Example of a flux integral, Multivariable Calculus

[@GLENNSCHEXNAYDER-n5c]

It appears that there was a sign error when factoring out the -2. The second (sin(v)) component should have a positive coefficient, but it ended up not mattering because the whole thing evaluated to zero in the end.

[@bevinmaultsby]

Oh indeed, well spotted. I'll add a note in the description.

[@rosskious7084]

Smart, does the problems, and explains the details, plus you don’t joke around but gets to the problem at hand. What more could you ask for. You should have a LOT more views. Even though you are extremely pretty, you do it with class and does not distract for the presentation.

[@bevinmaultsby]

Thank you for the very kind remarks!

[@moondxstq4297]

Hi could you explain more the differences between upward/downward normal, inward/outward, and positive/negative orientation?

[@bevinmaultsby]

Sure, please check out this video! I go over orientation, and the notion of being consistently oriented. kzbin.info/www/bejne/eKKwoGxngcx2fKM

[@eswyatt]

Ha! I just wrapped up Units 6 and 7 of this series, with notes coming in at 107 pages. Now I must add.

[@bevinmaultsby]

Wow, great job!

[@adeeba7425]

MANY MANY THANKS AND MANY MANY HEARTS AND ROSES TO YOU DR.BELVIN.SINCERELY YOURS ADEEB.ADEEB.

[@bevinmaultsby]

You're very welcome!

[@ussamafadili2580]

Thank you You helped us Thank you from the heart ❤️

[@bevinmaultsby]

You're so very welcome! I'm glad I was able to help.

[@FelipeHenrique-yq3bu]

Wonderful class, thank you!

[@bevinmaultsby]

You're so welcome! Thank you for watching :)

[@retiredaccount0]

This is an excellent video. Thank you.

[@bevinmaultsby]

Glad you enjoyed it!

[@sportmaster2586]

When I use Gauss' Divergence Theorem I obtain a different answer of 64pi, how come?

[@bevinmaultsby]

Great question. To use the Divergence Theorem, the region must be fully enclosed. So we would need to say the flux across S, the top lid and the bottom lid (three separate computations) equals the integral of the divergence across the solid 3D region. Did you account for both lids?

[@sportmaster2586]

@@bevinmaultsby Thanks for the reply. I didn't account for both lids which is where my error is, appreciate the quick response.

[@bevinmaultsby]

@@sportmaster2586 great, glad you got it sorted. Div theorem is a great way to solve this actually, since div F is just 3.

[@sportmaster2586]

@@bevinmaultsby Isn't F = (x,z,0) so isn't div(F) = 1?

[@bevinmaultsby]

@@sportmaster2586 yes, not sure what I was looking at!

[@amandaliu7409]

Hi, when I tried to do this question without parameterising the surface (I used dot product of F and the partial derivatives of z) I got -40π instead of 40π. Is it because I used upward pointing normal instead of downward pointing normal? Thank you so much

[@bevinmaultsby]

Hi Amanda, you should use a parametrization to compute a flux integral, in order to evaluate the surface differential dS. I'm worried that you replaced dS with the gradient of z(x,y), (2x,2y,1), which is not correct. Here is an approach which will probably mirror the terms you had/explain why yours came out almost correct. Based on you wrote, let's parametrize the surface as r(x,y) = (x,y,f(x,y)) (where z = f(x,y) = x^2+y^2). Then r_x = (1,0,2x), r_y = (0,1,2y), and r_x x r_y = (-2x,-2y,1) (

[@amandaliu7409]

@@bevinmaultsby Thank you so so much! This is very helpful!! I will watch more of your videos!!!

[@glennschexnayder3720]

I think I’m misunderstanding something. It looks like there’s no intersection of the vector field and the paraboloid, so I don’t understand how it can affect the flux across the surface. The vector field is completely contained in the x,y plane, is it not? What am I missing?

[@bevinmaultsby]

Good question: the vector field F exists at every point (x,y,z) in the whole 3D space (R3). For each point (x,y,z), F associates the vector whose coordinates are (x,z,0). So for example, at the point (2,3,13) on the paraboloid, we can imagine the vector (2,13,0). If you imagine the paraboloid like a cup on a table, to each point on the cup we would imagine a vector parallel to the table. Does that help?

[@glennschexnayder3720]

@@bevinmaultsby Aaaahhh, so essentially, the definition of the vector field ensures that every vector is parallel to the x,y plane, but not necessarily contained in it. That makes sense.