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anonymous
 4 years ago
Set up the double integral for the volume bounded between the surface z=xy^2 and the plane z3yx. 0<y<3^(1/2)
anonymous
 4 years ago
Set up the double integral for the volume bounded between the surface z=xy^2 and the plane z3yx. 0<y<3^(1/2)

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352988377508:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0hmmm, does the plane the the surface intersect ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0Shouldn't the formula for the plane have an equals somewhere?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352988979091:dw pfft, i cant get a clear idea of the shape

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0z=xy^2 and the plane z3yx. 0<y<3^(1/2) y = 0 to sqrt(3) z = 0 to 3x 3x3sqrt(3)x = 0 x =  3sqrt(3)/2 to  3sqrt(3)/2 z = plane to surface; (3yx) to (xy^2)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int_{3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~\int_{3y+x}^{xy^2} ~dz~dy~dx\] maybe

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int_{3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~{xy^2}{(3y+x)}~dy~dx\] \[\int_{3\sqrt3/2}^{3\sqrt3/2}~{\frac13x(\sqrt3)^3}{(\frac1233+x\sqrt3)}~dx\] \[\int_{3\sqrt3/2}^{3\sqrt3/2}~{x\sqrt3}{\frac92x\sqrt3}~dx\] \[\int_{3\sqrt3/2}^{3\sqrt3/2}~\frac92~dx\] its either zero or i might have my bounds mismathed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry that was actually suppose to be z=3yx.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So the curve would equal C = xy^2 3y+x
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