# How to convert a circle off origin to polar coordinates for integration?

I am trying to find the surface area of $x^2+y^2+z^2=a^2$ over the region $x^2 +y^2 \leq ax$. I rewrote the region as $\left(x-\frac{a}{2}\right)^2 + y^2 \leq \frac{a^2}{4}$. This is where I am stuck. I do not know how to write the $\mathbf{bounds}$ for this region (circle) in polar coordinates because the circle isn't centered at the origin. I tried converting the region equation to polar coordinates but that did not help and made the integration a huge mess. A hint would be much appreciated, I really want to solve the majority of this problem.

• If you're looking for surface area, does it matter all that much if you just move the whole thing to be centered at the origin? Jan 22, 2014 at 23:31
• @MartianInvader Sorry, I forgot to mention that the surface is a sphere with radius a; $x^2 + y^2 +z^2 = a^2$ Jan 22, 2014 at 23:33
• Then I think you have a typo in your question, did you mean for your region to be $x^2+y^2 \leq ax$? Otherwise you've described a "region" that's just a curve. Jan 22, 2014 at 23:37
• @MartianInvader Yes, sorry about that. Fixed. Jan 22, 2014 at 23:38

Draw your circle (the boundary of your region). It has diameter from the origin to $A=(a,0)$. The $\theta$ values in this region go from $-\pi/2$ to $\pi/2$, so that's the first part of your answer. To find the $r$ values, draw a line from the origin making an angle $\theta$ with the $x$-axis, where $-\pi/2<\theta<\pi/2$. Label the intersection of the line and circle $P$, and consider the segment $OP$. The minimum value of $r$ on the segment is obviously $0$. The maximum value of $r$ on the segment depends on the value of $\theta$; you need to find $r_{\rm max}$ in terms of $\theta$.
Hint: consider the triangle $\triangle OPA$.
$$x={a\over 2}+r\cos\theta,\qquad y=r\sin\theta.$$
• . . . where $r=a/2$. This is not quite polar coordinates, but it might be a good way of doing the integral, I haven't tried. Jan 23, 2014 at 0:02