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Studying for a test? Prepare with these 11 lessons on Green's, Stokes', and the divergence theorems.
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Video transcript
We're now ready to get into the meat of evaluating this surface integral. And we really need to re-express it in terms of a double integral in the domain of the parameters. And the first thing I'm going to do is rewrite this part right over here using our parameters. And we already know that n, our normal vector times our surface differential, can also be written as kind of a vector version of our surface differential that points in the same direction as our normal vector. And this is going to be the same thing. And we need to make sure that we get the order on the cross product right as the partial derivative-- and I'm going to confirm this in a second. The partial derivative of the parametrization with respect to one of the parameters crossed with the partial derivative of the parametrization with respect to the other parameter. And then that whole thing-- I'm not going to take the absolute value because I need a vector right over here-- times differentials of the parameters-- d theta dr. And we can swap these two things around depending on what will make our eventual double integral easier. But we can't swap these two things around because this would actually change the direction of the vector. So we need to make sure that this is popping us out in the right direction. So let's think about the direction that the partial with respect to r will take us. So as r increases, we're going to be moving radially outward from the center of our surface. Let me do this in a different color. As r increases, we'll be moving radially outwards. So this quantity will be a vector that looks something like that. And then as theta increases, we'll be going roughly in that direction. And so if we take the cross product of those two things-- and we could use the right-hand rule. Take your right hand, point your index finger in the direction of that yellow vector. Let me make it clear-- this is the orange vector right over here. Take your index finger in the direction of that yellow vector. So that's my index finger, my shakily drawn yellow index finger. Put your middle finger in the direction of the orange vector. So my middle, you bend it and put it in the direction of the orange vector. And we don't care what the other two fingers do. And then, your thumb will be in the direction of the cross product. So your thumb will point outward like that. That's my best attempt at drawing it. Which is exactly the direction we needed to point it. We need it to point upward here in order to be oriented properly with the direction that we are actually traversing the path. So this is actually the right order. If when we did this we got the thumb pointing into it or below the plane, then we would actually have to swap these orders. So with that out of the way, let's actually evaluate this cross product. So the cross product of the partial of our parametrization with respect to r crossed with the partial of our parametrization with respect to theta. I like to set up the matrix to take the cross product. So we'll put i, j, and k components, < like that. And then, first, I will write the partial with respect to r. So the i-component, if you take the derivative of this with respect to r, is just going to be cosine theta. The derivative of this with respect to r is just sine theta. And the derivative of this with respect to are r is just going to be a negative sine theta. And then we're going to cross it with this, so the derivative of this with respect to theta is going to be negative r sine of theta. The derivative of our j-component with respect to theta will be r cosine theta. And the derivative or our k-component-- or our z-component with respect to theta-- is going to be-- let's see-- negative r cosine theta. Is that right? Yeah. Derivative of sine theta's cosine theta. Yeah. So it's negative r cosine theta. And now we just evaluate this determinant over here. Our i-component is going to be-- so ignore that row and this column. And we get sine times negative r cosine theta. I'll just do this in a new color. So we're going to get negative r-- that wasn't a new color. I'll do it in purple. We will get negative r cosine theta sine theta minus-- well, this is going to be a negative number. So when you subtract a negative number, it's going to be plus. So it's going to be plus r cosine theta sine theta. It's always nice when things cancel out like this. This plus this is just 0. Negative of it plus the positive of it. So that all canceled out to 0. We don't have an i-component. Now, let's go to the j-component. And remember, we need to do our little checkerboard pattern. So it's going to be minus j. And it's going to be-- ignore this column, that row-- cosine theta times negative r. Cosine theta is negative r cosine squared theta. I just multiplied those two. And then, from that I'm going to subtract this times that. And so this times that, the negatives cancel out. We get r sine squared theta. So this is minus-- let me make sure. Yeah, I'm going to subtract the product of these two. The product is positive. It's going to be r sine squared theta. This is always a hard part. You can make a lot of careless mistakes here. And this looks like we might be able to simplify this in a second, but I'll wait. Actually, I'll just distribute this negative sign just for fun. So if we distribute the negative sign, these all become positive. Helps simplify things a little bit. And now let's worry about the k-component. The k-component I will be doing in purple-- well, I'll do it in blue. The k-component. Ignore this row, ignore this column. So plus k times cosine theta times r cosine theta is r cosine squared of theta. And then from that, I'm going to subtract negative r sine theta times sine theta. So that's going to be negative r sine squared theta, but I'm subtracting it. So it's going to be plus r sine squared theta. And this looks like we're going to simplify it as well. And so this piece right over here, we can factor out-- let me just rewrite it. This can be rewritten as r times cosine squared theta plus sine squared theta. Basic trig identity. That just evaluates to 1. So this is just r times j. And this over here simplifies for the exact same reason. This is r times cosine squared theta plus sine squared theta. This also is just 1. So this just simplifies to r times k. This whole cross product-- all of this business right over here-- simplified to, quite luckily, is equal to r times our j unit vector plus r times our k unit vector. And so now we can write our surface integral. Our original surface integral we can write as the double integral. Or we might want to change the order in which we integrate, but we'll give ourselves that option a little bit later. Double integral. Now, it's going to be over our parameter domain or the r theta domain. So it's the double integral of-- we still have the curl of f. And we're going to have to evaluate the curl of f. So I'll just write "curl of our vector field f" dotted with this business, rj plus rk. And then we have our two parameters. And we might want to switch the order. So maybe we could write d theta dr. And then if we do it in this order, theta goes from 0 to 2 pi, and r is going from 0 to 1. But if we swap these two, then obviously, we're going to have to swap these two as well. I'll leave you there. In the next video, we will evaluate the curl of f. And maybe in that video, if we have time, we will get to the finish line.