Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Evaluate the iterated integral by converting to polar coordinates:

$\large \int^2_0 \int^{\sqrt{2x-x^2}}_0 xy~dy~dx$

I successfully completed most of the problem; however, I had difficulty with the limits of integration for the variable r.

To properly describe the region with polar coordinates, I drew the graph; from this, I see that r takes on the values from 0 to the points on the graph $y= \sqrt{2x-x^2}$, including the values inside. In polar form, the function would be $r^2-2r\cos\theta=0$. Unsure of how to proceed, I looked at the answer key. I saw that the upper limit of integration for r was $2\cos\theta$. From this, suspected that they divided by r. Isn't this technically a fallacious way of proceeding, though? Doesn't r take on the value 0?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

The region of integration is a semicircle. It is the top half of a circle of radius $1$ centered at $(1,0)$. If I wanted to do this in polar coordinates, I would move the origin to the center of the circle by defining $u=x-1, du=dx$ Now I have $$\int_0^2\int_0^{\sqrt{2x-x^2}}xy\; dy\ dx=\int_0^2\int_0^{\sqrt{1-u^2}}(u+1)y\; dy \; du$$. Now the integral is over the top half of the unit circle centered at the origin, so $r$ goes from $0$ to $1$ and $\theta$ goes from $0$ to $\pi$

Without the shift, $\theta$ would range from $0$ to $\frac \pi 2$ and $r$ would range from $0$ to $2 \cos \theta$. You are correct that they divided by $r$ to get the upper limit, however $r$ is not zero at the upper limit (except at $\theta=\frac \pi 2$ where you should strictly make a limit argument). $r$ does go to zero in the region of integration, but not at the edge where the division was done.

share|improve this answer
    
So, since we integrate $dr$ from 0 to $2\cos\theta$, we don't have to worry about $2\cos\theta$ having to assume the value $r=0$, because the lower limit takes care of that? If this is correct, how can one explain this is fewer indefinite words, that is, explain it more mathematical. Also, what exactly do you mean by making a limit argument? –  Mack Apr 20 '13 at 17:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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