3 Week 16
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3.1 Riemann Sum and Arc Length of Polar Curves [In Progress]
Last time I wasn’t able to answer this so let’s try again this time.
\[ \begin{aligned} L&=\int_a^b \sqrt{r^2+\left(\frac{d r}{d \theta}\right)^2}\,\mathrm{d}\theta\\ &=\lim_{n\to\infty}\sum_{i=1}^n\left(\sqrt{r(\theta_i)^2+r'(\theta_i)^2}\cdot\Delta\theta\right) \end{aligned} \]
TODO
3.2 HW14.3 - WACalcTutBankLT1 7.5.003c.Tut.
Compute the surface area obtained by rotating the curve \(y = \sqrt[3]{x}\), for \(x\in [1, 125]\) about the y-axis.
3.2.1 Solution
Since the function is bijective within the domain, let us consider again \[\left\{\begin{aligned} y &= \sqrt[3]{x}\\ x &\in [1, 125] \end{aligned}\right.\;\implies\;\left\{\begin{aligned} x&=y^3\\ y&\in [1,5] \end{aligned}\right.\] by taking the inverse (i.e. swapping the variable \(x\) and \(y\) as before).
Now one may use the integral
\[ \begin{aligned} A&=\int_1^5 \underbrace{2\pi x(y)}_{\text{circumference}} \cdot \underbrace{\sqrt{1+x'(y)^2}}_{\text{\color{orange}height}\class{alref}{\cssId{alref1}{}}}\,\mathrm{d}y\\ &=\cdots=\frac{(-10 \sqrt{10} + 5626 \sqrt{5626}) \cdot \pi}{27}, \end{aligned} \]
- If you’re curious about the height part being an arc length, or that the band in the widget is a cone section instead of a cylinder, recall the discussion we had about Schwarz’s Lantern in Section 1.6.1.
as demonstrated by the plot and widget below.
3.3 Bijective Functions, Inverses, and Change of Coordinate System
This was asked multiple times in my MLC, so I’d also like to discuss a bit about it: Section 4.3.