W06 - Examples

CoM of a parabolic plate

CoM of a parabolic plate

Find the CoM of the region depicted:

Solution

(1) Compute the total mass:

Area under the curve with density factor $\rho$:

$$m={\int }_0^2\rho x^{2}dx\gg \gg \rho {\frac{x^3}{3}|}_0^2\gg \gg \frac{8\rho }{3}$$

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(2) Compute $M_y$:

Formula:

$$M_y={\int }_a^b\rho xdA$$

Interpret and calculate:

$$\begin{array}{c}{M}_y={\int }_0^2\rho xf(x)dx\gg \gg \rho {\int }_0^2{x}^{3}dx\\ \\ \gg \gg 4\rho =M_y\end{array}$$

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(3) Compute $M_x$:

Formula:

$$M_x={\int }_c^d\rho ydA$$

Width of horizontal strips between the curves:

$$w(y)=2-\sqrt{y}$$

Interpret $dA$:

$$dA=(2-\sqrt{y})dy$$

Calculate integral:

$$\begin{array}{c}{M}_x={\int }_c^d\rho ydA\gg \gg {\int }_0^4\rho y(2-\sqrt{y})dy\\ \\ \gg \gg {\int }_0^4\rho y(2-\sqrt{y})dy\gg \gg {\int }_0^4\rho 2ydy-{\int }_0^4\rho y^{3/2}dy\\ \\ \gg \gg \frac{16\rho }{5}=M_x\end{array}$$

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(4) Compute CoM coordinates from moments:

CoM formulas:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Insert data:

$$\begin{array}{c}\bar{x}=\frac{4\rho }{8\rho /3}\gg \gg \frac{3}{2},\bar{y}=\frac{16\rho /5}{8\rho /3}\gg \gg \frac{6}{5}\\ \\ \text{CoM}=(\bar{x},\bar{y})=\left(\frac{3}{2},\frac{6}{5}\right)\end{array}$$ Link to original

Computing CoM using only vertical strips

Computing CoM using only vertical strips

Find the CoM of the region:

Solution

(1) Compute the total mass $m$:

Area under the curve times density $\rho$:

$${\int }_0^{\pi /2}\rho \cos xdx=\rho \sin x{|}_0^{\pi /2}=\rho$$

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(2) Compute $M_x$, using vertical strips and the ‘midpoints’ method:

Plug $f_2(x)=f(x)=\cos x$ and $f_1(x)=0$ into formula:

$$M_x={\int }_0^{\pi /2}\rho \frac{1}{2}{f}_2^{2}dx\gg \gg {\int }_0^{\pi /2}\rho \frac{1}{2}{\cos }^{2}xdx$$

Integration by ‘power to frequency conversion’:

$$\begin{array}{c}{\int }_0^{\pi /2}\rho \frac{1}{2}{\cos }^{2}xdx\gg \gg \frac{\rho }{4}{\int }_0^{\pi /2}(1+\cos 2x)dx\\ \\ \gg \gg \frac{\rho }{4}x{|}_0^{\pi /2}+{\frac{\rho \sin 2x}{8}|}_0^{\pi /2}=\frac{\pi \rho }{8}\end{array}$$

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(3) Compute $M_y$ using vertical strips and the regular method:

Plug $f(x)=\cos x$ into formula:

$$M_y={\int }_a^b\rho xf(x)dx\gg \gg {\int }_0^{\pi /2}\rho x\cos xdx$$

Integration by parts. Set $u=x$, $v^{\prime }=\cos x$ and so $u^{\prime }=1$, $v=\sin x$:

$$\begin{array}{c}{\int }_0^{\pi /2}\rho x\cos xdx\gg \gg \rho x\sin x{|}_0^{\pi /2}-\rho {\int }_0^{\pi /2}\sin xdx\\ \\ \gg \gg \frac{\pi \rho }{2}\cdot 1-\rho (-\cos \frac{\pi }{2}--\cos 0)=\rho \left(\frac{\pi }{2}-1\right)\end{array}$$

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(4) Compute CoM:

CoM formulas:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Plug in data:

$$\begin{array}{c}\bar{x}=\frac{\rho (\pi /2-1)}{\rho }\gg \gg \frac{\pi }{2}-1\\ \\ \bar{y}=\frac{\pi \rho /8}{\rho }\gg \gg \frac{\pi }{8}\\ \\ \text{CoM}=(\bar{x},\bar{y})=\left(\frac{\pi }{2}-1,\frac{\pi }{8}\right)\end{array}$$ Link to original

Center of mass using moments and symmetry

Center of mass using moments and symmetry

Find the center of mass of the two-part region:

Solution

(1) Symmetry: CoM on $y$-axis

Because the region is symmetric in the $y$-axis, the CoM must lie on that axis. Therefore $\bar{x}=0$.

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(2) Additivity of moments:

Write $M_x$ for the total $x$-moment (distance measured to the $x$-axis from above).

Write $M_x^{\mathrm{tri}}$ and $M_x^{\mathrm{circ}}$ for the $x$-moments of the triangle and circle.

Additivity of moments equation:

$$M_x=M_x^{\mathrm{tri}}+M_x^{\mathrm{circ}}$$

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(3) Find moment of the circle $M_x^{\mathrm{circ}}$ using Symmetry:

By symmetry we know ${\bar{x}}^{\mathrm{circ}}=0$.

By symmetry we know ${\bar{y}}^{\mathrm{circ}}=5$.

Area of circle with $r=2$ is $A=4\pi$, so total mass is $M=4\pi \rho$.

Centroid-from-moments equation:

$${\bar{y}}^{\mathrm{circ}}=\frac{M_x^{\mathrm{circ}}}{m}$$

Solve the equation for $M_x^{\mathrm{circ}}$:

$$\begin{array}{c}{\bar{y}}^{\mathrm{circ}}=\frac{M_x^{\mathrm{circ}}}{m}\gg \gg 5=\frac{M_x^{\mathrm{circ}}}{4\pi \rho }\\ \\ \gg \gg M_x^{\mathrm{circ}}=20\pi \rho \end{array}$$

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(4) Find moment of the triangle $M_x^{\mathrm{tri}}$ using Symmetry and the “$h/3$ trick”:

By Symmetry, the CoM of the triangle must lie on the vertical line $x=0$.

By the “$h/3$ trick,” the CoM of the triangle must lie on the horizontal line at $y=h/3$. In this case $h=3$, so it lies on the line $y=1$.

Therefore the triangle CoM is:

$$\left({\bar{x}}^{\text{tri}},{\bar{y}}^{\text{tri}}\right)=(0,1)$$

To get moments, multiply by mass $m=\rho \cdot \frac{1}{2}(4)(3)=6\rho$:

$$M_y^{\text{tri}}=0,M_x^{\text{tri}}=(6\rho )(1)=6\rho$$

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(5) Apply additivity:

$$M_x=M_x^{\mathrm{tri}}+M_x^{\mathrm{circ}}\gg \gg \rho (20\pi +6)$$

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(6) Total mass of region:

Area of circle is $4\pi$. Area of triangle is $\frac{1}{2}\cdot 4\cdot 3=6$. Thus $m=\rho A=\rho (4\pi +6)$.

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(7) Compute center of mass $\bar{y}$ from total $M_x$ and total $M$:

We have $M_x=\rho (20\pi +6)$ and $M=\rho (4\pi +6)$. Plug into formula:

$$\begin{array}{c}\bar{y}=\frac{M_x}{m}\gg \gg \frac{\rho (20\pi +6)}{\rho (4\pi +6)}\approx 3.71\\ \\ \text{CoM}=(\bar{x},\bar{y})=(0,3.71)\end{array}$$ Link to original

Improper integral - infinite bound

Improper integral - infinite bound

Show that the improper integral ${\int }_2^{\infty }\frac{dx}{x^3}$ converges. What is its value?

Solution

(1) Improper integral definition:

$${\int }_2^{\infty }\frac{dx}{x^3}\gg \gg \underset{R\to \infty }{\lim }{\int }_2^R\frac{dx}{x^3}$$

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(2) Replace infinity with a new symbol $R$:

$${\int }_2^R\frac{dx}{x^3}\gg \gg -\frac{1}{2}{x}^{-2}{|}_2^R\gg \gg \frac{1}{8}-\frac{1}{2R^2}$$

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(3) Take limit as $R\to \infty$:

$$\begin{array}{c}\underset{R\to \infty }{\lim }{\int }_2^R\frac{dx}{x^3}\gg \gg \underset{R\to \infty }{\lim }\frac{1}{8}-\frac{1}{2R^2}\gg \gg \frac{1}{8}\\ \\ \gg \gg \text{converges}\end{array}$$ Link to original

Improper integral - infinite integrand

Improper integral - infinite integrand

Show that the improper integral ${\int }_0^9\frac{dx}{\sqrt{x}}$ converges. What is its value?

Solution

(1) Improper integral definition:

$${\int }_0^9\frac{dx}{\sqrt{x}}\gg \gg \underset{R\to 0^{+}}{\lim }{\int }_R^9\frac{dx}{\sqrt{x}}$$

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(2) Switch $0$ to $R$ and integrate:

$$\begin{array}{c}{\int }_R^9\frac{dx}{\sqrt{x}}\gg \gg {\int }_R^9{x}^{-1/2}dx\\ \\ \gg \gg 2x^{+1/2}{|}_R^9\gg \gg 6-2\sqrt{R}\end{array}$$

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(3) Take limit as $R\to 0^{+}$:

$$\begin{array}{c}\underset{R\to 0^{+}}{\lim }{\int }_R^9\frac{dx}{\sqrt{x}}\gg \gg \underset{R\to 0^{+}}{\lim }6-2\sqrt{R}\gg \gg 6\\ \\ \gg \gg \text{converges}\end{array}$$ Link to original

Improper integral - infinity inside the interval

Example - Improper integral - infinity inside the interval

Does the integral ${\int }_{-1}^{+1}\frac{1}{x}dx$ converge or diverge?

Solution

WRONG APPROACH:

It is tempting to compute the integral incorrectly, like this:

$${\int }_{-1}^{+1}\frac{1}{x}dx\gg \gg \ln |x|{|}_{-1}^{+1}\gg \gg \ln |2|-\ln |-2|=0$$

But this is wrong! There is an infinite integrand at $x=0$. We must break it into parts!

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RIGHT APPROACH:

(1) Break apart at $x=0$ discontinuity:

$${\int }_{-1}^{+1}\frac{1}{x}dx\gg \gg {\int }_{-1}^0\frac{1}{x}dx+{\int }_0^{+1}\frac{1}{x}dx$$

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(2) Improper integral definition, both terms separately:

$$\gg \gg \underset{R\to 0^{-}}{\lim }{\int }_{-1}^R\frac{1}{x}dx+\underset{R\to 0^{+}}{\lim }{\int }_R^{+1}\frac{1}{x}dx$$

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(3) Integrate:

$$\begin{array}{c}{\int }_{-1}^R\frac{1}{x}dx\gg \gg \ln |R|-\ln |-1|\gg \gg \ln |R|\\ \\ {\int }_R^{+1}\frac{1}{x}dx\gg \gg \ln |1|-\ln |R|\gg \gg -\ln R\end{array}$$

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(4) Limits:

$$\begin{array}{c}\underset{R\to 0^{-}}{\lim }\ln |R|=-\infty ,\underset{R\to 0^{+}}{\lim }-\ln R=+\infty \\ \\ \gg \gg \text{diverges}\end{array}$$

Neither limit is finite. For ${\int }_{-1}^{+1}\frac{1}{x}dx$ to exist we’d need both of these limits to be finite. So the original integral diverges.

Link to original