Evaluating Surface Integrals

Surface integrals are kind of like higher-dimensional line integrals, it's just that instead of integrating over a curve C, we are integrating over a surface S. This can be tricky, but it has lots of applications, so let's learn how to do these things!

Script by Howard Whittle

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[theblackstoneproject]

Bravo! You have done what no one else on the internet (or my calc 3 professor) had been able to do: explain the many different cases you encounter when you do surface integrals.

[feeneyko]

That works for me! I really appreciate how concise it is! Many awesome videos drill deep into these concepts, but what I really need is ready-to-use skills!

[infinitasalo]

Just in time for my Calc III final, thank you so much

[itsazizkazemi]

We did it guys. We've successfully removed the numbers from math.

[cameronsantiago]

In my opinion, this topic is the most difficult to grasp out of all the topics in Calculus. However, difficulty is subjective

[olgaseverdenko]

I have a vector calculus exam tomorrow and I got completely confused about all the formulas and how and where to use them. After watching your videos I finally figured it all out in a couple hours!!! Your videos are very direct and specific! The explanations are great! Thank you so much for the videos and your help! You are a lifesaver for students

[codatheseus]

Ooo I'm here to help with my understanding of exterior algebra, linear algebra, and geometric algebra. Thanks Dave, I've been loving your back and forth with narcissist James Tour and you're helping me out with my other passions too. I love it.

[justpaulo]

Thank you for the video!

I could actually calculate the comprehension problem in my head b/c it's quite intuitive.
Basically the surface S is an inclined plane (z=1+x) that given the x limits goes from z=1 up to z=2.
Since we want to integrate f=z on that plane, we can take the area of the plane (i.e. 2*√2), and multiply it by the average value of z (i.e. z=1.5), leading to the final answer of 3*√2.

NOTE: The area of the plane is 2*√2 b/c the length of one side of the plane is the y range (equal to 2) and the other is √2 b/c that other side of the plane goes across 1 unit (x=0 -> x=1) as it goes up 1 unit too (z=1 -> z=2).

[nlikesmath]

I've been looking for a derivation of dS. seems all the videos online like to skip this detail. thank you

[MrCEO-jwvm]

Thank you so much, sir! I thought I won't catch up in my vector calc class, but I'm getting ahead now!

[rajashruthim]

I love all ur lectures, they were cool and clear 🔥♥️💯👍

[sotirisbakas]

You just explained it better than my Calculus II did in 6 months. Thank you!

[mage]

This helps me with the book introduction to electrodynamics, by Griffiths. Also we can compute the integral for any kind of surface.

[Kiky_MedPhysicist]

Thank you sir for your dedication and for making this free! 🙏

[chokonma]

This is very well explained, thank you for this!

[kyleparsotan]

11:48 @Professor Dave Explains I got 3 as the answer. I think I did something wrong. Used double integral (-Pdelz/delx -Qdelz/dely + R)dxdy

[kosiscatharsis]

In the problem the f will come out as (0, 0, 1+x), rx=(1, 0, 1), ry=(0, 1, 0), rx*ry=(-1, 0, 1). And the answer after solving the double integral is coming as 3 not 3(2)^1/2.

[eriklam]

11:40 - I got 2 times square root 2 instead.
The length of Z is square root 2 between X=0 and 1, length of Y is 2 between Y=0 and 2.
Area = length of Z times length of Y.
Not sure if I missed something?

[eggxecution]

made it, it seems the problem uses the formula of 6:10 because we have a function or a scalar field f(x, y, z) an not a vector field F or so that's how I comprehend it

[sushimeng]

Thanks for your video, it helps me comprehend the main concepts of surface integrals! And one thing that confused me is that at 9:00, why there is a vector r instead of the unit normal vector n?