September 16, 2021 – Physicspupil

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.

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

September 16, 2021October 24, 2022

202109161046 Exercises 1.1.A (Q2)

For the points $P=(1,-1,1)$ , $Q=(2,-2,2)$ , $R=(2,0,1)$ , $S=(3,-1,2)$ , does $\overrightarrow{PQ}=\overrightarrow{RS}$ ?

Solution.

$\begin{aligned} \overrightarrow{PQ} & = (2,-2,2) - (1,-1,1) \\ & = (1,-1,1) \\ \overrightarrow{RS} & = (3,-1,2) - (2,0,1) \\ & = (1,-1,1) \\ \end{aligned}$

$\therefore\enspace \overrightarrow{PQ}=\overrightarrow{RS}$

September 16, 2021October 24, 2022

202109161039 Exercises 1.1.A (Q1)

Calculate the magnitudes of the following vectors:

(a) $\mathbf{v}=(2,-1)$
(b) $\mathbf{v}=(2,-1,0)$
(c) $\mathbf{v}=(3,2,-2)$
(d) $\mathbf{v}=(0,0,1)$
(e) $\mathbf{v}=(6,4,-4)$

Solution.

(a)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(2)^2+(-1)^2} \\ & = \sqrt{5} \end{aligned}$

(b)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(2)^2+(-1)^2+(0)^2} \\ & = \sqrt{5} \end{aligned}$

(c)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(3)^2+(2)^2+(-2)^2} \\ & = \sqrt{17} \end{aligned}$

(d)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(0)^2+(0)^2+(1)^2} \\ & = \sqrt{1} \\ & = 1 \end{aligned}$

(e)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(6)^2+(4)^2+(-4)^2} \\ & = \sqrt{68} \\ & = 2\sqrt{17} \\ \end{aligned}$

September 16, 2021October 24, 2022

202109160920 Exercises 5.3.B (Q21)

Show that for all $x>0$ ,

$\displaystyle{\int_{0}^{x}\frac{\mathrm{d}t}{1+t^2}+\int_{0}^{1/x}\frac{\mathrm{d}t}{1+t^2}}$

is independent of $x$ .

Solution.

$\begin{aligned} & \qquad\int_{0}^{x}\frac{\mathrm{d}t}{1+t^2}+\int_{0}^{1/x}\frac{\mathrm{d}t}{1+t^2} \\ \dots &\enspace\textrm{as }\sin^2u +\cos^2u =1 \enspace\dots\\ \dots &\enspace\textrm{so }\tan^2u +1 = \sec^2u\enspace\dots \\ \dots &\enspace\textrm{let }t=\tan u \textrm{ then }\mathrm{d}t=\sec^2u\,\mathrm{d}u\enspace\dots \\ & = \int_{0}^{\arctan x}\frac{\sec^2u\,\mathrm{d}u}{1+\tan^2u} + \int_{0}^{\arctan (\frac{1}{x})}\frac{\sec^2u\,\mathrm{d}u}{1+\tan^2u} \\ & = \int_{0}^{\arctan x}\frac{\sec^2u\,\mathrm{d}u}{\sec^2u} + \int_{0}^{\arctan (\frac{1}{x})}\frac{\sec^2u\,\mathrm{d}u}{\sec^2u} \\ & = [u]|_{0}^{\arctan x} + [u]|_{0}^{\arctan (1/x)}\\ & = \arctan x + \arctan\bigg(\frac{1}{x}\bigg) \\ & \stackrel{(*)}{=} \frac{\pi}{2} \end{aligned}$

Lemma (*)

Let $y=\arctan x$ and $w=\arctan \displaystyle{\bigg(\frac{1}{x}\bigg)}$ ,

then $x=\tan y$ and $\displaystyle{\frac{1}{x}}=\tan w$ .

Thus,

$\begin{aligned} \tan y & = \frac{1}{\tan w} \\ \tan y & = \cot w \\ \tan y & = \tan \bigg(\frac{\pi}{2}-w\bigg) \\ y & = \frac{\pi}{2} - w \\ y + w & = \frac{\pi}{2} \\ \end{aligned}$

$\therefore\enspace \displaystyle{\arctan x + \arctan\bigg(\frac{1}{x}\bigg) = \frac{\pi}{2}}$

(independent of $x$ )

September 16, 2021July 25, 2022

202109160908 Exercises 5.1.A (Q18)

Use the free fall motion equation for position to show that the maximum height reached by an object launched straight up from the ground with an initial velocity $v_0$ is $\frac{v_0^2}{2g}$ .

Proof.

$\begin{aligned} v^2 & = u^2+2as \\ \dots \textrm{ upward} & \textrm{ taken to be positive }\dots \\ 0 & = v_0^2 + 2(-g)s \\ s & = \frac{v_0^2}{2g} \end{aligned}$

September 16, 2021October 24, 2022

202109160857 Exercises 5.1.A (Q17)

Integrate both sides of the equation

$\displaystyle{\frac{\mathrm{d}P}{P}+\frac{\mathrm{d}M}{M}=\frac{\mathrm{d}T}{2T}}$

to obtain the ideal gas continuity relation:

$\displaystyle{\frac{PM}{\sqrt{T}}=\textrm{constant}}$ .

Attempts.

$\begin{aligned} \int\frac{\mathrm{d}P}{P} + \int\frac{\mathrm{d}M}{M} & = \int\frac{\mathrm{d}T}{2T} \\ \ln |P| + \ln |M| - \frac{1}{2}\ln |T| & = C \\ \ln \bigg|\frac{PM}{T^{1/2}}\bigg| & = C \\ \frac{PM}{\sqrt{T}} & = \textrm{Const.} \\ \end{aligned}$