Question

Use spherical coordinates to evaluate the triple integral(x^2 + y^2 + z^2) dV, where E is the ball:x^2 + y^2 + z^2 < or =81.

Answer 1

Doesn't x^2 + y^2 + z^2 = p^2? Then I'd just have p^4 sin.phi drho dphi dtheta

Answer 2

yes. So you integral will be:
\(\int_0^{81}\int_{-\pi/2}^{\pi/2 }\int_0^{2\pi} p^2 d\theta d\phi dp\)

Answer 3

∫x2 u2 z2 d x = 1/3 x3 u2 z2

Answer 4

\[ \int_0^9 \int_{0}^{\pi } \int_0^{2\pi} r^4 \sin \theta d\phi d\theta dr \]

Answer 5

Would my limits of integration be: 0-p-9, 0-theta-2pi, 0-phi-pi? and then would I still have to put my \[\rho^2 \sin \phi\] before the differentials?

Answer 6

yes ... looks like you you took theta for phi. usually phi is used for azimuthal coordinate.

Answer 7

I usually use phi as the angle of declination. Do you have a math background or physics background? Just curious

Answer 8

physics

Answer 9

Ahh, I heard that physics people flip theta and phi around for some pointless reason. :)

Answer 10

for the sake of ease ...