How to find the flux of the vector field v through the surface S

where S is the portion of the sphere x² + y² + z² = 36 lying between the planes z =√11 and z = √20. and v(x,y,z) = xi + yj + zk. Assume the surface S is oriented in the upward direction.

Thanks for the help!!

There are two ways to do this problem: by directly computing ∫∫S v · dS, and by using the Divergence Theorem. I will use the Divergence Theorem here.

Note that, by Divergence Theorem, we have:
∫∫S v · dS = ∫∫∫E div v dV,

where E is the part of the ball enclosed by S.

In spherical coordinates, we can represent E using:
x = p*cos(a)sin(b), y = p*sin(a)sin(b), z = p*cos(b).

Note that I am using a and b instead of θ and φ to make this easier to type.

We need to determine the limits for a and b. As usual, 0 ≤ a ≤ 2π. However, the fact that we are between z = √11 and z = √20 will restrict b. Since z = 6cos(b), we see that:
√11 < z < √20 ==> √11 < 6cos(b) < √20 ==> √11/6 < cos(b) < √5/3.

We can describe the part of the ball between the two planes using spherical coordinates:
E = {(p, a, b) | 0 ≤ p ≤ 6, 0 ≤ a ≤ 2π, √11/6 ≤ cos(b) ≤ √5/3}.

Note that we don't actually need to determine the range for b, only the range for cos(b). You will see why when the integral is computed.

Thus:
∫∫S v · dS = ∫∫∫E div v dV
= ∫∫∫E (1 + 1 + 1) dV
= 3 ∫∫∫E dV
= 3 ∫∫∫ p^2*sin(b) dp da db (from p=0 to 6) (from a=0 to 2π) (from cos(b)=√11/6 to √5/3)
= 3 ∫ p^2 dp (from p=0 to 6) ∫ da (from a=0 to 2π) ∫ sin(b) db (from cos(b)=√11/6 to √5/3)
= [p^3 (evaluated from p=0 to 6)][a (evaluated from a=0 to 2π)][-cos(b) (evaluated from cos(b)=√11/6 to √5/3)]
= (216 - 0)(2π - 0)(√11/6 - √5/3)
= 72π(√11 - 2√5).

I hope this helps!