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Math Help - change of coordinates

  1. #1
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    change of coordinates

    i tried to use Latex but i failed so here's the problem the double integral of (x^2 + y^2) ^1/2 dx dy the inner limit is from 0 to (3y-y^2)^1/2 and the outer limit is from 0 to 3

    i know how to change the coordinates but having problem figuring out the limits for this problem
    Last edited by sean123; April 12th 2010 at 02:10 AM. Reason: latex error
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  2. #2
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    Quote Originally Posted by sean123 View Post
    i tried to use Latex but i failed so here's the problem the double integral of (x^2 + y^2) ^1/2 dx dy the inner limit is from 0 to (3y-y^2)^1/2 and the outer limit is from 0 to 3

    i know how to change the coordinates but having problem figuring out the limits for this problem
    \int_{y= 0}^3\int_{x= 0}^{3y- y^2} \sqrt{x^2+ y^2} dxdy
    x= 3y- y^2 is a parabola with horizontal axis. It crosses the y-axis at (0, 0) and (0, 3). Completing the square, x= \frac{9}{4}- (y- \frac{3}{2})^2 so that y- \frac{3}{2}= \pm\sqrt{x- \frac{9}{4}}. That tells us that x ranges from 0 to \frac{9}{4} and that, for each x, y ranges from \frac{3}{2}- \sqrt{x- \frac{9}{4}} to \frac{3}{2}+ \sqrt{x- \frac{9}{4}}.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    \int_{y= 0}^3\int_{x= 0}^{3y- y^2} \sqrt{x^2+ y^2} dxdy
    x= 3y- y^2 is a parabola with horizontal axis. It crosses the y-axis at (0, 0) and (0, 3). Completing the square, x= \frac{9}{4}- (y- \frac{3}{2})^2 so that y- \frac{3}{2}= \pm\sqrt{x- \frac{9}{4}}. That tells us that x ranges from 0 to \frac{9}{4} and that, for each x, y ranges from \frac{3}{2}- \sqrt{x- \frac{9}{4}} to \frac{3}{2}+ \sqrt{x- \frac{9}{4}}.
    i'm sorry but the the limit of 3y-y^2 should be raised to 1/2
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  4. #4
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    Okay, correcting for that:

    \int_{y= 0}^3\int_{x= 0}^{\sqrt{3y- y^2}} \sqrt{x^2+ y^2} dxdy
    x= (3y- y^2)^{1/2} leads to x^2= 3y- y^2 or x^2+ y^2- 3y= 0 so  x^2+ y^2- 3y+ 9/4= 9/4. That is the same as x^2+ (y- 3/2)^2= 9/4, a circle with center at (0, 3/2) and radius 3/2. The original function, then, is the right semicircle.

    So when you said "changing the coordinates" you did NOT mean "reverse the order of integration" but rather "change to polar coordinates"!

    To stay on the right side of the y-axis, take \theta from -\frac{\pi}{2} to \frac{\pi}{2} and, of course, r from 0 to 3/2. The "differential of area" in polar coordinates is rdrd\theta so your integral becomes \int_{r=0}^{3/2}\int_{\theta= -\frac{\pi}{2}}^{\frac{\pi}{2}} r^2 d\theta dr.

    That should be easy!
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