I had one of the most productive weeks I’ve had working through KA in a long time! I made it through 3 sections in the unit Applications of Integrals which, in total, included 8 videos and 3 exercises. What I worked through this week was very similar to what I learned in Week 141 and things started to click for me in terms of using calc to determine the volume of objects and concepts started falling into place for me. Overall, what I learned this week was really helpful in understanding how calculus is used with volume, but I still need a lot of practice to get an intuitive understanding for how it all works. I’m definitely making progress though!  

The first section I worked on this week was titled Volume: Disc Method (Revolving Around X- and Y-axes). The general idea of what was taught in this section, which had similarities to what I learned last week, is that you can use the formula for a circle, πr², to find the volume of certain objects. Here are two pages from my notes that go over what was taught:

The key concept here is that you can think of a function being rotated 360° around either the x- or y-axis as the creation of a 3D object which can be calculated in part using the formula πr². When working through this section, there were two things that came to me that made it easier for me to understand these types of questions.

The first was that I began to think of the base of the object as the interval along whichever axis the object was being measured. In the example in my notes, since the trumpet-like object was being measured along the x-axis, I consider the base of the object, from the formula volume = base * width * height, to be the interval across the x-axis. When writing the integral, that means that the integral sign IS the base, i.e. base = ^­­₀∫². Using the disc method, the width * height part of the equation IS πr². 

The second concept that became more clear to me when working through this section is that, if the object is being measured along the x-axis, the equation inside the integrand HAS to be in terms of x and vice versa if the object is being measured along the y-axis. In the example from my notes, the radius of the trumpet is y. Since the object is being measured across the x-axis however, the integrand needs to be in terms of x. Since the equation for a circle is πr² and r = y, the equation could be thought of as π(y)², BUT since y = x² and because you NEED to have the integrand be in terms of x, you can rewrite the equation at π(x²)² which equals πx⁴ and it is literally the exact same thing as πr².

Here are two example questions from this section I worked through in my notes that utilize the disc-method:

Question 1

• Step 1 – Determine formula for the area of a ‘slice’ of the object:
  • (Note: Since the object is being rotated around the y-axis, the radius of the object needs to be calculated using y.)
  • Area_(circle) = πr²
    • (…)
      • r = (x)^1/2 = y
        • ((x)^1/2)² = y²
      • r = y²
    • = π(y²)²
  • Area_(circle) = πy⁴
• Step 2 – Determine the volume of the object using the area of each ‘slice’ and the interval across the y-axis:
  • Vol. = B * W * H = y * x * z
    • = ₂∫³ πy⁴ dy
    • = π ₂∫³ y⁴dy
    • = π (y^{4 + 1}/4 + 1)₂|³
    • = π (y⁵/5)₂|³
    • = π ((3)⁵/5 – (2)⁵/5)
    • = π (243/5 – 32/5)
  • Vol. = 211π/5

Question 2

• Area_(circle) = πr²
  • (…)
    • r = 1/x² = y
      • x² = 1/y
      • (x²)^1/2 = (1/y)^1/2
    • r = (1/y)^1/2
      • = y^-1/2
  • = π(y^-1/2)²
  • = πy^-2/2
  • = πy^-1
• Area_(circle) = πy^-1
• Vol. = B * W * H = y * x * z
  • = ₁∫⁶ πy^-1 dy

The next section I worked on was titled Volume: Disc Method (Revolving Around Other Axes). (Also, as a random side note, it seems weird to me that the plural for “axis” is “axes”. Makes me think of 🪓’s.) The concept taught in this section is that you can use the disc method even when rotating an object around a x- or y-coordinate other than x = 0 or  y = 0, you just need to incorporate the x- or y-coordinate the object is being rotated around into the radius. Here’s an example question from my notes:

Question 3

One thing that my notes don’t make clear is that, on the first page, r = +2 + y² – 2 but really the equation should read as r = |–2| + y² – 2 since the radius BEGINS where x = –2 and so the radius stemming from x = –2 will be positive.

The last section I worked through this week was titled Volume: Washer Method (Revolving Around x- and y-axes). The general idea here is that if you have two functions and rotate both of them around either the x- or y-axis, you can find the volume of the object if the object was hollowed out between the inside function and the x- or y-axis and the space between the two functions was filled in (i.e. the object would look sort of like an elongated washer). You do this using the formula:

• Area_(washer) = Area_{(outside circle)} – Area_{(inside circle)} = π(r₂)² – π(r₁)² = π((r₂)² – (r₁)²)

 Using the formula π((r₂)² – (r₁)²), you then use the disc method to calculate the radius of each respective circle and, depending on whether the object is being measured across the x- or y-axis, determine r₂ and r₁ in terms of either x or y. Here are two example questions from my notes:

Question 4 

Question 5

I know last week I said I would type out the math for each question, but I’ve been writing this post for more than an hour and, since my notes of the final 3 questions are relatively neat and easy to follow, I’m once again not going to bother typing out the math. Plus, I usually try to write ~1,000 words per post and this post is closer to 1,350 words so, ya, I’m not going to bother.

I now have two sections left to get through in Applications of Integrals (1,360/1,900 M.P.). The first section, Volume: Washer Method (Revolving Around Other Axes) has 4 videos and 1 exercise in it and the final section, Calculator Active Practice only has only one exercise. I’m not sure if I’ll be able to make it through all 4 videos, both exercises AND the unit test this week but I think I have a small, outside chance depending on how difficult the last two exercises are. I also think it’s highly unlikely that I’ll get through everything I just mentioned and the Calculus 1 Course Challenge before the end of May (i.e. 9 days from now) but, again, I’m going to try my best to make it happen. To be honest, considering I’ve been working on Calculus 1 since Week 82 (holy shit… I just looked that up and it blew my mind. I’ve been working on this stupid subject FOR 60 WEEKS….. 😳) I think there will be plenty of things I will have forgotten that I’ll need to go back and review before I can pass the Course Challenge with a decent score. I guess in the big picture, what’s another few more weeks on top of 60. 🤷🏻‍♂️