I had an absolute monstrous week on KA. I got through everything in the Surface Integrals unit, however I’ll need to go back and make notes on the final four videos in the series which all worked through a single question. Nonetheless, as you’ll see in this post, I CRUSHED this week and got a ton of work done. I definitely got a stronger understanding of how to solve surface integrals but unfortunately still haven’t quite wrapped my head around why the math works. I can tell I’m getting a lot closer to intuitively understanding it however, and I gained a few key insights this week on why it works so I’m hoping I’m not too far off from fully understanding surface integrals. It’s annoying that I’ve been working through this unit for SO long and still haven’t figured it out but on the bright-side, after ~4.5 years I’m finally starting to do math that seems like it had some real-world, practical applications which is pretty satisfying.

(That said, it seems a bit crazy that it’s literally taken me more than four years of studying math just to be able to describe a 3D shape using math. 😮‍💨 The glass-half-full way of looking at it is that I’m pretty sure this is what’s taught in second- or third-year university, so it took me about a third of the time as it would have by taking the traditional route of learning it through high-school and university. So, I have that going for me which is nice. 👍🏻)

The first video in the section was a big one and was 22 minutes long. Here are four screen shots from this video and some descriptions under each screen shot explaining what Sal was talking about:

It’s probably hard to visualize from the still image, but the diagram in the center of the bottom of the screen shot that has (x, y, z) axes is supposed to be a 3D surface mapped by a vector function r ⃗(s, t). The diagram in the bottom right corner shows a 2D plane (which I think of as some random piece of paper) which maps to different points on the 3D surface.  The white line across both surfaces indicate how they map onto each other.

In this image you can see that Sal added lines onto the (s, t) piece of paper to indicate tiny, tiny changes in the s and t directions. These tiny changes refer to the two white and orange arrows on the 3D surface. They’re supposed to indicate the partial derivatives of s and t that map onto the 3D surface.

Here Sal explained that the magnitude (i.e. the absolute value) of a cross product is equal to the parallelogram they create by sliding theirs tails to the heads of each other. Therefore, by finding the cross product of the partial derivatives of s and t, you’re able to find a tiny, tiny little patch of surface area on the 3D surface. This is denoted with “d-sigma”, dσ.

This final screen show shows the calculus for solving the surface area of a vector. (Not sure if I said that right or if that’s even correct.) I should mention that in the bottom line Sal added f(x, y, z) into the integrand which I think acts as a scaler which could, for example, be used to calculate the surfaces temperature at any given (x, y, z) point on the surface. But I’m not sure if that’s actually true.

The second, third and fourth videos in the section all worked on one question which had to do with finding the surface area of a torus. It took me until the end of Thursday to get through these videos. I spent some time reviewing cross products and vectors, in general, to try to better understand how the math works. Here’s one of the videos I watched:

After all my review, I got a slightly better grasp on what vectors represent and how they’re calculated, but I didn’t make too much progress on understanding why the cross product works or why linear algebra works, in general. I was pretty frustrated but decided to move forward not really understanding why it works remembering that after calculus I’ll be moving onto Linear Algebra, so I’ll hopefully be able to make more sense of it all then.

Although I found it confusing, I made a note that the basic steps to solve a surface integrals are:

1. Find the object’s “parameterization”. (In the example below, the torus’s parameterization was with respect to variables s and t. These variables are both angles that I think of as θ and Φ.)
2. Find the magnitude of the cross product of the partial derivatives of those parameters (which would tell you the surface area of a tiny, tiny parallelogram on the object).
3. Evaluate the magnitude of the cross product inside the double integral with respect to the ds and dt to find the sum of all the tiny, tiny parallelograms (a.k.a. the surface area).

Below this are six screen shots from those three videos and then my notes on the videos. As you can see, I took a TON of notes on these videos so I’m not going to reiterate how it all works. Hopefully my notes make it clear on their own:

So… Ya.

On Friday, I surprisingly got through the single exercise I worked through this week on my first attempt without getting any questions wrong. 😳 I was SUPER annoyed that I tried to screen shot the final question I answer (which was actually pretty hard and had me use a lot of trig) but I accidentally went back a page and then wasn’t able to screen shot the question. Here are two of the questions I worked through:

Question 1

Question 2

The fifth, sixth and seventh videos were another mini-series which all worked through the same question about finding the surface integral of a sphere. I’d already worked on a question like this last week so I didn’t make notes on the entirety of the videos, but I did make a not on why the x-coordinate of a sphere is calculated with x = cos(θ)sin(Φ) and why the y-coordinate is calculated with y = sin(θ)sin(Φ). Once again, I’m not going to bother explaining it, but here’s a screen shot from the first video of the series and my notes about it below:

(Clearly you can see in my notes that I mixed up the sin and cos for the z-axis. I believe the conventional notation use for that angle is “phi”, Φ, where the angle is measured from the origin ‘up’ the z-axis in the positive direction. This means that the distance away from the z-axis is the ‘opposite’ side of the right angle, a.k.a. sin(Φ).)

The eighth and ninth videos of the section once again worked through a single question. Here are two screen shots from that question and the notes I took working through it:

As I said at the start, I finished the week watching the final four videos in the section which, once again, all worked through a single question. I didn’t have time to make any notes on those videos, however, so will have to do that at the start of this coming week.

I’m hoping I can get through Surface Integrals quickly this coming week so that I can then also get through the following section, Surface Integrals (Articles), by the end of the week. If I can manage it, I’ll only have two more sections left in Integrating Multivariable Functions (1,280/1,600 M.P.). The two sections only have three videos and three articles in them respectively so I think I could potentially get through both sections within a week. If all goes according to plan, I will FINALLY be working on the unit test of two weeks from now. I started this unit in Week 213, so if I’m able to wrap it up in three weeks (in Week 228) that means it will have taken my 15 weeks to get through it. I suppose that’s not an unreasonable amount of time (especially considering how long it’s taken me to get through other past units), but this unit has felt like an absolute grind so I’ll be PUMPED to be moving forward when the time finally comes. 😮‍💨