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### Course: Multivariable calculus>Unit 4

Lesson 13: Flux in 3D

# Vector representation of a surface integral

Different ways of representing a flux integral. Created by Sal Khan.

## Want to join the conversation?

• at when Sal starts explaining that dr_u is just a 'differential' and not a 'partial derivative,' I am very confused by his explanation.
• The derivative is the ratio of the change in r (dr) with the change in u (du). The differential, r (dr) is simply the change in r. The derivative or partial, as is the case here, would be like Miles per Hour. The differential is just Miles traveled without without the concerned for the time it took to travel that distance.
• When Sal replaces the dS with |r_u X r_v| du dv, is he doing a change of basis there with the absolute value of the cross product vector being the Jacobian?
• Yes, as he explained explained earlier in the intro to surface integral video, when you do coordinate substitution for dS then the Jacobian is the cross-product of the two differential vectors r_u and r_v.

The intuition for this is that the magnitude of the cross product of the vectors is the area of a parallelogram. We take the infinite sum of these parallelograms (by taking the infinite sum of du and dv) and we get the surface area!
• Do you have any specific examples of calculating flux?
Thanks
• At Sal is talking about dr_u and dr_v but in the integral its just the vectors r_u and r_v.... are these equivalent?
• No, they are NOT equivalent. r_u = ∂r/∂u, and r_v = ∂r/∂v in the integral, these are the derivatives with respect to parameters u or v.

At in the video, the ∂r_u and ∂r_v came from the expression: (∂r/∂u)du × (∂r/∂v)dv. He simplifies it by saying that du cancels with ∂u, and dv cancels with ∂v. But if you do that, then you end up with something that looks like: ∂r×∂r. He just tacked on _u and _v so that you could tell them apart.
(1 vote)
• At , shouldn't it be over the region R? Also, at , what if the magnitude of the cross product is 0? How can we cancel them out?
(1 vote)
• If the magnitude was zero then you wouldn't have a normal. It was be zero giving us an integral of zero and the integral of zero is zero.
(1 vote)
• If we have a polynomial to define the surface, how would this change the problem?
(1 vote)
• I have watched the surface integral videos and the 3D flux videos. In Flux video at he cancels out partial dx with dx however in the surface integrals he does not cancel them out to derive the formula. Can someone please explain how this thing works??