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joannaj93

Set up the double integral for the volume bounded between the surface z=xy^2 and the plane z-3y-x. 0<y<3^(1/2)

  • one year ago
  • one year ago

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  1. amistre64
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    |dw:1352988377508:dw|

    • one year ago
  2. amistre64
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    hmmm, does the plane the the surface intersect ?

    • one year ago
  3. TuringTest
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    Shouldn't the formula for the plane have an equals somewhere?

    • one year ago
  4. amistre64
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    • one year ago
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  5. amistre64
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    |dw:1352988979091:dw| pfft, i cant get a clear idea of the shape

    • one year ago
  6. amistre64
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    z=xy^2 and the plane z-3y-x. 0<y<3^(1/2) y = 0 to sqrt(3) z = 0 to 3x 3x-3sqrt(3)-x = 0 x = - 3sqrt(3)/2 to - 3sqrt(3)/2 z = plane to surface; (3y-x) to (xy^2)

    • one year ago
  7. amistre64
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    \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~\int_{3y+x}^{xy^2} ~dz~dy~dx\] maybe

    • one year ago
  8. amistre64
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    \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~{xy^2}-{(3y+x)}~dy~dx\] \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~{\frac13x(\sqrt3)^3}-{(\frac1233+x\sqrt3)}~dx\] \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~{x\sqrt3}-{\frac92-x\sqrt3}~dx\] \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~-\frac92~dx\] its either zero or i might have my bounds mismathed

    • one year ago
  9. joannaj93
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    Sorry that was actually suppose to be z=3y-x.

    • one year ago
  10. joannaj93
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    So the curve would equal C = xy^2 -3y+x

    • one year ago
  11. joannaj93
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    Sorry :(

    • one year ago
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