Working through KA this week was similar to the past few in that I didn’t actually make any progress earning Mastery Points but did learn quite a bit and made sense of some complicated concepts. The only thing I did this week was work through two articles from the section Polar, Spherical, and Cylindrical Coordinates, and didn’t even finish the second article… I always found polar coordinates difficult to understand and thankfully this week I got a lot of clarity into how and, more importantly, why they work. By the end of the week, I was more-or-less even able to visualize them in 3D! (Though I still find it pretty confusing and tricky to fully wrap my head around.) Like the past few weeks, I’m once again a bit disappointed that I didn’t make much progress getting through exercises this week, but I’m happy that I actually learned quite a bit. As I’ve said before, in the bigger picture, that’s definitely more important.

The first article I worked through was titled Triple Integrals in Cylindrical Coordinates. The beginning of the article talked about double integrals using polar coordinates and it was at that point where I started to understand why you need to multiply the tiny, tiny change in the angle, dθ, by the radius, r, in the double integral formula when using polar coordinates. Below are two screen shots from the article where the first one goes through the reasoning behind (r)(dθ) and the second relates it to the 3D integral (r)(dθ)(dz). Below the screen shots are my notes on why you have to use (r)(dθ):

Thinking about it now, an easy way to understand why you need to multiply dθ by r is because if you changed the angle by 1° and were measuring the change in radians 1 unit away from the origin compared to 10 units away from the origin, the former change-in-radians should be 10 times smaller than the latter. If you don’t multiply by r, then the two lengths would be “equal” in the sense that they each only change by 1°, so you HAVE to multiply by r to get the proper change in radians. This is what the screen shots means when they say that the change-in-angle is not a measurement of length.

After working through the double integral formula using polar coordinates, I was then introduced to the triple integral formula. Below is the first example I worked through. I still find this stuff very hard to understand so I’m not going to try and pretend like I actually fully know what’s going on. But regardless, here’s the first example and my notes working through it:

The entire question is fairly difficult for me to understand, but, in particular, I have a hard to thinking through how to find the bounds of the z-coordinate. In this example we’re dealing with a unit sphere, i.e. the radius of the sphere is 1. Similar to a 2D circle’s radius being equal to x² + y² = r², the radius of a sphere could be thought of as x² + y² + z² = r² where x² + y² might be thought of as (r₁)² leading to the formula for the radius of a sphere being denoted as (r₁)² + z² = (r₂)². Although I can think through each step of this example in isolation, I have a hard time visualizing the entire thing as a whole. I should mention too that this example took me a day and a half to work through because I got stuck on Step 2 where I had to find the integral of 2r(1 – r²)^1/2 and couldn’t remember how to use u-substitution. (I didn’t actually end up using u-sub but instead used my weird alphabet-substitution method where I denote each factor with a’(b(x)) and b’(x). 🤷🏻‍♂️)

I’m not going to bother go through this question in depth, but here’s a third example from that article:

I started the second article on Friday which was titled Triple Integrals in Spherical Coordinates which, as the title suggests, was very similar to the first article. The difference between cylindrical coordinates and spherical coordinates (as far as I understand it) is that instead of using dz to calculate the ‘third’ dimension, (which I think of as the ‘height’ of the object) but instead you use dΦ, which apparently in certain circumstances makes integrals easier to solve (although I don’t really understand why). Here are some screen shots from the article that go through how spherical coordinates work:

Like I said earlier talking about cylindrical coordinates, I can more-or-less make sense of how each component of spherical coordinates works individually, but it’s hard for me to visualize how it all comes together at once. I’m not going reiterate the entire thing from above, but here’s my understanding of what each component of the formula represents:

• Formula for tiny, tiny change in volume:
  • dV = (dr)(r dΦ)(r sin(Φ) dθ)
• dV
  • Tiny, tiny change in volume.
• (dr)
  • Tiny, tiny change in the radius, a.k.a. tiny, tiny change in distance away from the origin.
• (r dΦ)
  • The radius (at any given point) multiplied by the tiny, tiny change in the angle in alignment – or parallel? Not sure what’s the right word to use – to the z-axis.
• (r sin(Φ) dθ)
  • The radius multiplied by a given point on the z-axis (sin(Φ)) multiplied by a tiny, tiny change in the angle across the (x, y) plane.

This explanation could be completely wrong. I’m pretty sure I have it correct, though. Also, I believe the above terms can measure a tiny, tiny change in volume at any point within the radius of the sphere.

On Saturday I started the second example from this article and made it through the example relatively easily:

Considering how slow my progress has been lately, it was definitely a confidence booster working through that integral and finding the how of how to solve it fairly straightforward. (Understanding why the math works is a different story. 🙄)

I started the third and final example in that article at the end of Saturday and was really hoping to get through it before the end of the week but got stuck right at the beginning of it. The question asked me to convert the (x, y, z) coordinates into polar coordinates and I couldn’t understand how to do it. The article actually gave me the answer (which you can see in the screen shot below) but I wanted to actually understand why each conversion worked so I came up with and worked through an example of my own (which you can also see from my notes below). In the example I came up with, I had set a point, P, to be point on a sphere with a radius of 12 that was 30° out from the z-axis and 60° around the z-axis on the (x, y) plane in the first quadrant. (Not sure if that made sense.) Here’s the screen shot which shows the conversion equations and then my example below:

I’m guessing I’ll be able to get through this final example early this coming week and will hopefully be able to make some progress through this unit, Integrating Multivariable Functions (960/1,600 M.P.), in terms of earning mastery points. I still have 19 videos, seven articles and four exercises left to get through in this unit, not to mention the unit test itself. Given that, it’s looking pretty unlikely that I’ll actually get through it before the end of the year, but that’s ok. I’m not really too worried about it anymore, although it would be nice to move a little quicker. Who knew when I started this ~4.25 years ago that learning calculus would be this difficult. 🤷🏻‍♂️

(I remember thinking that I’d get through the entire math section in a single year. 🤣 So naïve…)