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# Surface integral ex3 part 4

Evaluating the third surface integral and coming to the final answer. Created by Sal Khan.

## Want to join the conversation?

• Why don't you add another r before you start integrating? Isn't the Jacobian of polar coordinates r? Or is that just when you switch from rectangular to cylindrical, where here you're never started in rectangular?
• You add r only if you are in polar coordinates. In this video Sal is not using polar coordinates but he is using variables r and theta. You could name those variables s and t and nothing would change.
• after watching all the videos for this example, although we omit the bottom surface because of the parameter 'z', how do we incorporate the surface area for it? It seems as though we could use algebra for the majority of this surface, except for s2.
• If the objective of the problem was to obtain the surface area of the figure, then yes, you could probably have found the area of the upper ellipse using algebra. But the problem wasn't to find the surface area of the figure, but to find how the scalar field `f(x,y,z) = z` behaved over the surface of the figure, and I doubt there is a way to do that without using integration.
• When evaluating the last integral at time , shouldn't it be positive (1/3)sin(theta) because the negatives cancel out? I know that this doesn't change the answer because either way that term is zero but just checking.
(1 vote)
• Nope! You're integrating cos and not differentiating cos. d(sin(ø))/dø = cos(ø). d(cos(ø))/dø = -sin(ø). So there is no negative from the integration of cosine :)