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joannaj93
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Set up the double integral for the volume bounded between the surface z=xy^2 and the plane z3yx. 0<y<3^(1/2)
 2 years ago
 2 years ago
joannaj93 Group Title
Set up the double integral for the volume bounded between the surface z=xy^2 and the plane z3yx. 0<y<3^(1/2)
 2 years ago
 2 years ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1352988377508:dw
 2 years ago

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

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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
z=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)
 2 years ago

amistre64 Group TitleBest 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
 2 years ago

amistre64 Group TitleBest 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
 2 years ago

joannaj93 Group TitleBest ResponseYou've already chosen the best response.0
Sorry that was actually suppose to be z=3yx.
 2 years ago

joannaj93 Group TitleBest ResponseYou've already chosen the best response.0
So the curve would equal C = xy^2 3y+x
 2 years ago
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