INSTRUCTOR: Calculate the surface integral, the double integral, over the surface S of capital F dotted with dS, where F is equal to angle bracket 0 comma negative z comma y, close angle bracket, and S is the portion of the unit sphere in the first octant with outward orientation.

First, let's write our parameterization of the surface S. This will be r of phi, or phi, comma theta. And that is equal to angle bracket sine phi cosine theta, comma sine phi sine theta, comma cosine phi, close angle bracket, with 0 less than or equal to theta, less than or equal to pi over 2, and 0 less than or equal to phi, less than or equal to pi over 2 to keep us in the first octant.

From this, we can find t sub phi is equal to angle bracket cosine phi cosine theta, comma cosine phi sine theta, comma negative sine phi close angle bracket. And also t sub theta is equal to angle bracket negative sine phi sine theta, comma sine phi cosine theta, comma 0, close angle bracket.

Now, usually I don't do the cross products. But this one has a little bit of tricky things-- a few tricky things in it. So all the phis, the thetas, and the sines that we're dealing with.

So t sub phi cross t sub theta is equal to the determinant of a 3 by 3 matrix with a first row of i, then j, then k. Second row of cosine phi cosine theta, then cosine phi sine theta, then negative sine phi. And a third row of negative sine phi sine theta, then sine phi cosine theta, and then 0.

This will then be equal to angle bracket sine squared phi cosine theta, comma sine squared phi sine theta, comma sine phi cosine phi cosine squared theta, plus sine phi cosine phi sine squared phi, close angle bracket.

Now, we can simplify that last term because cosine squared plus sine squared-- actually, let me just double check that last term. Because something doesn't seem quite right. [INAUDIBLE].

Ah, that should be a sine squared theta on that last term. Because sine squared theta plus cosine squared theta is equal to 1, we can-- that would just simplify to be sine phi cosine phi. So I'm just going to go ahead and rewrite that. Sine phi, cosine phi.

Now, I want to simplify that because this next line is just as tricky. We're going to find F dotted with dS, which is, well, we're taking the dot product of the magnitude of that. Actually, first, we need to find the magnitude of that. No. No we don't. Not here.

So we're going to find F dotted with this vector that we just found the t sub phi crossed with t sub theta. So this will be F dot dotted with open parentheses t sub phi cross t sub theta, close parentheses. Open angle bracket, 0 comma negative cosine phi, comma sine phi sine theta, dotted with angle bracket sine squared phi cosine theta, comma sine squared phi sine theta, comma sine phi cosine phi, close angle bracket.

When we find the dot product of this, interestingly enough, it's equal to 0. A little bit anticlimactic. So the double integral over the surface S of F dot dS is the integral from 0 to pi over 2 with respect to theta of the integral from 0 to pi over 2 with respect to phi of function 0, which would be equal to 0.