1 Week 13–14

Since these homeworks would better be explained in terms of interactive figures, I’ve put them in a web format which you can use directly from browser.

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1.1 HW 10.7 - SCalcCC5 6.2.014.

Consider the solid obtained by rotating the region bounded by the given curves about the line \(y = 3\). \[y=\underbrace{3x}_{\substack{f_1(x)\\\uparrow\\\text{outer}}}\quad y=\underbrace{3\sqrt{x}}_{\substack{f_2(x)\\\uparrow\\\text{inner}}}\] Find the volume V of this solid.

1.1.1 Solution

Recall that the area of the annulus at \(x\) (where \(0\leq x\leq1\)) is \[A(x)=\pi\big[f_1(x)-y_0\big]^2-\pi\big[f_2(x)-y_0\big]^2\] and that the volume is \[V=\int_a^bA(x)\,\mathrm{d}x,\] which in our case are

\[ \begin{aligned} A(x)&=\pi(3 x-3)^2-\pi(3 \sqrt{x}-3)^2 \\ & =9 \pi\left[\left(1-2 x+x^2\right)-(1-2 \sqrt{x}+x)\right] \\ & =9 \pi\left(-3 x+x^2+2 \sqrt{x}\right)\text{; and}\\\\ V&=\int_0^{1} A(x)\,\mathrm{d}x\class{alref}{\cssId{alref1}{}}\\ &=9 \pi \int_0^1\left(-3 x+x^2+2 \sqrt{x}\right)\,\mathrm{d}x\\ & =9 \pi\left[-\frac{3}{2} x^2+\frac{1}{3} x^3+\frac{4}{3} x^{3 / 2}\right]_0^1\\ &=9 \pi\left(-\frac{3}{2}+\frac{5}{3}\right)=\frac{3}{2} \pi. \end{aligned} \]

WebAssign’s solution has a typo on this line.

1.1.2 Sketch

Sketch the region, the solid, and a typical disk or washer. (Do this on paper. Your instructor may ask you to turn in this work.)

Mathematicians usually call them annuli (sing. annulus), not washers.

Plot of \(y=3x\) and \(y=3\sqrt{x}\) with area between curves from \(x=0\) to \(x=1\)

1.2 HW 10.10 - SCalcCC5 6.3.025.

Consider the following. \[ x=4 \sqrt{\sin (y)}, \quad 0 \leq y \leq \pi, \quad x=0;\quad \text { about } y=4 \]

1.2.1 (a)

Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve about the specified axis.

1.2.1.1 Solution

\[ \begin{aligned} \class{alref}{\cssId{alref2}{\phantom{.}}} A(y)&=\underbrace{2\pi(4-y)}_{\text{circumference}} \cdot \underbrace{4\sqrt{\sin (y)}}_{\text{height}},\,\text{and thus}\\ V&=\int_0^{\pi} 2 \pi(4-y) 4 \sqrt{\sin (y)}\,\mathrm{d}y.\\\\ \end{aligned} \]

\(\leftarrow\) area of the red cylinder at y

When to use cylindrical shells or annuli?
- the line around which the region is rotated is parallel to the axis of the independent variable \(\rightarrow\) annuli (Section 1.1);
- the line is orthogonal to the axis \(\rightarrow\) cylinders (Section 1.2).

Plot of \(x=4\sqrt{\sin y}\) and \(y=4\) with area between curves from \(y=0\) to \(y=\pi\)

1.2.2 (b)

Use technology to evaluate the integral. (Round your answer to four decimal places.)

1.2.2.1 Solution

For the numerical integration, we can use scipy:

The anti-derivative is non-elementary. If you’re really curious, you can start from here; you may notice that it will be much more complicated than the \(x^n\tan x\) some of you asked a while ago.

import numpy
import scipy
round(scipy.integrate.quad(lambda y:
  2 * numpy.pi * (4 - y) *
    4 * numpy.sqrt(numpy.sin(y)),
  0, numpy.pi)[0],
4)

146.299

or whatever “technology” you’d prefer.

1.3 Recall - Arc Length

Beware of conditions for the existence of inverse.
The condition for this is that the parameterization \[r(t)=\Big(x(t),y(t)\Big)\] has to be injective and continuously differentiable.
In case you want to know, the norm \(\left\|{r}^{\prime}(t)\right\|\) here is given by a metric tensor, which, if you intend to study e.g. the theory of relativity, will be encountered again.

\[ \begin{array}{rcl} f(x(t)) & = & g(y(t))\\ \leadsto\quad x(t)=f^{-1}(t) &\text{and}& y(t)=g^{-1}(t)\\ \end{array} \] \[ L=\int_a^b\Big\|{r}^{\prime}(t)\Big\|\,\mathrm{d}t= \int_a^b \sqrt{\left(x^{\prime}(t)\right)^2+\left(y^{\prime}(t)\right)^2}\,\mathrm{d}t \]

1.4 HW11.6 - SCalcCC5 6.4.017.MI.

Find the exact length of the curve. \[y^2=16(x+1)^3, 0 \leq x \leq 3, y>0\]

1.4.1 Solution

\[\begin{aligned} y>0\implies\quad& y(t)=\sqrt{t}\\ 0\leq x\implies\quad&x(t)=\sqrt[3]{\frac{1}{16}t}-1\\ \,0 \leq x \leq 3\implies\quad&\text{with bounds from }t=16\text{ to }t=16\cdot4^3\\\\ \implies\quad&\left\{\begin{aligned} y'(t)&=\frac{t^{-½}}{2}\\x'(t)&=\frac{2^{-⅓}}{6}t^{-⅔}\\ \end{aligned}\right.\\\\ \implies\quad& \left\{\begin{aligned} \text{length}&=\int_{16}^{16\cdot4^3}\sqrt{\left(\frac{t^{-½}}{2}\right)^2+\left(\frac{2^{-⅓}}{6}t^{-⅔}\right)^2}\,\mathrm{d}t\\ &=\cdots=\frac{-37 \sqrt{37} + 145 \sqrt{145}}{54}. \end{aligned}\right. \end{aligned}\]

1.5 HW11.7 - SCalcCC5 6.4.021.

Find the exact length of the curve. \[ \begin{aligned} x &= 9 + 12t^2\\ y &= 6 + 8t^3\\ 0 &\leq t \leq 5 \end{aligned} \]

Again, \[ L=\int_a^b \sqrt{\left(x^{\prime}(t)\right)^2+\left(y^{\prime}(t)\right)^2}\,\mathrm{d}t; \] but sometimes, when the curve is given in the form of \(y=f(x)\), i.e. \(y(t)=t\), one may reduce the above to \[ L=\int_a^b \sqrt{1+\left(y'(x)\right)^2}\,\mathrm{d}x. \]

1.6 Miscellanies

Optionally, these were the excursions we briefly chatted about before the Thanksgiving…

On rectification, measuring, pathologies; probably no time to code something original, but wikipedia:

1.6.1 Schwarz’s Lantern

recall the discussion we had about Riemann sums; can you try to play with the above and see if you can make the same observation as the pioneers of measure theory did a century ago?
https://www.wikiwand.com/en/Lebesgue_measure
https://www.wikiwand.com/en/Minkowski_content
Hilbert’s curve above
Banach-Tarski
locales, cardinals
fractals, dimensions, GMT