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Math Help - Polar coordinates and double integrals

  1. #1
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    Polar coordinates and double integrals

    Evaluate the given integral by changing to polar coordinates.
    where D is the region in the first quadrant that lies between the circles x^2 + y^2 = 16 and x^2 + y^2 = 4x.

    I'm confused on how to find the boundaries, since the second equation doesn't equal a constant. I changed everything to polar coordinates and got r
    ^2 cos ^2 x+r^2 sin ^2 x=16 so, r^2 = 16, r =4 so one of the boundaries would be 0 to 4. Also, when I change the integral to polar coordinates, I got the double integral of rcosx r dr dtheta. How do I find the other boundaries?

    Thanks so much!
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  2. #2
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    Quote Originally Posted by juicysharpie View Post
    Evaluate the given integral by changing to polar coordinates.
    where D is the region in the first quadrant that lies between the circles x^2 + y^2 = 16 and x^2 + y^2 = 4x.

    I'm confused on how to find the boundaries, since the second equation doesn't equal a constant. I changed everything to polar coordinates and got r^2 cos ^2 x+r^2 sin ^2 x=16 so, r^2 = 16, r =4 so one of the boundaries would be 0 to 4. Also, when I change the integral to polar coordinates, I got the double integral of rcosx r dr dtheta. How do I find the other boundaries?

    Thanks so much!
    x^2 + y^2 = 4x \Rightarrow r^2 = 4 r \cos \theta \Rightarrow r = 4 \cos \theta.
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  3. #3
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    Hi,

    Thanks for your help.
    I did use 4cos theta as the upper boundary, and 4 as a lower boundary when integrating with respect to r. I also used pi/2 and 0 as the upper and lower boundaries, respectively when integrating with respect to theta. I got the wrong answer though.. \int_0^\frac{\pi}{2} \ \int_0^{4 \cos \theta} \ r^2 \cos \theta dr d \theta . What I got was 64pi-256/12, but the answer isn't right. Are my boundaries incorrect?

    Thanks again.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by juicysharpie View Post
    Hi,

    Thanks for your help.
    I did use 4cos theta as the upper boundary, and 4 as a lower boundary when integrating with respect to r. I also used pi/2 and 0 as the upper and lower boundaries, respectively when integrating with respect to theta. I got the wrong answer though.. \int_0^\frac{\pi}{2} \ \int_0^{4 \cos \theta} \ r^2 \cos \theta dr d \theta . What I got was 64pi-256/12, but the answer isn't right. Are my boundaries incorrect?

    Thanks again.
    the boundaries of r and of theta is not correct see this graph it maybe help you
    remember you need the area between them not the area of the small circle and theta see it
    if there anything dose not clear ask

    Polar coordinates and double integrals-hint.jpg
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  5. #5
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    Thank you for the visual aid!
    I'm not really sure how to find boundaries using the graph. Are the boundaries for "r" 0 to 4? Also, I thought that since the problem said that it was in the first quadrant, the boundaries for theta would be 0 to pi/2.
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  6. #6
    MHF Contributor Amer's Avatar
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    ohh sorry the boundary for theta is correct since he said it was in the first Q

    but r change from the small
    the small circle r=4cos(theta)
    the big circle r=4 so the boundary is from 4cos(theta) to 4 right
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