18.02 Problem Set 9 (due Friday, April 23, 1999)
With Solutions

These problems on multiple integration and surface integrals involve a lot of work. Get started early, even though it's not due until next week. Quite a lot of the work is in Part I.

Part I     (20 points)

Hand in the the underlined problems; the others are for practice.

Lecture 27 (Tues. April 13): Triple integrals in rectangular and cylindrical coordinates.

Read: EP, sect. 12.8 to p. 787. Read: SN I, pp. I.2, I.3; Read EP 14.6, pp. 926-929 (concentrate on Exs. 1,2,3); Read EN 14.7, pp. 934-6.

Problems: EP p.791 Probs 5, 9, 11, 15, 23, 25, 27, 55, 60 (give $z$ in terms of $r$), 61. (Solutions in back of book; no. 60: $0 \leq r \leq R, Hr/R \leq z \leq H, 0 \leq \Theta \leq 2 \pi$) More problems: I.3/11, 12, 13 (cylindrical coordinates) (I.5). Even more problems EP: 14.6/1, 5, 33, 39 ($\bar{z}$ only; use symmetry, half the region), 43 (S.23); 14.7/9, 12 (vol. only),19 (S.24).

Lecture 28 (Thurs. April 15): Triple Integrals in spherical coordinates.

Read: Notes I, p. I.4; EP 14.7, pp. 936-940.

Probs. SN: I.4/14, 16 (I.5); EP 14.7 p.941 21, 26, 29, 40 (S.24,25).

Lecture 29 (Fri. April 16): Gravitational attraction. Vector fields.

Read: SN sect. G, probs: G.3/3, 4 (S.25). Read: SN, Vector Calculus, section 8. Probs. SN 8.2, nos. 1, 3 (8.2).

Part II     (25 points)

Directions: Try each problem alone for 25 minutes. If you subsequently collaborate, solutions must be written up independently. It is illegal to consult old problem sets.

1.

(Tues. 3 pts: 2,1) (cf. EP 14.6/33 (S.23))

A rectangular solid has dimensions $a,b,c$. Its density is $1$.

(a)

Find its moment of inertia about an edge of length $c$. (Place the solid so this edge lies along the $z$-axis.)

(b)

Suppose the dimensions of the box are $1$, $2$ and $3$. About which edge will the moment of inertia be greatest? smallest? Predict the answer by physical intuition, then verify it by using the formula you found in part (a).

Solution:

(a)

Moment of inertia about $z$-axis: $\int^a_0 \int^b_0 \int^c_0 (x^2+y^2) \, dx \, dy \, dz$.

$$\begin{array}{ll} \hbox{Inner:} & (x^2 + y^2) z ]^c_0 = c (x^... ...rac{1}{3} b^3 x)]^a_0 = \frac{abc}{3} (a^2+b^2)\, . \end{array}$$

(Note that this is symmetric in $a,b$ as it should be.)

(b)

$$\begin{array}{ll} \hbox{Should be \emph{biggest} about axis $... ...axis $2$} & = \frac{1 \cdot 3 \cdot 2}{3} (1^2+3^2) \end{array}$$

2.

(Tues. 3 pts.) Work EP 14.7, p. 940 no. 15 (cf. 14.7/12 (S.24))

Problem: Find volume of the region $D$ bounded above by the spherical surface $x^2+y^2+z^2=2$ and below by the paraboloid $z=x^2+y^2$.

Solution: Certainly $D$ has circular symmetry. Its projection into the $xy$-plane (its shadow) is a disc of some radius $a.$ To determine $a$ set $x=0$ and solve

$$\begin{multline*} z^2+y^2=2 \text{ and }z=y^2\Rightarrow z^2+z-2=(z-1)(z+2)=0\Rightarrow\\ z=-2 \text{ (makes no sense) or }z=1. \end{multline*}$$

So $a=1.$

In cylindrical polar coordinates the volume is

$$\int^{2 \pi}_0 \int^1_0 \int^{\sqrt{2-r^2}}_{r^2} \, rdz\, dr \, d \theta.$$

$$\begin{array}{ll} \hbox{Inner:} & rz]^{\sqrt{2-r^2}}_{r^2} = ... ...) \quad (\hbox{Book: } \frac{\pi}{6}(8 \sqrt{2}-7)) \end{array}$$

3.

(Thurs. 2 pts: 1,1)

Definition: The average value of $f(x,y,z)$ over a region $D$ in $3$-space is

$$\frac{1}{V(D)} \iiint f(x,y,z) \, dV, \qquad V(D)= \hbox{volume of } D \, .$$

Find the average distance of a point in a solid sphere of radius $1$ from

(a)

the center of the sphere

(b)

an axis of the sphere

(c)

a point on the surface of the sphere

Solution:

(a)

$$\int^{2\pi}_0 \int^{\pi}_0 \int^1_0 \rho \cdot \rho^2 \sin \varphi\, d \rho \, d \varphi \, d \theta$$

$$\quad \frac{\rho ^4}{4} \sin \varphi ]^1_0 = \frac{\sin \varphi}{4}, \quad \frac{- \cos \varphi}{4}]^{\pi}_0 = \frac{1}{2}, \quad \frac{1}{2} \cdot 2 \pi = \pi$$

then divide by the volume of the sphere $4/3 \pi$ so

$$\quad \pi \cdot \frac{3}{4 \pi} = 3/4.$$

(b)

$$\int^{2\pi}_0 \int^{\pi}_0 \int^1_0 \rho \sin \varphi \cdot \rho^2 \sin \varphi \, d \rho \, d \varphi \, d \theta$$

$$\quad\frac{1}{4} \sin^2 \varphi, \quad \frac{1}{4} \left( \frac{\varphi}{2} - \frac{\sin 2 \varphi}{4} \right) ]^{\pi}_0 = \pi/8, \quad 2 \pi \cdot \pi/8 = \pi^2/4$$

So

$$\quad \frac{\pi^2}{4} \cdot \frac{3}{4 \pi} = \frac{3 \pi}{16}.$$

(c)

Take the point on the boundary of the solid sphere to be $(0,0,1).$ Then the distance squared from it is

$$x^2+y^2+(z-1)^2=\rho ^2-2\rho\cos\phi+1.$$

Thus the integral we wish to evaluate is

$$\int^{2\pi}_0 \int^1_0 \int^{\pi}_0 (\rho ^2-2\rho\cos\varphi+1)^{\frac12} \cdot \rho^2 \sin \varphi \, d \varphi \,d \rho \, d \theta.$$

$$\quad \frac13\rho(\rho ^2-2\rho\cos\varphi+1)^{\frac32}]_0^\pi=\frac13 \rho ((\rho -1)^3-(\rho +1)^3)$$

$$\quad \frac13 (\frac25\rho ^5+2\rho ^3)]^1_0=\frac45,\ \frac{8\pi}5$$

$$\frac65.$$

4.

(Fri. 3 pts) Prob SN, G.3/1b

Solution: Grav. attractions on vertex of solid icream cone with density $\rho$

$$=G \int^{2\pi}_0 \int^{\pi/6}_0 \int^a_0 \rho \cdot \frac{\cos \rho}{\rho^2} \cdot \rho^2 \sin\varphi \, d \rho \, d \varphi \, d \theta$$

the integrand being density$\times$   grav.$\times dV.$ Dropping $G$ for the moment

$$\frac{\rho^2}{2} \sin \varphi \cos \varphi ]^a_0 = \frac{a^2}{2} \sin \varphi \cos \varphi$$

$$\frac{a^2}{4} \sin^2 \varphi ]^{\pi/6} = \frac{a^2}{4} \cdot \frac{1}{4} \quad = \frac{a^2}{16}$$

$$2 \pi \cdot \frac{a^2}{16} = \frac{\pi a^2}{8}$$

$$\quad G \cdot \frac{\pi a^2}{8}$$

5.

(Fri. 3 pts) Prob SN, G.3/5

Solution: Put the hemisphere with its flat surface up and its pole at the origin. Then do the integral in two pieces: the inner solid cone and the outer shell.

$$=G \int^{2\pi}_0 \int^{\pi/4}_0 \int^{a/\cos \varphi} \sin \varphi \cos \varphi \, d \rho \, d \varphi \, d \theta$$

(since the top surface is $z=a$, $\therefore \rho \cos \varphi =a$ or $\rho = a/ar \varphi$)

(Drop $G$ during the integration)

Evaluating Cone part:

$$\sin \varphi \cos \varphi \cdot \rho ]^{a/ar \varphi}= a \sin \varphi,$$

$$- a ar \varphi ]^{2 \pi/4} = a\left(1- \frac{\sqrt{2}}{2}\right),$$

$$2 \pi \frac{\sqrt{2}}{6}a = \frac{\sqrt{2} \pi a}{3} G$$

So the total is

$$= \pi a \left( 2- \sqrt{2} + \frac{\sqrt{2}}{3} \right) G = 2 \pi a G \left( 1- \frac{\sqrt{2}}{3} \right)$$

6.

(Fri. 3 pts)

(a)

Write down the most general vector field, all of whose vectors are perpendicular to the plane $2x-y+3z=5$.

(b)

The force of a positive unit dork is radically outward, with magnitude equal to the distance of the dork. Show that the force field in three-space resulting from two positive unit dorks, place on the $x$-axis at $x=1$ and $x=-1$, is the same as the force field of a single positive two-unit dork at the origin.

(c)

Write down the magnetic field in $3$-space arising from an infinite wire along the $y$-axis, carrying a unit positive current in the negative direction.

(The field is tangent to a circle centered on the $y$-axis, lying in a plane perpendicular to the $y$-axis, with direction given by the right hand rule, and magnitude inversely proportional to the distance from the $y$-axis.)

Solution:

(a)

If the vectors are perpendicular to $2x-y+3z=5$, they are parallel to the normal vector to this plane: ${\bf i} - {\bf j} + 3{\bf k}$ and so must be of the form

$${\bf F} = f(x,y,z) (2 {\bf i}) - {\bf j} + 3 {\bf k}$$

(b)

For a unit dork at the origin, ${\bf F} = x{\bf i}+ y{\bf j} + z {\bf k}$. For a unit dork at $(-1,0,0)$ and $(1,0,0)$ therefore the sum of the fields is

$$(x+1){\bf i}+ y{\bf j}+ z{\bf k} + (x-1){\bf i} + y{\bf j} + z{\bf k} = 2x{\bf i} + 2y{\bf j} + 2z{\bf k}$$

which is the same as a two unit dork at the origin.

(c)

Looking at the $xz$-plane ${\bf F} = - \frac{z{\bf i} + x{\bf k}}{(x^2 + z^2)}.$