Calc 3: Surface Integral Question?

Question

How would I go about solving

Evaluate double integral of (F dS) if F=(4xy)i(–6x^2)j+(4yz)k and the surface S is given by z=xsin(3y) for 0 double integral of (F dS)=

I started out geting P,Q,R
P = 4*x*y
Q = -6*x^2
R = 4*y*z = 4*y*x*sin(3y)

Now I wanted to use the forumula
double integral of (F dS) equal Double integral over D of (-P dg/dx -Q dg/dy + R) dA
(note: d = partial)

but I can't figure what dg/dx and dg/dy mainly because I don't know what g would be.

Answer 1

Toko me a little while to figure out your notation and procedure, but I have it now. Here g is defined from the surface as z = g(x, y), in this case g(x, y) = x sin 3y. So dg/dx = sin 3y and dg/dy = 3x cos 3y. You should be able to take it from here. Note that two terms of the integrand should cancel out and it should all be pretty easy from there.

Where does this formula come from? In the special case where the surface is defined as z = g(x, y) [more generally it could be defined as h(x, y, z) = 0; then this approach wouldn't work as is], then at a given point two vectors tangential to the surface are (1, 0, dg/dx) and (0, 1, dg/dy). Taking the cross product of these gives you (-dg/dx, -dg/dy, 1) as the normal vector (with upward orientation; otherwise you take the cross product in the reverse order and get the negative of this). So F . dS becomes F . (-dg/dx, -dg/dy, 1) dA and if you let F = (P, Q, R) you get (-P dg/dx - Q dg/dy + R) dA as the integrand, as you wrote.

Answer 2

Question 1: What is the z doing in the integrand, when you have an x = y + 2z^2 as a bound? (really, you're integrating "under" the surface"). Or Question 2: What is z doing in the bounds of integration? (the 0 ≤ z ≤ 3) I am fairly certain one of these two questions should help you on your way.