# Triple integrals

evaluate $$\int \int \int _E \sqrt{x^2+y^2} dV$$, where E is the solid bounded by the circular parabola $$z=9-4(x^2+y^2)$$ and the xy-plane

so here's what I did, i tried to set this up in cylindrical coordinates.

is when $$z=9-4(x^2+y^2)$$ equals with the xy-plane

so this means that z=0 and x=y

$$0=9-4(2x^2)$$
$$r=\frac{3}{\sqrt{8}}$$

the z-height:

$$z=9-4r^2$$

the angle:

theta should rotate in a ciricle so it should be 2 pi

the setup:

$$\int_0 ^{2 \pi} \int_0 ^{\frac{3}{\sqrt{8}}}\int _0 ^ {9-4r^2} r rdzdrd \theta$$

i evaluated this twice but it seems not to be the answer, where did I go wrong?

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HallsofIvy
Homework Helper
UrbanXrisis said:
evaluate $$\int \int \int _E \sqrt{x^2+y^2} dV$$, where E is the solid bounded by the circular parabola $$z=9-4(x^2+y^2)$$ and the xy-plane

so here's what I did, i tried to set this up in cylindrical coordinates.

is when $$z=9-4(x^2+y^2)$$ equals with the xy-plane

so this means that z=0 and x=y

$$0=9-4(2x^2)$$
$$r=\frac{3}{\sqrt{8}}$$
Here's your error. I don't know why you decided that x= y but in polar (cylindrical) coordinates, r is not x2, r2= x2+ y2. Your formula for the intersection of the paraboloid and z= 0 plane should be 0= 9- 4r2 so r= 3/2.

the z-height:

$$z=9-4r^2$$

the angle:

theta should rotate in a ciricle so it should be 2 pi

the setup:

$$\int_0 ^{2 \pi} \int_0 ^{\frac{3}{\sqrt{8}}}\int _0 ^ {9-4r^2} r rdzdrd \theta$$

i evaluated this twice but it seems not to be the answer, where did I go wrong?
Try
$$\int_{/theta= 0}^{2\pi}\int_{r=0}^{\frac{3}{2}}\int_{z=0}^{9- 4r^2} rdzdrd\theta[/itex]. I assume the "rr" was a misprint. HallsofIvy said: Try [tex]\int_{/theta= 0}^{2\pi}\int_{r=0}^{\frac{3}{2}}\int_{z=0}^{9- 4r^2} rdzdrd\theta[/itex]. The boundaries look ok to me, but shouldn't therebe a [tex]r^2$$ in stead of just $$r$$. I mean, one r comes from the transformation of the given integrand to polar coordinates, but we also have the Jacobian being equal to r, no ?

Marlon