I had one of those weeks where I felt like in some ways I didn’t make much progress at all but, in other ways, I feel like I learned and worked through a ton of new things. I only made it through 3 videos and 3 exercises this week which feels a bit disappointing, but everything I went through was relatively new for me and a relatively difficult to wrap my head around. The final section I worked on, Volume: Squares and Rectangles Cross Sections, introduced how integrals can be used to find the volume of shapes. I found this section really hard to understand and have a tenuous understanding of what was covered, at best. The good news is that the material that was covered seems like it won’t be too difficult for me to figure out with a bit more practice. The remainder of this unit is made up of sections that all have to do with volume so I guess I’ll find out!

I began the week working on the only exercise from the section Area: Horizontal Area Between Curves. In this exercise, I was given questions where I was asked to find the area between a function and the y-axis or the area between two ‘vertical’ (so to speak) functions where, in both types of questions, the bounds were given as y-coordinates. It took me a while to figure out the process to solve these questions but, after a bit of practice, they were pretty easy in the end. Here are two example questions:

Question 1

(Note: On this question you simply integrate the function x = 3cos(y) + 12/π with respect to y and using the bounds 2π and π/2.)

• x = 3cos(y) + 12/π
  • _π/2∫ ^2π 3*cos(y) + (12/π)y⁰ dy = 3*sin(y) + 12y/π _π/2|^2π
    • = (3*sin(2π) + 12(2π)/π) – (3*sin(π/2) + 12(π/2)/π)
    • = (0 + 24) – (3 + 6)
    • = 24 – 9
    • = 15

Question 2

• Givens:
  • f(y) = x = 7y + 18
  • g(y) = x = 2y² + 19y +18
• Question: What is the area bounded by the two functions?
  • Step 1 – Find function for difference between functions:
    • Difference = f(y) – g(y)
      • = (7y + 18) – (2y² + 19y + 18)
      • = 7y + 18 – 2y² – 19y – 18
      • = –2y² – 12y
  • Step 2 – Find the points where the functions cross, i.e. the upper and lower bounds:
    • f(y) = g(y)
      • 7y + 18 = 2y² + 19y + 18
      • 0 = 2y² + 19y + 18 – 7y – 18
      • 0 = 2y² + 12y
      • 0 = 2y(y + 6)
        • Therefore, 0’s @ [y = 0] and [y = –6]
• Conclusion:
  • The bounded area between f(y) and g(y) can be calculated with the integral _–6∫⁰ (–2y² – 12y) dy.

I began the next section, Area: Curves That Intersect at More Than Two Points, on Thursday. There weren’t any videos in this section, just a single exercise which seemed a bit weird. I found the exercise hard to get through not because the questions themselves were difficult, but because I had to learn a bunch of new techniques on Desmos which took me forever to figure out. Every question in this section would show me two functions that crossed over each other and then ask me to find the sum of the bounded areas of the functions. I would graph the functions on Desmos to find the (x, y) coordinate where they crossed (which was easy) but then I had to figure out how to enter the integral of each section to find the area. This is what I got stumped on because I didn’t know the correct way to write the integral so that it would output the integral’s value. I spent about an hour writing out the integrals in different ways and looking up videos to figure it out how to write them so that they’d spit out the answer. I eventually realized that what I was doing wrong was writing the integral with square brackets around the two functions when I should have been using curved brackets. I was using square brackets because that’s how the KA answer showed it so, when I finally figure out what I was doing wrong, I was a bit annoyed at KA. Once I figure it out, it took me about 15 minutes to get through the exercise. Below is an example question with the Desmos graph I created at the bottom which shows all the functions I had to write:

Once I got through that exercise, I then started the section Volume: Squares and Rectangles Cross Sections. Right now, I don’t have a great understanding of what was covered in this section. Below are two example questions from the one exercise I made it through. As shown in the KA answer, I found that the key to understand these questions was to think of the volume of each shape as many thin, ‘2d’ slices of area added together. The volume of any object is base x height, a.k.a. length x width x height, so by visualizing it like this, I find it much easier to simply calculate the area of a ‘2d object’ and then find the sum of all the little thin, 2d slices by taking their integral. (I don’t think that made much sense but hopefully the example questions will make it more clear. Also, I know that it should be “clearer” instead of “more clear” but I hate how “more clear” sounds so I refuse to write that.)

Question 1

(Note: Since I don’t understand the math that well and since, as you can see from my notes, there’s not much math to go through in either of these questions as it is, I’m not going to bother typing out the solutions.)

Question 2

I’m now halfway through the unit Applications of Integrals (960/1,900 M.P.). There are 5 more sections left to get through all of which, as I mentioned at the beginning of this post, have to do with calculating volume. I’m hoping it becomes a bit easier to understand how to calculate the volume of objects using integrals and am pretty optimistic/confident that it will. In a nerdy type of way, I think it will be kind of cool to be able to use calculus to find the volume of objects. People are definitely going to think I’m pretty sick once I figure it out. 😎