} y dy dx$

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 $\phi = \frac {\pi}{2} - \theta$

$\int_{0}^{\tan^{-1}\frac2{\pi}}\int_{0}^{\pi\sec \phi} (r \cos\phi) r\;dr\;d\phi$

• Okay, could you explain how you got two integrals out of that. Commented Nov 29, 2016 at 0:18
• The region is a rectangle. Plot it on paper. You see that there is a corner, and the equation that describes $r$ changes on either side of that corner. Commented Nov 29, 2016 at 0:22
• Finally had some time to work on it, and yes it took nearly 2 pages of work but the answer came out to $\pi ^2/3$ for the left integral and $2\pi ^2/3$, and that totals to $\pi ^2$ Commented Nov 29, 2016 at 1:55
• 2 pages? $\int_{0}^{\tan^{-1}\frac{\pi}2} \int_{0}^{2\sec \theta} (r \sin\theta) r\;dr\;d\theta = \int_{0}^{\tan^{-1}\frac{\pi}2} \frac {8 \sin\theta}{3 \cos^3\theta}d\theta = \frac {4}{3} (\cos \theta)^{-2}|_0^{\tan^{-1}\frac{\pi}2} = \frac {4}{3} (\tan^2 \theta + 1 )|_0^{\tan^{-1}\frac{\pi}2} = \frac {\pi^2}{3}$ and $\int_{0}^{\tan^{-1}\frac2{\pi}}\int_{0}^{\pi\sec \phi} (r \cos\phi) r\;dr\;d\phi = \int_{0}^{\tan^{-1}\frac2{\pi}}\frac 13 \pi^3 sec^2 \phi d\phi= \frac 13 \pi^3 \tan\phi|_{0}^{\tan^{-1}\frac2{\pi}} = \frac 23 \pi^3$ Commented Nov 29, 2016 at 3:09
• Well yes, but in my case I had to lay it all out and show every single step of every integrating trick I used, which is why it ended up being extensive. Commented Nov 29, 2016 at 8:29

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.

Even thought this increases the number of integrals, the form of the integrand suggests integrating w.r.t. $$\theta$$ first. This gives us three integrals, as visualized by this handy plot from Desmos:

The boundaries for the circles are obvious, leaving us to solve for the boundaries of the lines

$$\begin{cases}y=0 \\ x=0 \\ y=2 \\ x=\pi \end{cases} \implies \begin{cases}\theta=0 \\ \theta=\frac{\pi}{2} \\ r\sin\theta=2 \implies \theta = \sin^{-1}\left(\frac{2}{r}\right)\\ r\cos\theta=\pi \implies \theta = \cos^{-1}\left(\frac{\pi}{r}\right)\end{cases}$$

which means we can rewrite the integral as

$$I = \int_0^2\int_0^{\frac{\pi}{2}}r^2\sin\theta\:d\theta\:dr + \int_2^\pi\int_0^{\sin^{-1}\left(\frac{2}{r}\right)}r^2\sin\theta\:d\theta\:dr + \int_\pi^{\sqrt{\pi^2+4}}\int_{\sin^{-1}\left(\frac{2}{r}\right)}^{\cos^{-1}\left(\frac{\pi}{r}\right)}r^2\sin\theta\:d\theta\:dr$$

$$\equiv I_R + I_B + I_G$$

to designate integral red, blue, and green, respectively. The red integral is the most straight forward, so I will show how the calculation for integrals blue and green are done.

$$I_G = \int_\pi^{\sqrt{\pi^2+4}} r^2\left[-\cos\theta\right]_{\sin^{-1}\left(\frac{2}{r}\right)}^{\cos^{-1}\left(\frac{\pi}{r}\right)}\:dr = \int_\pi^{\sqrt{\pi^2+4}} r^2\left(\frac{\pi}{2}-\frac{(\pi+2)}{r}\right)\:dr$$

$$= \int_\pi^{\sqrt{\pi^2+4}} \frac{\pi}{2}r^2-(\pi+2)r\:dr$$

Similarly, $$I_B$$ also only gives you a polynomial in $$r$$ rather than powers of trig. While this integral was more easily integrated in cartesian, keeping this trick in mind for more exotic boundaries or integrands would be useful.