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Example of calculating a surface integral part 1

Video transcript

We saw several videos ago that we can parameterize a torus or a doughnut shape as a position vector-valued function of two parameters. And this is the outcome that we had. I think I did it over several videos because it was a bit hairy. And I'll write our position vector-valued function first. So we have r as a function of our two parameters s and t. And then I'll review a little bit of what all the terms-- what the s, the t, and the a's and the b's represent. But it's equal to b plus a cosine of s. And once again, we saw this several videos ago. So you might want to watch the videos on parameterizing surfaces with two parameters to figure out how we got here. Times the sine of t. I'm going to put the s terms and the t terms in different colors. Times our i unit vector. I'll put the vectors, or the unit vectors in this orange color. Plus-- I'll do it in the same yellow. Plus b plus a cosine of s times cosine of t times the j unit vector-- the unit factor in the y direction. Plus a sine of s times the k unit vector or the unit vector in the z direction. And in order to generate the torus or the doughnut shape, this is true for our parameters-- so we don't wrap multiple times around the torus-- for s being between 0 and 2 pi and for t being between 0 and 2 pi. And just as a bit of review, we're all of this came from-- and I'm going to have to do what my plan is for this video over several videos. But let's review where all of this came from. Let me draw a doughnut. My best effort at a doughnut right here. That looks like a doughnut or a torus. And you can imagine a torus, or this doughnut shape is kind of the product of two circles. You have the circle that's kind of the cross section of the doughnut at any point. You could take it there. You could take it over there. And then you have the circle that kind of wraps around all of these other circles or these other circles wrap around it. And so, when we derived this formula up here or this parameterization, a was the radius of these cross sectional circles. That's a. That's what these a terms were. And b was the distance from the center of our torus out to the center of these cross sections. So this was b. So you can imagine that b is kind of the radius of the big circle up to the midpoint of the, I guess, cross section. And a is the radius of the cross sectional circles. And when we parameterized it, the parameter s was essentially telling us how far-- s was telling us how far or where are we wrapping around this circle. So it's an angle from 0 to 2 pi to say where we are on that circle. And t tells us how much we've rotated around the larger circle. So if you think about it, you can specify any point on this doughnut or on this surface or on this torus by telling you an s or a t. And so that's why we picked that as the parameterization. Now, the whole reason why I'm even revisiting this stuff that we saw several videos ago is we're going to actually use it to compute an actual surface integral. And the surface integral we're going to compute will tell us the surface area of this torus. So this surface right here is sigma, like that and it's being represented by this position vector-valued function. That is parameterized by these two parameters right there. And if we wanted to figure out the surface area, if we just kind of set it as the surface integral we saw in, I think, the last video at least the last vector calculus video I did that this is a surface integral over the surface. Here this capital Sigma does not represent a sum, it represents a surface of a bunch of the little d sigmas-- a bunch of the little chunks of the surface. And just as a review, you can imagine each d sigma is a little patch of the surface right there. That is a d sigma. It's a double integral here because we want to add up all of the d sigmas in 2 directions. You can imagine one kind of rotating this way around the torus and then the other direction is going in the other direction around the torus. So that's why it's a double integral. And this is just going to give you the surface area, which is the whole point of this video and probably the next one or two videos. But if you wanted to also multiply these sigmas times some other value-- there's some scalar field that this is in that you cared about-- you could put that other value right there. But here we're just multiplying it by 1. And we saw in the last video that it's a way of expressing an idea, but you really can't do much computation with this. But a way that you can express this so that you can actually take the integral, you say this is the same thing-- and we saw this in the last several videos. This is the same thing as the double integral over the region over which our parameters are defined. So it's this region over here where s and t go from 0 to 2 pi of whatever function this is. We just have a 1 here, so we could just write a 1 if we like; it doesn't change much. Times-- and this is what we learned. Times the magnitude of the partial derivative of r with respect to s. The magnitude of that crossed with the partial derivative of r with respect to t ds. You could take it in either order, but ds dt. So we saw this in the last video. What we're going to do now is actually compute this. That's the whole point of this video. We're going to take the cross product of these two vectors. So let's figure out these vectors. Then in the next video we're going to take the cross product. And then the video after that we'll actually evaluate this double integral. And you're going to see it's a pretty hairy problem and this is probably the reason that very few people ever see an actual surface integral get computed. But let's do it anyway. So the partial derivative of r with respect to s-- so this term right here. We'll do the cross product in the next video. This term is what? We just want to hold t constant and took the partial with respect to just s. So this up here, if you distributed the sine of t times b-- that's just going to be a constant in terms of s, so we can ignore that. Then you have sine of t times this over here. So sine of t and a is a constant. And you take the derivative of cosine of s. That's negative sine of s. So the derivative of this with respect to s or the partial with respect to s is going to be minus a-- I'll write in green the sine of t, so you know that's where it came from. Sine of t and then sine of s. The derivative of this is negative sine of s. That's where that negative came from. And then I'll write the sine of s right there. Times the unit vector i. That's the partial of just this x term with respect to s. And then we'll do the same thing with the y term or the j term. So plus-- same logic-- b times cosine of t with respect to s. When you take the partial [INAUDIBLE] becomes 0 so you're left with a-- well, it's going to be a minus a again because when you take the derivative of the cosine of s it's going to be negative sine of s. Let me do it. You're going to have a minus a. This cosine of t. Minus a cosine of t. That's the constant terms. Sine of s. Just taking partial derivatives. Sine of s j. And then finally, we take the derivative of this with respect to s. And that's pretty straightforward. It's just going to be a cosine of s. So plus a cosine of s k. Now hopefully you didn't find this confusing. The negative sines because the derivative of cosines are negative sines. So negative sine of s. That's why it's negative sine of s times the constant. Negative sine of s times the constant-- the constant cosine of t sine of t. So hopefully this makes some sense just as a review of taking a partial derivative. Now let's do the same thing with respect to t. And I'll do that in a different color. So we're now going to take the partial of r with respect to t. So the partial of r with respect to t is equal to-- so now this whole term over here is a constant, and so it's going to be that whole term times the derivative of this with respect to t, which is just cosine of t. So it's going to be b plus a cosine of s times cosine of t i. And then, plus-- and it's actually going to be a minus because when you take the derivative of this with respect to t it's going to be minus sine of t. So it's going to be negative and then let me leave some space for this term right here. Negative sine of t. And you're going to have this constant out there. That's a constant in t. b plus a cosine of s. That's just that term right there. Derivative of cosine t is negative sine of t times j. And then the partial of this with respect to t-- this is just a constant in t. So the partial's going to be 0. So I'll write plus 0k. Let me do all my vectors in that same color. Plus 0 times the unit vector k. So that gives us our partial derivatives. Now we have to take their cross product, then find the magnitude of the cross product, and then evaluate this double integral. And I'll do that in the next couple of videos.