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.
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asked Jan 22 '14 at 23:28
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$\begingroup$ 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? $\endgroup$ – MartianInvader Jan 22 '14 at 23:31
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$\begingroup$ @MartianInvader Sorry, I forgot to mention that the surface is a sphere with radius a; $x^2 + y^2 +z^2 = a^2$ $\endgroup$ – zerosofthezeta Jan 22 '14 at 23:33
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$\begingroup$ 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. $\endgroup$ – MartianInvader Jan 22 '14 at 23:37
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$\begingroup$ @MartianInvader Yes, sorry about that. Fixed. $\endgroup$ – zerosofthezeta Jan 22 '14 at 23:38
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Sorry, don't know how to insert diagrams in my post so will have to ask you to draw your own.
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 $\triangle OPA$.
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$$x={a\over 2}+r\cos\theta,\qquad y=r\sin\theta.$$
answered Jan 22 '14 at 23:32
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
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$\begingroup$ . . . 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. $\endgroup$ – David Jan 23 '14 at 0:02
