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Math Help - Figuring out the bounds of an integral

  1. #1
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    Figuring out the bounds of an integral

    I understand how to solve double integrals, but I am having trouble figuring out what the bounds are. This is a problem we did in class:

    Evaluate the integral (assume S = integral)

    SS (4 - x^2 - y^2) dxdy where R is the first quadrant sector of the circle x^2 + y^2 = 4 between the lines y = 0 and x = 0. For reference there is supposed to be an R under the double integral.

    The teacher came up with these polar coordinates

    0 <= r <= 2 and 0 <= theta <= pi/4

    I understand how he got the bounds for r but what about for theta? Thanks in advance
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  2. #2
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    Quote Originally Posted by pakman View Post

    The teacher came up with these polar coordinates

    0 <= r <= 2 and 0 <= theta <= pi/4
    I think he made a mistake it should have been,
    0<=theta<=pi/2
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    Quote Originally Posted by ThePerfectHacker View Post
    I think he made a mistake it should have been,
    0<=theta<=pi/2
    But how exactly did he derive those bounds? Or you I mean.
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    Quote Originally Posted by pakman View Post
    But how exactly did he derive those bounds? Or you I mean.
    Because you start at the positive x-axis and move in the positive direction (counterclockwise) on the circle to create a quater-circle. You need to rotate 1/4 of an angle to reach 1/4 of the circle (in the first quadrant) which is pi/2 because a full circle is 2*pi radians.
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  5. #5
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    Quote Originally Posted by pakman View Post
    I understand how to solve double integrals, but I am having trouble figuring out what the bounds are. This is a problem we did in class:

    Evaluate the integral (assume S = integral)

    SS (4 - x^2 - y^2) dxdy where R is the first quadrant sector of the circle x^2 + y^2 = 4 between the lines y = 0 and x = 0. For reference there is supposed to be an R under the double integral.

    The teacher came up with these polar coordinates

    0 <= r <= 2 and 0 <= theta <= pi/4

    I understand how he got the bounds for r but what about for theta? Thanks in advance
    The integrand is about dx*dy. So the boundaries should be for the dx and the dy. How did dr and d(theta) come into the computations?

    No wonder you are lost.
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    Quote Originally Posted by ticbol View Post
    The integrand is about dx*dy. So the boundaries should be for the dx and the dy. How did dr and d(theta) come into the computations?

    No wonder you are lost.
    Sorry, I forgot to mention that the problem was supposed to be converted from dxdy to r dr dtheta.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    Because you start at the positive x-axis and move in the positive direction (counterclockwise) on the circle to create a quater-circle. You need to rotate 1/4 of an angle to reach 1/4 of the circle (in the first quadrant) which is pi/2 because a full circle is 2*pi radians.
    Thank you, now I understand (finally!)
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