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# Surface integral example, part 1

Visualizing a suitable parameterization. Created by Sal Khan.

## Want to join the conversation?

• what does it mean to not just integrate the surface but time it with x^2? • In the videos before, Sal calculated the surface area. You can think of this as summing up the number of the tiny surface elements – which is the same as assigning each surface element a value of one and then summing up over all ones.

In the current video, Sal assigns each surface element a different value, namely x^2, depending on the surface element's x position (he could have also chosen a value that depends on x, y, and z, but this makes the example more simple). You can think of the surface integral as summing up all the different values of all surface elements.
• At , why does he multiply cos(t) with cos(s)? I don't understand the reasoning behind this? • Despite being a very old comment:
On the xy plane, cos(s) and sin(s) would give us the border of the circle of unit 1 (there's an implicit "1" multiplying "cos(s)" and "sin(s)"). But looking to the z plane (the left Sal's drawing at ), you can see that the vector of unit 1, that's making a "t" angle with the xy plane, projects a vector on the xy plane (and the size of that vector is the cos(t)). Then, this projection on the xy plane (cos(t)) turned out to be the radius of the circle on the xy plane. So now, instead of an implicit "1" multiplying "cos(s)" and "sin(s)", we use "cos(t)" as the radius.

And knowing this, generalizing for any sphere of "a" radius, besides the cos(t) we can just multiply each component of the parameterization by "a", like this:
r(s, t) = (a*cos(t)*sin(s), a*cos(t)*cos(s), a*sin(t))
• @ https://www.youtube.com/watch?v=E_Hwhp74Rhc#t=583 , why doesnt the angle not need to go in the "other dirction as sal says" but is only constrained @ -pi/2 and pi/2? • Sal didn't get this wrong. The way he set this up works. Here he has made the t kind of like a latitude coordinate. The s is set up like a longitude coordinate. So for t, it need only give how much north or south the point is above or below the equator. So using the standard orientation with N at the point (0,0,1), we have t is pi/2, and so z = sin t = sin (pi/2) = 1, and likewise at S = (0,0,-1), where t = -pi/2, and z = sin t = sin (-pi/2) = -1.
• Is it possible to solve the surface integral of a sphere without first parameterizing it? • If the portion of the sphere you're integrating over is a function (like a hemisphere), then yes; you can use a geometric shortcut or the formula for functions (not recommended).

If it's the whole sphere, then you should parametrize it in spherical coordinates. The surface area element is related to the Jacobian (I can't remember exactly how this instant though), so you don't have to calculate the entire cross product.

You may also be able to use Gauss's Divergence Theorem, though then you'll probably end up switching to spherical coordinates anyway.
• Is it possible to integrate over 1dσ? • It seems that Sal comes up with a strategy on the fly to parameterize a surface with two variables. Is there a method that can be used that works every time or nearly every time, that can give you a parameterization? • You could actually "parametrize" this surface by
x=s
y=t
z=f(x,y) = f(s,t).
You'd have to divide the surface into a top part and a bottom part for positive and negative z values, so that z is a function of x and y.

In that case, you probably wouldn't even bother changing the names of the variables - you'd just stick with calling them x and y, and you'd wind up integrating over dxdy. And it would work, and it is perfectly fine to do it that way, and in some cases it works out nicely. In some cases it is the best "parametrization", even if it seems too simple to bother calling it that.

But if you try to do this particular example that way, it gets messy with square roots all over the place.

Using the geometry of the surface to choose the parametrization leads to simpler math. That's the main purpose. But just going with a parametrization like x=x and y=y and z=f(x,y) does work.
• Curious, will I get a different answer if evaluating the surface integral with a different parameterization? • At , Sal says that the radius will be smaller than it was before, but I'm confused because I thought that the radius of a unit sphere is always constant. Does he mean the z-coordinate or something like that instead of the radius?   