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Math Help - Find the double integral

  1. #1
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    Find the double integral

    So I'm confused on how to take the first integral because of the radical... please explain how you came to your answer. (S=integral, so S:0,1 means the integral from 0 to 1, just to explain my notation)

    S:0,1(S:0,1 (xy((x^2+y^2)^(1/2)))dy)dx
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  2. #2
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    Quote Originally Posted by Rhode963 View Post
    So I'm confused on how to take the first integral because of the radical... please explain how you came to your answer. (S=integral, so S:0,1 means the integral from 0 to 1, just to explain my notation)

    S:0,1(S:0,1 (xy((x^2+y^2)^(1/2)))dy)dx
    I think this is what you mean

    \int_{0}^{1}\int_{0}^{1}\frac{xy}{\sqrt{x^2+y^2}}d  ydx

    It is just a u sub. let u=x^2+y^2 \implies du=2ydy Remember x is constant with respect to the inside integral.

    This is exactly like integrating

    \int_{0}^{1}\frac{3y}{\sqrt{3^2+y^2}}dy
    where three is playing the role of x!

    I hope this helps.
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  3. #3
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    Great help, but it is actually the double integral of xy*sqrt(x^2+y^2) from 0 to 1. Sorry, I don't know how to get the integral symbol to work.
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  4. #4
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    Would u-sub still work? So for the inner integral I could pull out an x, and get x times the integral of sqrt(u)du?
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  5. #5
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    Quote Originally Posted by Rhode963 View Post
    Would u-sub still work? So for the inner integral I could pull out an x, and get x times the integral of sqrt(u)du?
    It doesn't change anything the substituion still works.
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  6. #6
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    So for the inside integral, I got the answer to be (3/2)x, does that sound right to you? It seems like a strange amount of canceling. And when plugging into the final integral, I got the final answer to be 3.
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  7. #7
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    Quote Originally Posted by Rhode963 View Post
    Great help, but it is actually the double integral of xy*sqrt(x^2+y^2) from 0 to 1. Sorry, I don't know how to get the integral symbol to work.
    \int_0^1 \int_0^1 xy \, \sqrt{x^2+y^2} dy \, dx =\int_0^1 x \int_0^1 y \, \sqrt{x^2+y^2} dy \, dx.

    Just apply the suggested substitution on the inner integral.
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