Calculus 3 Integral

Question

Evaluate ∬S​F⋅dS where F=−xi+2yj​−zk and S is the portion of y=3x2+3z2 that lies behind y=6 oriented in the positive y-axis direction

Answer 1

24π

Work:

1) Consider using divergence theorem:

Let S = the region we want the answer to.

Let S₂ = the region of the "circular cap" at y = 6, 3x² +3z² = y, so x² +z² = 2

Let E = the 3d region bounded by S and S₂.

By divergence theorem, - ∫∫∫_E div(F)dV = ∫∫_S FdS + ∫∫_S2 F dS₂

Or: - ∫∫_S2 F dS₂ -∫∫∫_E div(F)dV = ∫∫_S FdS

Note the negative on the divergence integral: This is because of the orientation of S; in the positive y direction, this points inward to the region E, which is considered negative. Note that this will affect how we do ∫∫_S2 F dS₂

2) Now take div(F) = -1+2-1 = 0. A very nice result.

3) Now take ∫∫_S2 F dS_{2 =} ∫∫_D F •nˆ dA

n^ˆ is easy to find because it's just a circle: n^ˆ =<0,-1,0> (-1 because we need it to point inward to E to stay consistent).

Now F •nˆ = -12.

So we need to find: -12∫∫_DdA . But the integral is just the area of a circle with radius √2

So the integral = -24π

Now plug into the divergence theorem: -(-24π) - 0 = answer = 24π.

This problem is also not so hard to do by just surface integral without divergence theorem.