- Introduction to the surface integral
- Find area elements
- Example of calculating a surface integral part 1
- Example of calculating a surface integral part 2
- Example of calculating a surface integral part 3
- Surface integrals to find surface area
- Surface integral example, part 1
- Surface integral example part 2
- Surface integral example part 3: The home stretch
- Surface integral ex2 part 1
- Surface integral ex2 part 2
- Surface integral ex3 part 1
- Surface integral ex3 part 2
- Surface integral ex3 part 3
- Surface integral ex3 part 4
Evaluating the surface integral. Created by Sal Khan.
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- Why does dS = r_u cross r_v? Is it a formula, or do you choose?(8 votes)
- Recall, dS is the small surface area element. The reason that a cross product is used has to do with the fact that taking the cross product between two vectors gives the area of the parallelogram formed between those two vectors. if we add all these parallelograms up, we get the total surface area. Sal goes into a good intuitive explanation of why this formula comes about in previous videos as well.(19 votes)
- why is it not r(u,v)= ui + (v^2)j - (u+v^2)k(6 votes)
- r is the position vector - i.e. it is a vector from the origin to each point on the surface S
(note that I've not used u and v since they are exactly equivalent to x and y)
The formula for the coordinates of each point on the surface is
z = x + y².
You can think of this as z being a function in terms of x and y —
z(x,y) = x + y².
In other words, each point in the x-y plane maps onto the surface and the height of that surface above the x-y plane will be
x + y².
Thus, to reach any point from the origin you can do the following:
move x units along the x-axis
move y units parallel to the y-axis
move z units parallel to the z-axis (where z = x + y²)
r(x,y) = x i + y j + (x+y²) k
NB: if you replace x with u and y with (v²) in the equation for r, then you would need to find the z above the point x=u, y=v² on the x-y plane which would be z = u + (v²)²
— i.e. z = u + v⁴!
It might be helpful to review Sal's videos on parameterization:
- What does the result mean when you evaluate a surface integral? Especially when the function f(x,y,z) is not 1. Is is volume? Because you are multiplying all these areas of these parallelograms by the function f(x,y,z)?(2 votes)
- Can you please tell when we will get the mission problems for the Multivariable calculus Differential equations and linear algebra??(3 votes)
- Why do we do integration in physics for any question to get a whole surface ,and how to know whether to use the integration in finding any surface in a question ?(2 votes)
- Hello Sal and Khan
At the very end of #67, surface integral, example 2 part 2 (this video I hope), Sal evaluates the integral of the square root of (1+2v^2) as equaling 2/3(1+2v^2)^3/2 or the integral of (1 + 2v^2)^1/2 = 2/3 (1 +2v^2)^3/2 . This seems to be incorrect. Isn't this evaluation actually a rather complex trig substitution or some other substitution?
Kenny Goldman(2 votes)
- A little bit late, but no, try it out, if you diff 2/3(1+2v^2)^3/2 you get exactly 4v(1+2v^2)^1/2 (reduce power by one, 3/2*2/3 is one, and then by chain rule inner diff of 1+2v^2 is just 4v). No trig necessary.(1 vote)
- I do not understand when to use the module of product, or product only of the partial derivatives of R.(2 votes)
- I've never heard of the module of a product, but I'll try to answer what I think you are trying to ask.
dS (the unit area) is equal to the magnitude of the vector that results from taking the crossproduct of the two surface vectors.
Therefore, if you are trying to evaluate a function over a surface,
(in this video that function is:
f(x,y,z) = y)
then you would always want the magnitude of the crossproduct.
I would highly recommend (re-)watching the first videos in this playlist.
- Why do we have to write x,y, and z in terms of u and v for the parametrization?(1 vote)
- Since the formula calls for parametrization. If you do some more online research, there are more formulas for general functions in terms x, y, z. I think Sal and Grant should have noted those formulas as well.(3 votes)
- At0:20the cross product of two partial derivatives is taken, why is the notation not
|dr(u) X dr(v)| with partial d's in front?(1 vote)
- Why do we need to have x=u etc? In other words why do we need to parametrize this if its not necessarily a simpler calculation?(1 vote)
- For almost all of these problems you have to think of the parameter space separately from the x,y,z space, so it's best to get used to the idea before you have to do more difficult problems.(1 vote)
Now that we've set up the parametrization, let's try to evaluate the integral. And the next thing we'll do is essentially try to express ds right over here in terms of du and dv. And we've seen this before. ds is going to be equal to the magnitude of the cross product of the partial of r with respect to u crossed with the partial of r with respect to v du dv. So first, let's take the cross product, and we'll do that with a 3 by 3 matrix. I'll do that right over here. We set up, and I'll just kind of fill in what r sub u and r sub v is in the actual determinant right over here. So first we have our components, i, j, and k. And now first, let's think about what r sub u is, the partial of r with respect to u. Well, its i component is going to be 1. The partial of u with respect to u is just 1, so its i component is 1. Its j component is going to be 0, partial of v with respect to u is 0. v does not change with respect to u, so this is going to be 0. This should be parentheses around here. Partial of this with respect to u is just going to be 1 again. The partial of v squared with respect to u is just 0. So this is just 1 again. And then r sub v, the partial of r with respect to v, the i component is going to be 0, j component is going to be 1, and the partial of u plus v squared with respect to v is going to be 2v. So it's a pretty straightforward determinant, so let's try to evaluate it. So the i component, it's going to be i. So we cross out this column, this row. It's going to be 0 times 2v minus 1 times 1. So essentially, it's just going to be negative 1 times i. So we're going to have negative i, so this is equal to negative i. And then the j component-- and we're going to have to put a negative out front, because, remember, we do that checkerboard pattern. So cross out that row, that column. 1 times 2v is 2v. Let me make sure I got that. 1 times 2v is 2v minus 0 times 1. So it's just going to be 2v. But since it was the j component, which is going to be negative, it's going to be negative 2vj. Let me make sure I did that right. That column, that row, 1 times 2v is 2v minus 0 is 2v. The checkerboard pattern, you'd have a negative j. So you have negative 2vj. And then we have find the k component. Cross out that row, that column, 1 times 1 minus 0. So it's going to be plus k. So if we want the magnitude of this, this whole thing right over here is just going to be the square root-- I'm just taking the magnitude of this part right over here, the actual cross product. It's going to be negative 1 squared, which is just 1, plus negative 2v squared, which is 4v squared, plus 1 squared, which is just 1. So this whole thing is going to evaluate to-- we have 2 plus 2v squared du dv. Or actually, let me-- I almost made a mistake. That would have been a disaster. 2 plus 4v squared du dv. And if we want, maybe it'll help us a little bit if we factor out a 2 right over here. This is the same thing as 2 times 1 plus 2v squared du dv. If you factor out the 2, you get this is equal to the square root of 2 times the square root of 1 plus 2v squared du dv. And I now think we are ready to evaluate the surface integral. So let's do it. All right. So let me just write this thing down here. So I'm going to write everything that has to do with v, I'm going to write in purple. So I'm just going to write the ds part right over here. It is the square root of 2 times the square root of 1 plus 2v squared. And then we're going to have du, which I'll write in green, du, and then dv. So this is just the ds part. This is just the ds. Now, we have the y right over here. y is just equal to v. All right, so I'll write that in purple. So y is just equal to v. That is y-- let me make it very clear-- and all of this is ds. And now I can write the bounds in terms of u and v. And so the u part, u is the same thing as x. It goes between 0 and 1. And then v, v is the same thing as y, and y goes between, or v goes between, 0 and 2. And I now think we're ready to evaluate. And the u and v variables are not so mixed up, so we can actually separate out these two integrals, make this a product of two single integrals. The first thing, if we look at it with respect to du, all this stuff in purple is just a constant with respect to u, so we can take it out of the du integral. We can take all this purple stuff out of this du integral. And so this double integral simplifies to the integral from 0 to 2 of-- I'll write it as the square root of 2v times the square root of 1 plus 2v squared. So I factored out all of this stuff. And then you have times the integral from 0 to 1 du, and then times d, and then you have dv. And now, if this was really complicated, I could say, OK, this is just going to be a function of u. It's a constant with respect to v. And you could factor this whole thing out and separate the integrals. But this is even easier. This integral is just going to evaluate to 1. So this whole thing just evaluates to 1. And so we've simplified this into one single integral. So this simplifies to-- and I could even take the square root of 2 out front-- the square root of 2 times the integral from v is equal to 0 to v is equal to 2 of v times the square root of 1 plus 2v squared dv. And so we really are in the home stretch of evaluating this surface integral. And here, this is basic. This is actually a little bit of u substitution that we can do in our head. If you have a function, or kind of this embedded function right over here, 1 plus 2v squared, what's the derivative of 1 plus 2v squared? Well, it would be 4v. And we have something almost 4v here. We can make this 4v by multiplying it by 4 here, and then dividing it by 4 out here. This doesn't change the value of the integral. And so now this part right over here, it's pretty straightforward to take the antiderivative. The antiderivative of this is going to be-- we have this embedded function's derivative right over here, so we can kind of just treat it like an x or a v. Take the antiderivative with respect to this thing right over here, and we get this is essentially 1 plus 2v squared to the 1/2 power. We increment it by 1, so it's 1 plus 2v squared to the 3/2 power. And then you divide by 3/2, or you multiply by 2/3, so times 2/3. So this is the antiderivative of that. And then, of course, you still have all of this stuff out front-- square root of 2 over 4 And we are going to evaluate this from 0 to 2. And actually, just to simplify it, let me factor out the 2/3. We don't have to worry about that at 2 and 0. So I'm going to factor out the 2/3, so times 2 over 3. And actually, this will cancel out. That becomes a 1. That becomes a 2. And so we are left-- this stuff over here is 1/6 times. And now if you evaluate this at 2, you have 2v squared. That's going to be 2 times 4 is 8 plus 1 is 9, 9 to the 3/2 power. So 9 to the 1/2 is 3, 3 to the 3rd is 27. So it's going to be equal to 27. And then minus this thing evaluated at 0. Well, this evaluated at 0 is just going to be 1. 1 to the 3/2 is just 1, so minus 1. And so this gives us-- oh, I almost made a mistake. There should be a square root of 2 up here, square root of 2 over 6 times 27 minus 1. So drum roll. This gives us the value of our surface integral is, let's see, 27 minus 1 is 26. So we get 26 times the square root of 2 over 6. And we can simplify it a little bit more. Divide the numerator and denominator by 2, and you get 13 square roots of 2 over 3. And we are done.