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Math Help - Double Integration - polar co-ordinates

  1. #1
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    Double Integration - polar co-ordinates

    I have to evaluate the integral between 0 and 2a of the integral between 0 and sqrt(2ax-x^2) of x^2 dxdy

    I am supposed to draw a diagram and draw a radial slice and I have to convert to polar co-ordinates.

    The only thing I know how to do is x becomes rcos(theta) and dxdy becomes r dr dtheta

    Any help would be great thanks.
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    Quote Originally Posted by jpquinn91 View Post
    I have to evaluate the integral between 0 and 2a of the integral between 0 and sqrt(2ax-x^2) of x^2 dxdy

    I am supposed to draw a diagram and draw a radial slice and I have to convert to polar co-ordinates.

    The only thing I know how to do is x becomes rcos(theta) and dxdy becomes r dr dtheta

    Any help would be great thanks.
    The region of integration is the top half a circle of radius a and center (a,0). In terms of polar coords

     <br />
\int_0^{\frac{\pi}{2}} \int_0^{2a \cos \theta} r^2 \cos^2 \theta r dr d \theta<br />
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    Thanks for the help, could you please tell me how you know what to change the limits to and what the equation/centre of the circle is?
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    Quote Originally Posted by jpquinn91 View Post
    Thanks for the help, could you please tell me how you know what to change the limits to and what the equation/centre of the circle is?
    You are going from a botton curve y = 0 to a top curve y = \sqrt{2ax-x^2}. For the latter, squaring both sides gives

    y^2 = 2ax-x^2 \;\text{or}\; x^2 -2ax + a^2 + y^2 = a^2 \; \text{or}\; (x-a)^2 + y^2 = a^2. (Just the top half).

    In polar coords this becomes

    r^2 = 2ar \cos \theta or just r = 2a \cos \theta.

    So we have a 1/2 circle centered at (a,0) with radius r = a. In order to find limits we ask, "for what values of \theta and r will sweep out this half circle. Here \theta will go from 0 to \frac{\pi}{2}and r from r = 0 to the outside curve r = 2a \cos \theta. That's how I got the limits.
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    OK, i think i've got the answer. Is 4a^4 right? I just don't understand where r^2 = 2arcos(theta) comes from.
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    Quote Originally Posted by jpquinn91 View Post
    OK, i think i've got the answer. Is 4a^4 right? I just don't understand where r^2 = 2arcos(theta) comes from.
    Not sure how you got your answer. You might want to provide some more details so we can check. As for the the r^2 = 2a r \cos \theta we have  <br />
x^2 +y^2 = 2 a x<br />
. Now x^2+y^2 = r^2 and x = r \cos \theta. Just substitute.
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    I evaluated the integral that you wrote down after changing to polar co-ords and got:

    Int between Pi/2 & 0 of (cos^2(theta)*(2acos(4theta)))^4 /4 d theata

    which is: Int between Pi/2 & 0 of (16a^4(cos^6(theta))) /4 d theta

    = 4a^4(sin^6(theta)) , subs in values of theat

    which is: 4a^4 - 0 = 4a^4 i think ?????
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  8. #8
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    Quote Originally Posted by jpquinn91 View Post
    I evaluated the integral that you wrote down after changing to polar co-ords and got:

    Int between Pi/2 & 0 of (cos^2(theta)*(2acos(4theta)))^4 /4 d theata

    which is: Int between Pi/2 & 0 of (16a^4(cos^6(theta))) /4 d theta

    = 4a^4(sin^6(theta)) , subs in values of theat

    which is: 4a^4 - 0 = 4a^4 i think ?????
    Problem \int \cos^6 \theta d \theta \ne \sin^6 \theta + c
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    is this right? Int cos^6(theta) = (9*sin(4*x)-4*sin(2*x)^3+48*sin(2*x)+60*x)/192
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    Quote Originally Posted by jpquinn91 View Post
    is this right? Int cos^6(theta) = (9*sin(4*x)-4*sin(2*x)^3+48*sin(2*x)+60*x)/192
    Yep - Maple agrees.
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    Does evaluate mean find the area or just write down the integral seeing as I don't think I would be expected to use maple?
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    Quote Originally Posted by jpquinn91 View Post
    Does evaluate mean find the area or just write down the integral seeing as I don't think I would be expected to use maple?
    I would say evaluate the integral.
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    So is 5/8 *Pi*a^4 correct? Can you do these double integrals on maple?
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  14. #14
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    Quote Originally Posted by jpquinn91 View Post
    So is 5/8 *Pi*a^4 correct? Can you do these double integrals on maple?
    Yes, and your answer is spot on!
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