vollume with spherical cordinates

[aron101782]

problem #21 in stewart 5E calculus

Evaluate:
[MATH]\int\int\int xyz dV[/MATH]where V lies between the spheres [MATH]\rho=2[/MATH] and [MATH]\rho=4[/MATH] and above the cone [MATH]\phi = \frac{\pi}{3}[/MATH][MATH]\int\limits_{0}^{2\pi}\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{2}^{4} \rho^5 \sin^3\phi\cos\phi \cos\theta\sin\theta d\rho d\phi d\theta[/MATH]if somebody could reason with me geomentrically why this evaluates to 0. It seems to me visually to exist.

 

[tkhunny]

Why [math][0,2\pi][/math]? Are you SURE xyz = |xyz| at all times?

 

[aron101782]

I don't know much into the analytics of the problem but the solution manual has the integral set up that way and it evaluates to 0. It seems to me that pi/3 would cut a section within 2<=p<=4.

 

[LCKurtz]

Remember, you are not calculating a volume here. That would be [MATH]\iiint_V 1~dV[/MATH], and it would be positive. But your integrand isn't positive in the region. See tkhunny's note. You might even see a reason to expect an answer of [MATH]0[/MATH].

 

[aron101782]

oh.. that's right the integrand.

 

[aron101782]

maybe one of you can answer my previous post

volume using triple integrals in spherical cordinates

here is a wedge cut out of a unit sphere \int_{0}^{\frac{\pi}{6}}\int_{0}^{\pi}\int_{0}^{1} \kappa^2\sin\theta d\kappa d\phi d\theta=\frac{\pi}{9} I would like to know how to get this same wedge only rotated 90 degrees such that theta is along the axis and phi is along the positive y axis. I...

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