W06 Regular

Due date: Sunday 2/22, 11:59pm

Moments and CoM

01

03

FlatCoMMan

Find the center of mass of FlatCoMMan. Assume a constant mass density $\rho$. Use additivity of moments.

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Solution

Solutions - 0110-03

Assume $\rho =1$ for all of these.

The value of $\rho$ does not affect the CoM point if $\rho$ is a constant.

Region 1:

$$\begin{array}{c}{m}_1=\frac{1}{2}(2)(1)\gg \gg 1\\ \\ M_{1y}=0\\ \\ M_{1x}={\int }_8^{9}y(2+\frac{0-2}{1}y)dy\gg \gg \frac{25}{3}\end{array}$$

Region 2:

$$\begin{array}{c}{m}_2=\pi {(1)}^2\gg \gg \pi \\ \\ M_{2y}=0\\ \\ M_{2x}=m_2{\bar{y}}_2\gg \gg \pi (7)\gg \gg 7\pi \end{array}$$

Region 3:

$$\begin{array}{c}{m}_3={\int }_{-2}^2(6-(2+x^2))dx\gg \gg \frac{32}{3}\\ \\ M_{3y}=0\\ \\ M_{3x}={\int }_{-2}^2\frac{1}{2}(6^2-{(2+x^2)}^2)dx\gg \gg \frac{704}{15}\end{array}$$

Region 4:

$$\begin{array}{c}{m}_4=2\left(\frac{1}{2}\right)\gg \gg 1\\ \\ M_{4y}=m_4{\bar{x}}_4\gg \gg -1\left(\frac{9}{4}\right)\gg \gg -\frac{9}{4}\\ \\ M_{4x}=m_4{\bar{y}}_4\gg \gg 1(5)\gg \gg 5\end{array}$$

Region 5:

$$\begin{array}{c}{m}_5=2\left(\frac{1}{2}\right)\gg \gg 1\\ \\ M_{5y}=m_5{\bar{x}}_5\gg \gg 1\left(\frac{9}{4}\right)\gg \gg \frac{9}{4}\\ \\ M_{5x}=m_5{\bar{y}}_5\gg \gg 1(5)\gg \gg 5\end{array}$$

Region 6:

$$\begin{array}{c}{m}_6=3\left(\frac{1}{2}\right)\gg \gg \frac{3}{2}\\ \\ M_{6y}=m_6{\bar{x}}_6\gg \gg -\frac{3}{2}\left(\frac{5}{4}\right)\gg \gg -\frac{15}{8}\\ \\ M_{6x}=m_6{\bar{y}}_6\gg \gg \frac{3}{2}\left(\frac{3}{2}\right)\gg \gg \frac{9}{4}\end{array}$$

Region 7:

$$\begin{array}{c}{m}_7=3\left(\frac{1}{2}\right)\gg \gg \frac{3}{2}\\ \\ M_{7y}=m_7{\bar{x}}_7\gg \gg +\frac{3}{2}\left(\frac{5}{4}\right)\gg \gg \frac{15}{8}\\ \\ M_{7x}=m_7{\bar{y}}_7\gg \gg \frac{3}{2}\left(\frac{3}{2}\right)\gg \gg \frac{9}{4}\end{array}$$

FlatCoMMan:

$$\begin{array}{c}m=\frac{50}{3}+\pi ,M_x=\frac{2093+210\pi }{30},M_y=0\\ \\ \gg \gg (\bar{x},\bar{y})=(0,\frac{2093+210\pi }{500+30\pi })\approx 4.63\end{array}$$Link to original

02 $★$

04

CoM from Simpson’s

Use Simpson’s rule (with 6 subintervals) to estimate the CoM of this region:

You will need to estimate $M_x$ and $M_y$ and $m$ with three separate integrals. You can use a calculator for your arithmetic.

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Solution

Solutions - 0110-04

(1) Simpson’s Rule formula:

$${\int }_a^{b}f(x)dx\approx S_n=\frac{\mathrm{\Delta }x}{3}(f(x_0)+4f(x_1)+2f(x_2)+\cdots +f(x_n))$$

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(2) Simpson’s for total mass $m$:

$$\mathrm{\Delta }x=\frac{b-a}{n}\gg \gg \frac{8-2}{6}\gg \gg 1$$

Therefore:

$$\begin{array}{c}{S}_6\gg \gg \frac{1}{3}(0+4(1)+2(3)+4(3)+2(2)+4(3)+0)\\ \\ \gg \gg \frac{38}{3}\end{array}$$

So:

$$m\gg \gg \rho {\int }_2^{8}f(x)dx\gg \gg \approx \frac{38}{3}\rho$$

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(3) Simpson’s for moment to $x$-axis:

Integral formula:

$$M_x={\int }_2^8\frac{\rho }{2}f{(x)}^{2}dx$$

Approximate with $S_6$:

$$\begin{array}{c}{M}_x\approx S_6\\ \\ \gg \gg \frac{1}{3}\rho (\frac{1}{2}{0}^2+4\frac{1}{2}\left(1^2\right)+2\frac{1}{2}\left(3^2\right)+4\frac{1}{2}\left(3^2\right)+2\frac{1}{2}\left(2^2\right)+4\frac{1}{2}\left(3^2\right)+\frac{1}{2}{0}^2)\\ \\ \gg \gg \frac{102\rho }{6}\end{array}$$

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(4) Simpson’s for moment to $y$-axis:

Integral formula:

$$M_y={\int }_2^8\rho xf(x)dx$$

Approximate with $S_6$:

$$\begin{array}{c}{M}_y\approx S_6\\ \\ \gg \gg \frac{1}{3}\rho (1(2)(0)+4(3)(1)+2(4)(3)+4(5)(3)+2(6)(2)+4(7)(3)+1(8)(0))\\ \\ \gg \gg 68\rho \end{array}$$

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(5) Compute CoM:

$$\begin{array}{cc}\bar{x}& =\frac{M_y}{m}\gg \gg \frac{68}{38/3}\gg \gg \frac{102}{19}\approx 5.37\\ & \\ \bar{y}& =\frac{M_x}{m}\gg \gg \frac{17}{38/3}\gg \gg \frac{51}{38}\approx 1.34\end{array}$$Link to original

Improper integrals

03 $★$

02

Proper vs. improper

For each integral below, determine whether it is proper or improper, and if improper, explain why.

(a) ${\displaystyle {\int }_0^2\frac{dx}{x^{1/3}}}$ $$ (b) ${\displaystyle {\int }_0^1{e}^{-x}dx}$ $$ (c) ${\displaystyle {\int }_0^{\pi /2}\sec xdx}$

(d) ${\displaystyle {\int }_0^{\infty }\sin xdx}$ $$ (e) ${\displaystyle {\int }_1^{\infty }\ln xdx}$ $$ (f) ${\displaystyle {\int }_0^3\ln xdx}$

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Solution

Solutions - 0130-02

(a) Improper: integrand $\to \infty$ as $x\to 0^{+}$.

Note: this converges too, since it’s a $p$-integral to zero with $p<1$.

(b) Proper: no source of infinity.

Note: automatically converges.

(c) Improper: $\sec x\to \infty$ as $x\to \pi /2$.

Note: this diverges. Antiderivative is $\ln |\tan x+\sec x|\to \infty$ as $x\to \pi /2$.

(d) Improper: infinite upper bound.

Note: this diverges. Antiderivative is $-\cos x$ which has no limit as $x\to \infty$.

(e) Improper: infinite upper bound.

Note: this diverges. Antiderivative is $x\ln x-x\to -\infty$ as $x\to \infty$. (L’Hopital’s rule, indeterminate form.)

$$x\ln x-x=\frac{\ln x-1}{1/x}\stackrel{x\to \infty }{⟶}\frac{1/x}{-1/x^2}=-x\to -\infty$$

(f) Improper: infinite integrand at $x=0$.

Note: this converges. Antiderivative is $x\ln x-x\to 0$ as $x\to 0^{+}$. (Same indeterminate form as (e).)

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04

03

Gabriel’s Horn - Volume and surface of revolution

The curve $y=\frac{1}{x}$ for $x>1$ is rotated about the $x$-axis. The resulting shape is Gabriel’s Horn.

(a) Find the volume enclosed by the horn by evaluating a convergent improper integral.

(b) Show that the surface area of the horn is infinite by applying comparison to a $p$-integral which is divergent.

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Solution

Solutions - 0130-03

Volume:

$$\begin{array}{c}{V}_G={\int }_1^{\infty }\pi {\left(\frac{1}{x}\right)}^{2}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }{\int }_1^R\pi {\left(\frac{1}{x}\right)}^{2}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }\pi (-\frac{1}{R}+1)\gg \gg \pi \end{array}$$

Surface area:

$$S_G={\int }_1^{\infty }2\pi \left(\frac{1}{x}\right)\sqrt{1+{(-1/x^2)}^2}dx$$

But notice this:

$$\left(\frac{1}{x}\right)\sqrt{1+{(-1/x^2)}^2}>\frac{1}{x}$$

But ${\displaystyle {\int }_1^{\infty }2\pi \frac{1}{x}dx}$ diverges!

So by the comparison test, $S_G$ diverges as well.

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05

01

Comparison test

Use the comparison test to determine whether the integral converges:

$${\int }_1^{\infty }\frac{dx}{x^4+\sqrt{x}}$$

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Solution

Solutions - 0130-01

(1) Find comparable integrand:

Higher power dominates for large $x$:

$$\frac{1}{x^4+\sqrt{x}}⟶\frac{1}{x^4}\text{as }x\to \infty$$

Therefore, compare to $1/x^4$.

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(2) Make comparison:

$$\frac{1}{x^4+\sqrt{x}}<\frac{1}{x^4}\text{for all }x\ge 1$$

And:

$${\int }_1^{\infty }\frac{1}{x^4}dx\text{converges}$$

because it is a $p$-integral with $p=4>1$.

By the Comparison Test, we conclude that:

$${\int }_1^{\infty }\frac{1}{x^4+\sqrt{x}}dx\text{converges}$$Link to original

06

05

Computing improper integrals

For each integral below, give the limit interpretation of improper integral and then compute the limit. Based on that result, state whether the integral converges. If it converges, what is its value?

(a) ${\displaystyle {\int }_0^1\ln xdx}$ $$ (b) ${\displaystyle {\int }_{-\infty }^{+\infty }x{e}^{-x^2}dx}$ $$ (c) ${\displaystyle {\int }_1^2\frac{1}{{(x-1)}^2}dx}$

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Solution

Solutions - 0130-05

(a)

(1) Definition of improper integral:

$${\int }_0^1\ln xdx=\underset{R\to 0^{+}}{\lim }{\int }_R^1\ln xdx$$

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(2) Antiderivative and limit:

$$\begin{array}{ccc}& \gg \gg \underset{R\to 0^{+}}{\lim }x\ln x-x{|}_R^1& \\ & & \\ & \gg \gg \underset{R\to 0^{+}}{\lim }0-1-(R\ln R-R)\gg \gg -1\text{(converges)}& \text{(Note A)}\end{array}$$

Note A: Use L’Hopital:

$$R\ln R=\frac{\ln R}{1/R}{\gg \gg }^{d/dR}\frac{1/R}{-1/R^2}=-R\to 0$$

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(b)

(1) Definition of improper integral:

$${\int }_{-\infty }^{+\infty }x{e}^{-x^2}dx=\underset{R\to +\infty }{\lim }{\int }_0^{R}x{e}^{-x^2}dx+\underset{R\to -\infty }{\lim }{\int }_R^{0}x{e}^{-x^2}dx$$

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(2) Antiderivative and limit:

$$\begin{array}{c}\gg \gg \underset{R\to +\infty }{\lim }{-\frac{1}{2}{e}^{-x^2}|}_0^R+\underset{R\to -\infty }{\lim }{-\frac{1}{2}{e}^{-x^2}|}_R^0\\ \\ \gg \gg -\frac{1}{2}+-\frac{1}{2}\gg \gg 0\text{(converges)}\end{array}$$

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(c)

(1) Definition of improper integral:

$${\int }_1^2\frac{1}{{(x-1)}^2}dx=\underset{R\to 1^{+}}{\lim }{\int }_R^2\frac{1}{{(x-1)}^2}dx$$

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(2) Antiderivative and limit:

$$\begin{array}{c}\gg \gg \underset{R\to 1^{+}}{\lim }{-\frac{1}{x-1}|}_R^2\\ \\ \gg \gg \underset{R\to 1^{+}}{\lim }-1--\frac{1}{R-1}\gg \gg +\infty \text{(diverges)}\end{array}$$Link to original