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amistre64

  • 5 years ago

double integral of f(x,y) = 4x^2 y -y^2; x,y bounds: y=x^3, and y=cbrt(x) ??

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  1. amistre64
    • 5 years ago
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    \[\int\limits_{a}^{b} \int\limits_{h(x)}^{f(x)} (4x^2 y-y^2)dydx\]

  2. amistre64
    • 5 years ago
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    1st Quadrant restriction

  3. anonymous
    • 5 years ago
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    \[= \int_a^b 2x^2f(x)^2 - \frac{f(x)^3}{3} - 2x^2h(x)^2 +\frac{h(x)^3}{3} dx\] \[= \int_a^b[ 2x^2(f(x) - h(x))(f(x) + h(x)) - \frac{1}{3}(f(x)^3 - h(x)^3) ]dx\] Can't do much more without knowing f(x) and h(x)

  4. amistre64
    • 5 years ago
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    the bounds of y=cbrt(x) and y=x^3 are 1 and 0 so I believe a=0, b=1; h(x)=0; f(x)=1 prolly should a used c and d for those eh...

  5. anonymous
    • 5 years ago
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    Oh. ok

  6. anonymous
    • 5 years ago
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    What are your bounds on x? Or what is the domain you're integrating the function over?

  7. amistre64
    • 5 years ago
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  8. anonymous
    • 5 years ago
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    That is the domain? or are you trying to find the area?

  9. amistre64
    • 5 years ago
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    the shaded area is the D part, im assuming that meant Domain in the book :)

  10. anonymous
    • 5 years ago
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    yeah, ok. So you are integrating over that domain.

  11. anonymous
    • 5 years ago
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    So you should set it up as either \[\int\limits_{0}^{1}\int\limits_{x^3}^{\sqrt[3]{x}}{4x^2 y -y^2}\ dydx\] or \[\int\limits_{0}^{1}\int\limits_{\sqrt[3]{y}}^{y^3}{4x^2 y -y^2}\ dxdy\]

  12. amistre64
    • 5 years ago
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    is that [sqrt(y),y] or [sqrt(y),y^3]

  13. anonymous
    • 5 years ago
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    The second version?

  14. amistre64
    • 5 years ago
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    i saw them changing stuff in the book, but at 3am it was hard to stay focuesd lol

  15. anonymous
    • 5 years ago
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    The limits on x are x=cubedRoot(y) to x=cubed(y)

  16. anonymous
    • 5 years ago
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    You can do either version

  17. amistre64
    • 5 years ago
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    thnx :)

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