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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)
|dw:1352988377508:dw|
hmmm, does the plane the the surface intersect ?
Shouldn't the formula for the plane have an equals somewhere?
|dw:1352988979091:dw| pfft, i cant get a clear idea of the shape
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)
\[\int_{-3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~\int_{3y+x}^{xy^2} ~dz~dy~dx\] maybe
\[\int_{-3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~{xy^2}-{(3y+x)}~dy~dx\] \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~{\frac13x(\sqrt3)^3}-{(\frac1233+x\sqrt3)}~dx\] \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~{x\sqrt3}-{\frac92-x\sqrt3}~dx\] \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~-\frac92~dx\] its either zero or i might have my bounds mismathed
Sorry that was actually suppose to be z=3y-x.
So the curve would equal C = xy^2 -3y+x