Triple Integral: Evaluating ∫∫∫E sqrt(x2+y2) dV

[UrbanXrisis]

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.

the radius:

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?

 

[HallsofIvy]

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.

the radius:

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].<br /> <br /> I assume the &quot;rr&quot; was a misprint.

 

[marlon]

Try
\int_{/theta= 0}^{2\pi}\int_{r=0}^{\frac{3}{2}}\int_{z=0}^{9- 4r^2} rdzdrd\theta[/itex].<br />

<br /> <br /> The boundaries look ok to me, but shouldn&#039;t therebe a 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 ?<br /> <br /> <br /> Marlon