## anonymous 3 years ago 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)

1. amistre64

|dw:1352988377508:dw|

2. amistre64

hmmm, does the plane the the surface intersect ?

3. TuringTest

Shouldn't the formula for the plane have an equals somewhere?

4. amistre64

5. amistre64

|dw:1352988979091:dw| pfft, i cant get a clear idea of the shape

6. amistre64

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)

7. amistre64

$\int_{-3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~\int_{3y+x}^{xy^2} ~dz~dy~dx$ maybe

8. amistre64

$\int_{-3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~{xy^2}-{(3y+x)}~dy~dx$ $\int_{-3\sqrt3/2}^{3\sqrt3/2}~{\frac13x(\sqrt3)^3}-{(\frac1233+x\sqrt3)}~dx$ $\int_{-3\sqrt3/2}^{3\sqrt3/2}~{x\sqrt3}-{\frac92-x\sqrt3}~dx$ $\int_{-3\sqrt3/2}^{3\sqrt3/2}~-\frac92~dx$ its either zero or i might have my bounds mismathed

9. anonymous

Sorry that was actually suppose to be z=3y-x.

10. anonymous

So the curve would equal C = xy^2 -3y+x

11. anonymous

Sorry :(