I once again had an abysmal week working through KA. I only managed to get through a single exercise which was the same one I was working on LAST week from the section Volume: Triangles and Semicircles Cross Sections. I finished it on Sunday after something like 7-8 attempts. I barely passed the 5-hour mark this week of time working on KA which I’m also disappointed about. I hate to make excuses, but this week I had likely the busiest/most stressful work week that I’ve had in a few years. Without going into details, everything work sorted itself out and I’m now on the other side of what was taking up so much of my time and causing me so much stress so I should be in the clear going forward in terms of being able to work on KA and hopefully get through this goddamn unit sooner rather than later. (Longest sentence of all time.)

The major breakthrough I had this week was getting a stronger grasp on how the volume equation, vol. = B * W * H = x * y * z, relates to integrals. Here’s a really messy page from my notes that explains how I think about it now:

One of the key things to be aware of (at least from the questions I was working on this week, anyways) is the direction that the area of each ‘slice’ of volume is being measured from, i.e. if that area is perpendicular to the x-axis or y-axis. You need to know that because if the area is being measured across the x-axis then the bounds of the integral will be x-coordinates and vice versa if the area is being measured along the y-axis.

Just like last week, I’m don’t have much time to spare (next week should be better!) so I’m going show 3 example questions but, once again, I’m not going to type out the math below each question.

Question 1

Question 2

(Note: In Step 5 of KA’s explanation, they change the bounds of the integral from [(–2), 2] to [0, 2] and also changed the fraction in front of the integral from 9/4 to 9/2. This took me awhile to understand. It seemed to me that if you were going to ‘cut the integral in half’ (i.e. in my mind multiply it by ½) then the fraction in front, 9/4, would need to be multiplied by ½ leading to 9/8. Since everything in the integral IS the volume of the object, and since the object is symmetrical across the y-axis, if you cut the shape in half the volume of the whole shape is actually 2 * everything in the integral with the new bounds, therefore the fraction in front is 9/4 * 2 = 18/4 = 9/2.)

Question 3

I’m SLOWLY getting close to finishing this unit, Applications of Integral (1,120/1,900 M.P.), and am hoping to make some decent progress through it this coming week. The next 3 sections all have to do with calculating volume using a disc method (whatever that is 🤔) so I’m hoping that method isn’t too difficult to understand, especially since I now have a better grasp on the volume equation and how it relates to integrals, in general. Between the 3 sections, there are 8 videos and 3 exercises. I would be PUMPED if I could get through all 3 sections this coming week and it would make me feel a lot better for making such little progress over the past two weeks. As I said, I think I’m in the clear at work and should have a lot more time on my hands so, even though it’s been a LONG time since I managed to get through 3 sections in one week, I think I might be able to do it! 💪🏼😤