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# Surface integral ex2 part 2

Evaluating the surface integral. Created by Sal Khan.

## Want to join the conversation?

• Why does dS = r_u cross r_v? Is it a formula, or do you choose?
• 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.
• why is it not r(u,v)= ui + (v^2)j - (u+v^2)k
• 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)

That is: 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:

HTH
• 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)?
• It is the mass of the surface. If y is a density function, then denisty*area=mass.
• Can you please tell when we will get the mission problems for the Multivariable calculus Differential equations and linear algebra??
• 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?
regards
Kenny Goldman
• 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.
• 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.

HTH
(1 vote)
• 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.