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Math Help - describe the region of integration (double integrals)

  1. #1
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    describe the region of integration (double integrals)

    describe the region of integration of the folowing integrals

    int(between 0 and 2) int (between sqrt(4-x^2) and 0) (x^2 + y^2) dy dx

    int(2 and -1) int(x+2 and x^2) (x+y) dy dx

    how would I go about describing integrals like this?

    sorry if it appears I havent thought about this - I have - I just have no idea how to think about it!

    many thanks
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  2. #2
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    Quote Originally Posted by James0502 View Post
    describe the region of integration of the folowing integrals

    int(between 0 and 2) int (between sqrt(4-x^2) and 0) (x^2 + y^2) dy dx

    int(2 and -1) int(x+2 and x^2) (x+y) dy dx

    how would I go about describing integrals like this?

    sorry if it appears I havent thought about this - I have - I just have no idea how to think about it!

    many thanks
    The region is defined by the limits of the integral.

    The first integral is :

    \int_0^2 \int_{\sqrt{4-x^2}}^{0} (x^2+y^2)dydx

    The y limits state that  \sqrt{4-x^2} < y < 0

    Now, the region  y = \sqrt{4-x^2} can be expressed  y^2+x^2 = 2^2 , which is a circle of radius 2 with its center at the original, and the region  y = 0 is a straight line in line with the x axis! So in the y axis, the region is bounded between the x axis, and the circle described.

    And the x axis limits are x = 0 and x = 2. So I'm sure you can understand what this means!
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  3. #3
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    Quote Originally Posted by James0502 View Post
    describe the region of integration of the folowing integrals

    int(between 0 and 2) int (between sqrt(4-x^2) and 0) (x^2 + y^2) dy dx

    int(2 and -1) int(x+2 and x^2) (x+y) dy dx

    how would I go about describing integrals like this?

    sorry if it appears I havent thought about this - I have - I just have no idea how to think about it!

    many thanks
    That says that x must lie between 0 and 2. On a graph, draw vertical lines x= 0 and x= 2 to bound that region. For every x, y must be between \sqrt{4- x^2} and 0 so you want to draw y= 0 and \sqrt{4- x^2}. The region is that bounded by those three lines and one curve. If you recognize that x^2+ y^2= 4 as a circle of radius
    2, you should see that y= \sqrt{4- x^2} is, since the square root is positive, the upper half of that circle. Since x lies between 0 and 2, this is the fourth of the circle in the first quadrant.
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  4. #4
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    yes, thankyou all for the help - I have the left half of a circle - is this correct?

    I'm struggling with the second one though.. I have drawn the limits but I cant see how the x + y relates?

    many thanks
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  5. #5
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    x+y is just what you have to integrate that's all.

    You just need to graph where the double integral is taken and that's all, it's quite easy, not hard.
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