Open Textbook Library – Physicspupil

June 8, 2022October 24, 2022

202206081501 Exercise 4.1 (Q15)

In an electric circuit with a supplied voltage (emf) $E$ , a resistor with resistance $r_0$ , and an inductor with reactance $x_0$ , suppose you want to add a second resistor. If $r$ represents the resistance of this second resistor then the power $P$ delivered to that resistor is given by

$\displaystyle{P=\frac{E^2r}{(r+r_0)^2+x_0^2}}$

with $E$ , $r_0$ , and $x_0$ treated as constants. For which value of $r$ is the power $P$ maximized?

_{extracted from Michael Corral. (2020). Elementary Calculus}

Lemma. (quotient rule)

Let $f(x)=\displaystyle{\frac{g(x)}{h(x)}}$ , where both $g$ and $h$ are differentiable and $h(x)\neq 0$ . The quotient rule states that the derivative of $f(x)$ is

$f'(x)=\displaystyle{\frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}}$

_{Wikipedia on Quotient Rule}

Roughwork.

Defining

$\begin{aligned} g(r) & = E^2r\\ h(r) & = (r+r_0)^2+x_0^2\\ \end{aligned}$

and

$\begin{aligned} g'(r) & = E^2\\ h'(r) & = 2(r+r_0)\\ \end{aligned}$

then

$\displaystyle{\frac{\mathrm{d}P}{\mathrm{d}r}}=0\Longrightarrow g'(r)h(r)-g(r)h'(r)=0$

Computing as follows

$\begin{aligned} (E^2)\big( (r+r_0)^2+x_0^2 \big) & = (E^2r)\big( 2(r+r_0) \big) \\ (r+r_0)^2+x_0^2 & = 2r(r+r_0) \\ r^2+2rr_0+r_0^2 +x_0^2 & = 2r^2+2rr_0 \\ r & = \sqrt{r_0^2+x_0^2} \\ \end{aligned}$

The mathematically formal way is to show that

$\displaystyle{\frac{\mathrm{d}^2P(r)}{\mathrm{d}r}\bigg|_{r=\sqrt{r^2_0+x_0^2}}}<0$

But from a physical point of view, assume that the electric currents $i$ passing through every components in series are the same, and the potential difference across each total up to the supplied voltage, namely,

$\begin{aligned} i(t) & = I_{\textrm{peak}}\sin (\omega t) \textrm{ where }I_{\textrm{peak}}\textrm{= const.} \\ E & = ir_0+x_0\frac{\mathrm{d}i}{\mathrm{d}t}+ir \\ \end{aligned}$

The power $P$ delivered to resistor $r$ is given by the formula

$\boxed{P=Vi=\displaystyle{\frac{V^2}{r}}=i^2r}$

and the only way to maximize $P$ , is to maximize either or both $i$ and $V$ .

Solving for a first-order ordinary differential equation:

$f(r,i,i')= ir_0+x_0i'+ir-E=0$

deriving current $i$ wrt $\omega t$ , we have

$\begin{aligned} 0 & = i'r_0+x_0i''+i'r \\ 0 & = -x_0I_{\textrm{peak}}\sin (\omega t)+(r_0+r)I_{\textrm{peak}}\cos (\omega t) \\ 0 & = -x_0\tan (\omega t) + r_0+r \\ \tan (\omega t) & = \frac{r_0+r}{x_0} \\ \sin (\omega t)&=\frac{r_0+r}{\sqrt{(r_0+r)^2+x_0^2}} \\ \end{aligned}$

Thus $i(r)=\displaystyle{I_{\textrm{peak}}\frac{r_0+r}{\sqrt{(r_0+r)^2+x_0^2}}}$

$\begin{aligned} P & = i^2r \\ & = \bigg(\frac{I_{\textrm{peak}}(r_0+r)}{\sqrt{(r_0+r)^2+x_0^2}}\bigg)^2 \cdot r \\ & = \frac{I_{\textrm{peak}}^2(r_0+r)^2r}{(r_0+r)^2+x_0^2} \\ \end{aligned}$

(to be continued)

Recall the relation between root mean square (rms) values and peak values:

$\begin{aligned} V_{\textrm{rms}} & = \frac{V_\textrm{peak}}{\sqrt{2}} \\ I_{\textrm{rms}} & = \frac{I_\textrm{peak}}{\sqrt{2}} \\ \end{aligned}$

Recall also that the resistance $R_L$ of an ideal inductor is zero ( $=0$ ), and that after the circuit has shortly reached steady state (i.e., constant current $i$ anywhere/anytime), the potential difference ( $\textrm{p.d.}$ ) or voltage drop ( $\Delta V$ ) across the inductor will become zero ( $=0$ ) before long.

Try again,

$\begin{aligned} I_{\textrm{peak}} & = i \\ \frac{I_{\textrm{peak}}^2(r_0+r)^2r}{(r_0+r)^2+x_0^2} & = \frac{i^2(r_0+r)^2r}{(r_0+r)^2+x_0^2} \\ & = \frac{(ir_0+ir)^2r}{(r_0+r)^2+x_0^2} \\ & = \frac{E^2r}{(r_0+r)^2+x_0^2} \\ \end{aligned}$

The derivative test seems inevitable. Maybe you could show that resistance $r=\sqrt{r_0^2+x_0^2}$ maximizes power $P$ , simply by drawing a phasor diagram?

(to be continued)

June 8, 2022October 24, 2022

202206081349 Exercise 4.1 (Q12)

The phase velocity $v$ of a capillary wave with surface tension $T$ and water density $p$ is

$\displaystyle{v=\sqrt{\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi}}}$

where $\lambda$ is the wavelength. Find the value of $\lambda$ that minimizes $v$ .

_{extracted from Michael Corral. (2020). Elementary Calculus.}

Setup.

Let $v=v(u(\lambda ))$ s.t.

$\begin{aligned} v(u) &=\sqrt{u} \\ u(\lambda ) & =\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi } \\ \end{aligned}$

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}u}\big( v(u)\big) & = \frac{1}{2}u^{-1/2}=\frac{1}{2}\Bigg(\sqrt{\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi}}\Bigg)^{-1} \\ \frac{\mathrm{d}}{\mathrm{d}\lambda}\big( u(\lambda )\big) & = -\frac{2\pi T}{\lambda^2p}+\frac{g}{2\pi} \\ \end{aligned}$

Hence,

$\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}\lambda} & = \bigg(\frac{\mathrm{d}v}{\mathrm{d}u}\bigg) \bigg(\frac{\mathrm{d}u}{\mathrm{d}\lambda}\bigg) \\ & = \frac{1}{2}\Bigg(\sqrt{\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi}}\Bigg)^{-1}\bigg(-\frac{2\pi T}{\lambda^2p}+\frac{g}{2\pi}\bigg) \\ \end{aligned}$

Working.

For $\displaystyle{\frac{\mathrm{d}v}{\mathrm{d}\lambda}}=0$ requires:

$\begin{cases} \displaystyle{\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi}} \neq 0 \\ \displaystyle{-\frac{2\pi T}{\lambda^2p}+\frac{g}{2\pi}}= 0 \\ \end{cases}$

Answer.

We have the satisfying stationary point:

$\lambda_0 =\displaystyle{2\pi\sqrt{\frac{T}{pg}}}$ .

It remains to be verified that this is a global minimum indeed.

(to be continued)

May 27, 2022July 24, 2022

202205271015 Parametrization 001

equation $y=mx+c$ describes a straight line with slope $m$ and $y$ -intercept $c$ , as shown below:

With some scalar parameter $t$ parametrize the equation, in vector representation of units $\hat{\imath}$ and $\hat{\jmath}$ , by

$\mathbf{s}(t)=t\,\hat{\mathbf{i}} + (mt+c)\,\hat{\mathbf{j}}\quad\textrm{where } t\in (-\infty ,\infty)$

whereas for a quadratic equation $y=ax^2+bx+c$ which describes a parabolic curve, its parametric representation is

$\mathbf{s}(t)=t\,\hat{\mathbf{i}} + (at^2+bt+c)\,\hat{\mathbf{j}}\quad\textrm{where } t\in (-\infty ,\infty)$

and similarly for a circle of radius $r$ centered at the origin $O(0,0)$ , its locus is parametrized as

$\mathbf{s}(t)=r\cos t\,\hat{\mathbf{i}}+r\sin t\,\hat{\mathbf{j}}\quad\textrm{where } t\in [-2\pi ,2\pi ]$ .

(to be continued)

November 22, 2021October 24, 2022

202111221615 Exercises 9.1.A (Q1)

For Exercises 1-8, determine if the given sequence is convergent. If so then find its limit.

$\{ ne^{-n}\}_{n=0}^{\infty}$

Solution.

By L’Hôpital’s Rule, treating an integer $n \ge 0$ as a real-valued variable $x$ ,

$\begin{aligned} & \quad \lim_{n\to\infty} \frac{n}{e^n} \\ & = \lim_{n\to\infty} \frac{0}{e^n} \\ & = 0 \\ \end{aligned}$

Thus the sequence is convergent and its limit is $0$ .

October 15, 2021October 24, 2022

202110150945 Exercises 8.1.A (Q1)

For Exercises 1-6, find the area of the region bounded by the given curves.

$y=x^2$ and $y=2x+3$

Roughwork.

$\begin{aligned} y& = x^2 = 2x+3 \\ 0 &= x^2-2x-3 \\ 0 &= (x-3)(x+1) \\ x &=3,-1 \end{aligned}$

$\therefore$ $\{(3,9),(-1,1)\}$ are the points of intersection.

The area $A$ of the shaded region is

$\begin{aligned} A & = \int_{-1}^{3}|(2x+3)-(x^2)|\,\mathrm{d}x \\ & = \bigg[ x^2+3x-\frac{x^3}{3}\bigg]\bigg|_{-1}^{3} \\ & = \bigg( (3)^2+3(3)-\frac{(3)^3}{3}\bigg) - \bigg( (-1)^2+3(-1)-\frac{(-1)^3}{3}\bigg) \\ & = (9) - \bigg(-\frac{5}{3}\bigg) \\ & = \frac{32}{3}\quad \textrm{(squared units)} \\ \end{aligned}$

(alternatively)

$\begin{array}{cl} {A & = \displaystyle\int_{1}^{9}\bigg( \sqrt{y} - \frac{y-3}{2}\bigg)\,\mathrm{d}y + \int_{0}^{1}\Big( (\sqrt{y})-(-\sqrt{y})\Big)\,\mathrm{d}y }\\ & = \displaystyle\bigg( \frac{2y^{3/2}}{3}-\frac{y^2}{4}+\frac{3y}{2}\bigg)\bigg|_{1}^{9} + \bigg( \frac{4y^{3/2}}{3}\bigg)\bigg|_{0}^{1}} \\ & = \displaystyle\Bigg\{ \bigg[ \frac{2(9)^{3/2}}{3}-\frac{(9)^2}{4}+\frac{3(9)}{2}\bigg] -\bigg[\frac{2(1)^{3/2}}{3}-\frac{(1)}{4}+\frac{3(1)}{2}\bigg]\Bigg\} }\\ &\displaystyle\qquad \quad + \Bigg\{\bigg[\frac{4(1)^{3/2}}{3}\bigg] - \bigg[ \frac{4(0)^{3/2}}{3} \bigg]\Bigg\} }\\ & =\displaystyle \bigg( 18-\frac{81}{4}+\frac{27}{2}\bigg) - \bigg( \frac{2}{3}-\frac{1}{4}+\frac{3}{2} \bigg) + \bigg(\frac{4}{3}\bigg) - (0)} \\ & = \displaystyle\frac{32}{3}\quad \textrm{(squared units)}} \end{array}$

October 12, 2021October 24, 2022

202110121413 Exercise 8.5.B (Q21)

This exercise is related to Einstein’s famous law $E=mc^2$ . The relativistic momentum $p$ of a particle of mass $m$ moving at a speed $v$ along a straight line (say, the $x$ -axis) is

$\displaystyle{p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}}$ ,

where $c$ is the speed of light. The relativistic force on the particle along that line is

$\displaystyle{F=\frac{\mathrm{d}p}{\mathrm{d}t}}$ ,

which is the same formula as Newton’s Second Law of motion in classical mechanics. Assume that the particle starts at rest at position $x_1$ and ends at position $x_2$ along the $x$ -axis. The work done by the force $F$ on the particle is:

$\displaystyle{W=\int_{x_1}^{x_2}F\,\mathrm{d}x=\int_{x_1}^{x_2}\frac{\mathrm{d}p}{\mathrm{d}t}\,\mathrm{d}x}$

(a) Show that

$\displaystyle{\frac{\mathrm{d}p}{\mathrm{d}v}=\frac{m}{\bigg( \displaystyle{1-\frac{v^2}{c^2}}\bigg)^{3/2}}}$ .

(b) Use the Chain Rule formula

$\displaystyle{\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}p}{\mathrm{d}v}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}}$

to show that

$\displaystyle{F\,\mathrm{d}x=v\frac{\mathrm{d}p}{\mathrm{d}v}\,\mathrm{d}v}$ .

(c) Use parts (a) and (b) to show that

$\displaystyle{W=\int_{0}^{v}\frac{\mathrm{d}p}{\mathrm{d}v}\, v\,\mathrm{d}v=\int_{0}^{v}\frac{mv}{\bigg(\displaystyle{ 1-\frac{v^2}{c^2}}\bigg)^{3/2}}\,\mathrm{d}v}$ .

(d) Use part (c) to show that

$\displaystyle{W=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2}$ .

(e) Define the relativistic kinetic energy $K$ of the particle to be $K=W$ , and define the total energy $E$ to be

$\displaystyle{E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}}$ .

So by part (d), $K=E-mc^2$ . Show that

$E^2=p^2c^2+(mc^2)^2$ .

(Hint: Expand the right side of that equation.)

(f) What is $E$ when the particle is at rest?

Solution.

(a)

$\begin{aligned} \frac{\mathrm{d}p}{\mathrm{d}v} & = \frac{\mathrm{d}}{\mathrm{d}v}\bigg( \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\bigg) \\ & = \frac{\bigg(\sqrt{1-\displaystyle{\frac{v^2}{c^2}}}\bigg)(m)-(mv)\bigg(\displaystyle{\frac{1}{2}}\Big( 1-\displaystyle{\frac{v^2}{c^2}}\Big)^{-1/2}\Big( -\displaystyle{\frac{2v}{c^2}}\Big)\bigg)}{\bigg(\sqrt{1-\displaystyle{\frac{v^2}{c^2}}}\bigg)^2} \\ & = \frac{\bigg( 1-\displaystyle{\frac{v^2}{c^2}}\bigg) (m)-(mv)\bigg( \displaystyle{-\frac{v}{c^2}}\bigg)}{\bigg( 1-\displaystyle{\frac{v^2}{c^2}}\bigg)^{3/2}} \\ & = \frac{m}{\bigg( 1-\displaystyle{\frac{v^2}{c^2}}\bigg)^{3/2}} \end{aligned}$

(b)

$\begin{aligned} \int_{x_1}^{x_2}F\,\mathrm{d}x & = \int_{x_1}^{x_2}\frac{\mathrm{d}p}{\mathrm{d}t}\,\mathrm{d}x \\ \int_{x_1}^{x_2}F\,\mathrm{d}x & = \int_{x_1}^{x_2}\frac{\mathrm{d}p}{\mathrm{d}v}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}x \\ F\,\mathrm{d}x & = \frac{\mathrm{d}p}{\mathrm{d}v}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}x \\ & = v\,\frac{\mathrm{d}p}{\mathrm{d}v}\,\mathrm{d}v\\ \end{aligned}$

(c)

$\begin{aligned} W & =\int_{0}^{v}\frac{\mathrm{d}p}{\mathrm{d}v}\, v\,\mathrm{d}v \\ & = \int_{0}^{v}\Bigg( \frac{m}{\big( 1-\frac{v^2}{c^2}\big)^{3/2}}\Bigg) (v)\,\mathrm{d}v\\ & =\int_{0}^{v}\frac{mv}{\bigg(\displaystyle{ 1-\frac{v^2}{c^2}}\bigg)^{3/2}}\,\mathrm{d}v \\ \end{aligned}$ .

(d)

$\begin{aligned} W & = \int_{0}^{v}\frac{mv}{\bigg(\displaystyle{1-\frac{v^2}{c^2}}\bigg)^{3/2}}\,\mathrm{d}v \\ \dots\enspace & \textrm{let }u=\frac{v}{c}\enspace\dots \\ \dots\enspace & \textrm{then }\mathrm{d}v=c\,\mathrm{d}u\enspace\dots \\ & = \int_{0}^{\frac{v}{c}}\frac{mc^2u}{(1-u^2)^{\frac{3}{2}}}\,\mathrm{d}u \\ \dots\enspace & \textrm{let }u=\sin\theta\enspace\dots \\ \dots\enspace & \textrm{then }\mathrm{d}u=\cos\theta\,\mathrm{d}\theta\enspace\dots \\ & = \int\frac{mc^2\sin\theta}{\cos^3\theta}\cdot\cos\theta\,\mathrm{d}\theta \\ & = \int\frac{mc^2\sin\theta}{\cos^2\theta}\,\mathrm{d}\theta \\ \dots\enspace & \textrm{let }w=\cos\theta\enspace\dots \\ \dots\enspace & \textrm{then }\mathrm{d}w=-\sin\theta\,\mathrm{d}\theta\enspace\dots \\ & \quad -\int \frac{mc^2}{w^2}\,\mathrm{d}w \\ & = \frac{mc^2}{w} \\ & = \frac{mc^2}{\cos\theta} \\ & = mc^2\sec\theta \\ & = \frac{mc^2}{\sqrt{1-u^2}} \\ & = \enspace\dots \\ & = \bigg[\frac{mc^2}{\sqrt{1-u^2}}\bigg]\bigg|^{\frac{v}{c}}_{0} \\ & = \frac{mc^2}{\sqrt{1-\displaystyle{\frac{v^2}{c^2}}}}-mc^2 \\ \end{aligned}$

Parts (e) and (f) are left to the readers.

October 12, 2021October 24, 2022

202110120936 Exercises 13.3.6 (Q43)

Graph each function by algebraically determining its key features. Then state the domain and range of the function.

$f(x)=x^2-7x+12$

Solution.

$\begin{aligned} f(x) & = x^2-7x+12 \\ & = (x-3)(x-4) \end{aligned}$

We see that $y=f(x)=0$ when $x=\{ 3,4\}$ .

i. $\therefore$ The $x$ -intercepts of $f(x)$ are thus $(3,0)$ and $(4,0)$ .

Plugging in $x=0$ will give the $y$ -intercept:

$\begin{aligned} y\textrm{-intercept}& = f(0) \\ & = (0)^2-7(0)+12 \\ & = 12 \end{aligned}$

ii. $\therefore$ The $y$ -intercept is thus $(0,12)$ .

Differentiating $y=f(x)$ with respect to $x$ ,

$\begin{aligned} & \quad f'(x) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f(x)\big) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(x^2-7x+12) \\ & = 2x-7 \\ \end{aligned}$

When $x=3.5$ and $f(3.5)=-0.25$ , the slope of $f(x)$ is zero, i.e., $f'(x)=0$ .

iii. $\therefore$ $(\frac{7}{2},-\frac{1}{4})$ is a turning point (/extreme point/vertex).

$\begin{aligned} y & =f(x) \\ & = x^2-7x+12 \\ & = x^2-7x+\bigg( \frac{7}{2}\bigg)^2 - \bigg( \frac{7}{2}\bigg)^2 + 12 \\ & = \bigg( x-\frac{7}{2}\bigg)^2 - \frac{1}{4}\\ \end{aligned}$

iv. $\therefore$ The axis of symmetry of the graph is $x=\frac{7}{2}$ .

Differentiating $y=f(x)$ twice with respect to $x$ ,

$\begin{aligned} &\quad f''(x)\\ & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f'(x)\big) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(2x-7) \\ & = 2 \quad (>0)\\ \end{aligned}$

We see that the slope is increasing with $x$ .

v. $\therefore$ The graph of $f(x)$ is concave upward (/convex downward).

vi. $\therefore$ The domain is $\mathbb{R}$ and the range $\{ y\in\mathbb{R}:y\ge -\frac{1}{4}\}$ .

September 16, 2021October 24, 2022

202109161207 Exercises 3.5.A (Q1)

Find the volume $V$ inside the paraboloid $z=x^2+y^2$ for $0\le z\le 4$ .

Background. (Triple integral in cylindrical coordinates)

$\displaystyle{\iiint\limits_{S}f(x,y,z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z=\iiint\limits_{S'}f(r\cos\theta ,r\sin\theta ,z)\, r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z}$ where the mapping $x=r\cos\theta$ , $y=r\sin\theta$ , $z=z$ maps the solid $S'$ in $r\theta z$ -space onto the solid $S$ in $xyz$ -space in a one-to-one manner.

_{pg.122, Chapter 3.5, Michael Corral. (2008). Vector Calculus}

Setup.

hence,

Solution.

Using vertical slices, we see that

$\begin{aligned} V & =\iint\limits_{R}(4-z)\,\mathrm{d}A \\ & = \iint\limits_{R}\big( 4-(x^2+y^2)\big)\,\mathrm{d}A\end{aligned}$

where $R=\{ (x,y):x^2+y^2\le 4\}$ is the disc in $\mathbb{R}^2$ . In polar coordinates $(r,\theta )$ we know that $x^2+y^2=r^2$ and that $R'=\{ (r,\theta ): 0\le r\le 2, 0\le\theta\le 2\pi\}$ .

Thus,

$\begin{aligned} V & = \int_{0}^{2\pi}\int_{0}^{2} (4-r^2)\, r\,\mathrm{d}r\,\mathrm{d}\theta \\ & = \int_{0}^{2\pi}\int_{0}^{2} (4r-r^3)\,\mathrm{d}r\,\mathrm{d}\theta \\ & = \int_{0}^{2\pi} \bigg[ 2r^2-\frac{r^4}{4}\bigg]\bigg|_{0}^{2}\,\mathrm{d}\theta \\ & = 4\int_{0}^{2\pi} \mathrm{d}\theta \\ & = 4(2\pi )\\ & = 8\pi\quad \textrm{(cubic units)} \end{aligned}$

September 16, 2021October 24, 2022

202109161150 Exercises 3.3.B (Q9)

Let $M$ be a constant. Show that

$\displaystyle{\int_{z_1}^{z_2}\int_{y_1}^{y_2}\int_{x_1}^{x_2}M\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z}=M(z_2-z_1)(y_2-y_1)(x_2-x_1)$ .

Solution.

$\begin{aligned} & \qquad \int_{z_1}^{z_2}\int_{y_1}^{y_2}\int_{x_1}^{x_2}M\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\ & = M\int_{z_1}^{z_2}\int_{y_1}^{y_2}\big[ x\big]\big|_{x_1}^{x_2} \,\mathrm{d}y\,\mathrm{d}z \\ & = M(x_2-x_1)\int_{z_1}^{z_2}\int_{y_1}^{y_2} \,\mathrm{d}y\,\mathrm{d}z \\ & = M(x_2-x_1)\int_{z_1}^{z_2}\big[ y\big] \big|_{y_1}^{y_2}\,\mathrm{d}z \\ & = M(y_2-y_1)(x_2-x_1)\int_{z_1}^{z_2}\,\mathrm{d}z \\ & = M(y_2-y_1)(x_2-x_1)\big[ z\big] \big|_{z_1}^{z_2} \\ & = M(z_2-z_1)(y_2-y_1)(x_2-x_1) \\ \end{aligned}$

September 16, 2021October 24, 2022

202109161118 Exercises 3.3.A (Q3)

For Exercises 1-8, evaluate the given triple integral.

$\displaystyle{\int_{0}^{\pi}\int_{0}^{x}\int_{0}^{xy}x^2\sin z\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x}$

Solution.

$\begin{aligned} &\qquad \int_{0}^{\pi}\int_{0}^{x}\int_{0}^{xy}x^2\sin z\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{\pi}\int_{0}^{x}\big( -x^2\cos z\big)\big|_{z=0}^{z=xy}\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{\pi}\int_{0}^{x}(-x^2)(\cos xy-1)\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{\pi} (-x^2)\bigg(\frac{\sin xy}{x}-y\bigg)\bigg|_{y=0}^{y=x}\,\mathrm{d}x \\ & = \int_{0}^{\pi} (-x^2)\bigg( \frac{\sin x^2}{x}-x \bigg)\,\mathrm{d}x \\ & = \int_{0}^{\pi} (x^3 - x\sin x^2)\,\mathrm{d}x \\ & = \int_{0}^{\pi}x^3\,\mathrm{d}x - \int_{0}^{\pi} x\sin x^2\,\mathrm{d}x \\ & = \bigg( \frac{x^4}{4} \bigg)\bigg|_{0}^{\pi} - \bigg[ \cdots \bigg] \\ & = \enspace \dots \textrm{ (to be continued) }\dots \end{aligned}$

Roughwork. (Integration by parts)

$\begin{aligned} & \qquad \int x\sin x^2\,\mathrm{d}x \\ & = \int \sin x^2\,\mathrm{d}\bigg(\frac{x^2}{2}\bigg) \\ & = \frac{1}{2}\cdot\int \sin x^2\,\mathrm{d}(x^2) \\ & = - \frac{\cos x^2}{2} \\ \end{aligned}$

$\begin{aligned} & = \dots \textrm{ (to continue) }\dots \\ & = \bigg(\frac{x^4}{4}\bigg)\bigg|_{0}^{\pi} - \bigg( -\frac{\cos x^2}{2} \bigg)\bigg|_{0}^{\pi} \\ & = \bigg(\frac{\pi^4}{4}\bigg) + \bigg(\frac{\cos\pi^2}{2} - \frac{1}{2}\bigg) \\ & = -\frac{1}{2} + \frac{\pi^4}{4} + \frac{\cos\pi^2}{2} \\ \end{aligned}$