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Math Help - Double Integral

  1. #1
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    Double Integral

    Please help with the following integral calculation:
    \int\!\!\!\!\!\int_S(x+2y)dxdy where S is defined by the curves: y=2\cdot x^2 and y=1+x^2.
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  2. #2
    Senior Member Spec's Avatar
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    Start with dy first: 2x^2 \leq y \leq 1+x^2

    To get the limits for dx: 2x^2=1+x^2 \Longleftrightarrow x^2=1
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  3. #3
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    Is this correct ?
    \int_{-1}^1\int_{2x^2}^{1+x^2}(x+2y)dydx=\int_{-1}^1(y^2+xy)|_{2x^2}^{x^2+1}dx=\int_{-1}^1(1+x+2x^2-x^3-3x^4)dx=
    =(x+\frac{x^2}{2}+\frac{2x^3}{3}-\frac{x^4}{4}-\frac{3x^5}{5})|_{-1}^1=\frac{32}{15}
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by sillyme View Post
    Is this correct ?
    \int_{-1}^1\int_{2x^2}^{1+x^2}(x+2y)dydx=\int_{-1}^1(y^2+xy)|_{2x^2}^{x^2+1}dx=\int_{-1}^1(1+x+2x^2-x^3-3x^4)dx=
    =(x+\frac{x^2}{2}+\frac{2x^3}{3}-\frac{x^4}{4}-\frac{3x^5}{5})|_{-1}^1=\frac{32}{15}
    it is correct
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