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# Surface integral ex3 part 1

Breaking apart a larger surface into its components. Created by Sal Khan.

## Want to join the conversation?

• Why are we taking the integral of z?
• the question is just set up that way. We are taking the integral of z because it is the function in the given equation- if the z were zy^2 we would take the surface integral of zy^2.
• But the surface area of surface 1 is πr²=π, not 0, right?
• It gets multiplied by z, which is 0 for surface 1, so you end up with 0.
• Can we evaluate S2 by using a line integral concept since the surface is like an area of curtain between the x-y plane and z = 1 - x plane ?
• I was wondering how you would go about calculating the surface area of an intersection of two surfaces. Say you want to find the surface area of the surface cut out by x² + y² = 2x and x² + y² = z². You then know that z is the square root of 2x, but how do you find boundaries for x and y?
• What does taking a surface integral do? Does it give you the surface area?
(1 vote)
• It gives you the function for the surface area by throwing in a function in the formula.
• what is the meaning of taking the surface integral of z? Surface integral is the area of surface defined by r(s,t) right? whats z doing there?
(1 vote)
• at i could have done also the inverse? r_v x r_u?
(1 vote)
• Hi, the way to solve S2 is really interesting. At , I just don't understand why z= v, with the domain of 0<=v<= 1-cosu. Shouldn't there be 2 symmetrical triangle-like things wrapping around the chopped cylinder? Could anyone please tell me how these 2 parts are both included in this method?
(1 vote)
• I will try to make it clear that indeed both of those "triangle-like things" are included in this method. The parameter u is the angle made by r with the positive x-axis, and as u goes from 0 to pi, the x value goes from 1 to -1 on this interval, and z = 1 - x = 1 - cos(u) goes from 0 to 2 on this interval. This corresponds to the right side of the S2 as Sal drew it in the video, or you could call it the first "triangle-like thing." The second one comes as u completes it's turn, going from pi to 2pi. Here x goes from -1 back to 1 where it began, and z goes from 2 back to 0 where it began. Each "triangle-like thing" has one half the unit circle as a base, and 2 as a height. Hope this helps picture it correctly.
(1 vote)