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how would you go about using translated polar coordinates $y= rsin(theta) +5$ and $x=rcos(theta)$ and to evaluate double integrals with $f(x,y)= x^2+y^2$ bound by the circle $x^2+y^2 = 2x$? Please and thanks. I just need a general idea.

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  • $\begingroup$ Please do not keep changing your posted question. That becomes very frustrating for those who are trying to help you (who have already answered given the (original, then modified once, information)! $\endgroup$ – Namaste Jan 28 '14 at 16:07
  • $\begingroup$ You should take the time to make it complete from the start, when you first post a question. It is very tiresome to try to answer a moving target. $\endgroup$ – Namaste Jan 28 '14 at 16:10
  • $\begingroup$ I think I got it now thanks to you guys and especially @John Habert $\endgroup$ – user124473 Jan 28 '14 at 16:11
  • $\begingroup$ Glad to hear you figured it out. $\endgroup$ – John Habert Jan 28 '14 at 16:13
  • $\begingroup$ sorry @amWhy my bad I'll be more clear next time :D! $\endgroup$ – user124473 Jan 28 '14 at 16:17
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Based on original information (including first modification of question)

$x^2 + y^2 = r^2,\quad x = r\cos \theta,\quad y = r\sin\theta$

If bounded by $x^2 + y^2 = 2x$, that means the function is bounded by $$r^2 = 2r\cos \theta \iff r^2 - 2\cos\theta = 0 \iff r(r-2\cos\theta) = 0 \iff r = 0,\; r=2\cos \theta$$ So your bounds of integration with respect to $r$ are from $r = 0$ to $r = 2\cos \theta$

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  • $\begingroup$ NicetoseeyouonlineAmy:+) $\endgroup$ – mrs Jan 28 '14 at 15:59
  • $\begingroup$ $r=\pm\sqrt{2}\cos\theta$ is not the correct equation of the boundary circle in polar coordinates. $\endgroup$ – John Habert Jan 28 '14 at 16:05
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Hint: $x^2+y^2 = r^2$ and $dx\, dy = r\, dr\,d\theta$ (same is true for $dy\,dx$)

For a boundary of $x^2+y^2=2x$, transforming this into polar yields the equation $r^2=2r\cos\theta$. Solving this for $r$ gives two answers $r=0$ and $r=2\cos\theta$ (since

$r^2=2r\cos\theta \Longleftrightarrow r^2-2r\cos\theta = 0 \Longleftrightarrow r(r-2\cos\theta)=0$.

These solutions provide your limits for $r$.

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  • $\begingroup$ use translated polar coordinates given by x=rcos(theta) + 1 and y = rsin(theta) to evaluate the integral function f(x,y)= x^2+y^2 bounded by x^2+y^2 = 2x $\endgroup$ – user124473 Jan 28 '14 at 15:47
  • $\begingroup$ and what i did was $r= sqrt(2rcos(theta) + 1)$ and r = 0 for my inner integral and the outer integral was from 0 to pi/2. But evaluating the inner integral I dont what to do with the r attached to the cos(theta) $\endgroup$ – user124473 Jan 28 '14 at 15:50
  • $\begingroup$ Having the boundary in the original problem would have been more helpful than saying translated polar coordinates. $\endgroup$ – John Habert Jan 28 '14 at 15:52
  • $\begingroup$ sorry that's my bad on my part $\endgroup$ – user124473 Jan 28 '14 at 15:53
  • $\begingroup$ If you need more info on how to proceed, I can edit my answer. $\endgroup$ – John Habert Jan 28 '14 at 15:56

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