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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)
 one year ago
 one year 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)
 one year ago
 one year ago

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amistre64Best ResponseYou've already chosen the best response.0
dw:1352988377508:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
hmmm, does the plane the the surface intersect ?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
Shouldn't the formula for the plane have an equals somewhere?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
dw:1352988979091:dw pfft, i cant get a clear idea of the shape
 one year ago

amistre64Best 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)
 one year ago

amistre64Best 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
 one year ago

amistre64Best 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
 one year ago

joannaj93Best ResponseYou've already chosen the best response.0
Sorry that was actually suppose to be z=3yx.
 one year ago

joannaj93Best ResponseYou've already chosen the best response.0
So the curve would equal C = xy^2 3y+x
 one year ago
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