The Physics Language – Page 10

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

202110121137 Exercise 1.1.1

(a) Show that $x^2\in\langle x-y^2,xy\rangle$ in $k[x,y]$ ( $k$ any field).

(b) Show that $\langle x-y^2,xy,y^2\rangle = \langle x,y^2\rangle$ .

(c) Is $\langle x-y^2,xy\rangle = \langle x^2,xy\rangle$ ? Why or why not?

Definition.

Let $f_1,\dots ,f_s\in k[x_1,\dots ,x_n]$ . We let $\langle f_1,\dots ,f_s\rangle$ denote the collection $\langle f_1,\dots ,f_s\rangle = \{ p_1f_1+\cdots +p_sf_s:p_i\in k[x_1,\dots ,x_n]\}$ for $i= 1,\dots ,s$ .

(a)

$\begin{aligned} x^2 & = (x)(x-y^2) + (y)(xy) \\ & \in \langle x-y^2,xy\rangle \\ \end{aligned}$

(b)

First, we want to show

$\langle x-y^2,xy,y^2\rangle \in \langle x,y^2\rangle$ .

For any $p_1,p_2,p_3\in k[x,y]$ ,

$\begin{aligned} & \quad \langle x-y^2,xy,y^2\rangle \\ & = (p_1)(x-y^2)+(p_2)(xy)+(p_3)(y^2) \\ & = p_1x-p_1y^2+p_2xy+p_3y^2 \\ & = (p_1+p_2y)(x)+(p_3-p_1)(y^2) \\ & \in \langle x,y^2\rangle \\ \end{aligned}$

Next, we want to show

$\langle x,y^2\rangle\in\langle x-y^2,xy,y^2\rangle$ .

For any $p_1,p_2\in k[x,y]$ ,

$\begin{aligned} &\quad \langle x,y^2\rangle \\ & = (p_1)(x)+(p_2)(y^2) \\ & = (p_1)(x-y^2)+(0)(xy)+(p_1+p_2)(y^2) \\ & \in \langle x-y^2,xy,y^2\rangle \\ \end{aligned}$

All in all,

$\langle x-y^2,xy,y^2\rangle = \langle x,y^2\rangle$ .

(c) I guess $\textrm{\scriptsize{NOT}}$ .

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 28, 2021November 29, 2021

202109281636 SQL Table 001

CREATE TABLE SQL_Table_001
    (ID int,
    Subject varchar(255),
    Object varchar(255),
    Category varchar(255)
    );
INSERT INTO SQL_Table_001
    (Subject, Object, Category)
    VALUES
    ('algebraic geometry', 'regular (polynomial) functions', 'algebraic varieties')
    ('topology', 'continuous functions', 'topological spaces')
    ('differential topology', 'differentiable functions', 'differentiable manifolds')
    ('complex analysis', 'analytic (power series) functions', 'complex manifolds');

CREATE TABLE SQL_Table_001

(ID int,

Subject varchar(255),

Object varchar(255),

Category varchar(255)

);

INSERT INTO SQL_Table_001

(Subject, Object, Category)

VALUES

('algebraic geometry', 'regular (polynomial) functions', 'algebraic varieties')

('topology', 'continuous functions', 'topological spaces')

('differential topology', 'differentiable functions', 'differentiable manifolds')

('complex analysis', 'analytic (power series) functions', 'complex manifolds');

Output.
+------+-------------------------+-------------------------------------+----------------------------+
|  ID  |  Subject                |  Object                             |  Category                  |
+------+-------------------------+-------------------------------------+----------------------------+
|  1   |  algebraic geometry     |  regular (polynomial) functions     |  algebraic varieties       |
|  2   |  topology               |  continuous functions               |  topological spaces        |
|  3   |  differential topology  |  differentiable functions           |  differentiable manifolds  |
|  4   |  complex analysis       |  analytic (power series) functions  |  complex manifolds         |
+------+-------------------------+-------------------------------------+----------------------------+
(J.S. Milne on Algebraic Geometry (2017), pg. 7)

Output.

+------+-------------------------+-------------------------------------+----------------------------+

+------+-------------------------+-------------------------------------+----------------------------+

+------+-------------------------+-------------------------------------+----------------------------+

(J.S. Milne on Algebraic Geometry (2017), pg. 7)

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)

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

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

September 16, 2021October 24, 2022

202109161108 Exercises 3.3.A (Q2)

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

$\displaystyle{\int_{0}^{1}\int_{0}^{x}\int_{0}^{y}xyz\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x}$

Solution.

$\begin{aligned} &\qquad \int_{0}^{1}\int_{0}^{x}\int_{0}^{y}xyz\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{1}\int_{0}^{x} \bigg( \frac{xyz^2}{2}\bigg|_{z=0}^{z=y} \bigg)\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{1}\int_{0}^{x} \frac{xy^3}{2}\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{1} \bigg( \frac{xy^4}{8}\bigg|_{y=0}^{y=x} \bigg) \,\mathrm{d}x \\ & = \int_{0}^{1} \frac{x^5}{8}\,\mathrm{d}x \\ & = \bigg[ \frac{x^6}{48} \bigg]_{0}^{1} \\ & = \frac{1}{48} \\ \end{aligned}$

September 16, 2021October 24, 2022

202109161055 Exercises 3.3.A (Q1)

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

$\displaystyle{\int_{0}^{3}\int_{0}^{2}\int_{0}^{1}}xyz\,\mathrm{d}x\, \,\mathrm{d}y\,\mathrm{d}z$

Solution.

$\begin{aligned} & \qquad \int_{0}^{3}\int_{0}^{2}\int_{0}^{1}xyz\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\ & = \int_{0}^{3}\int_{0}^{2} \bigg(\frac{x^2yz}{2}\bigg|_{x=0}^{x=1}\bigg)\,\mathrm{d}y\,\mathrm{d}z \\ & = \int_{0}^{3}\int_{0}^{2}\bigg(\frac{yz}{2}\bigg)\,\mathrm{d}y\,\mathrm{d}z \\ & = \int_{0}^{3}\bigg( \frac{y^2z}{4}\bigg|_{y=0}^{y=2} \bigg) \,\mathrm{d}z \\ & = \int_{0}^{3} z\,\mathrm{d}z \\ & = \bigg[\frac{z^2}{2}\bigg]_{0}^{3} \\ & = 4\frac{1}{2} \\ \end{aligned}$