18.02 Problem Set 10 (due Thursday, April 29, 1999)

Part I     (9 points)

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

Lecture 30 (Thurs. April 22): Surface integrals and flux.

Read: SN, vector calculus section 9.

Problems: SN p.9.7 nos. 1, 2, 3, 4, 6, 8 (S. 36,37).

Lecture 31 (Thurs. April 22): Applications of flux, divergence theorem.

Read: EP pp. 995-998, 1000-1001.

Lecture 32 (Tues. April 27): Divergence theorem.

Read: SN, Vector Calculus, section 10; EP pp. 1006-1008.

Problems: SN, p. 10.5 nos. 1a, 2, 3, 5; p. 10.6 6, 7i, 8 (S.38,39).

Part II     (20 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.

(Thurs. 8 pts: 2, 3, 3) Take the surface bounded below by the right-angled cone $z=(x^2 + y^2)^{1/2}$, and above by the unit sphere centered at the origin. The upper and lower surfaces intersect in a circle.

Let $S$ be the disc having this circle as its boundary; $T$ the spherical cap forming the upper surface, and $U$ the cone forming the lower surface. Orient all three surfaces so that the normal vector points generally upward (i.e, has a positive $k$-component).

Calculate the flux of the vector field $F=z{\bf k}$ over each of these three surfaces. In each case, do the calculation directly from the definition of the surface integral for flux, as in the Notes and Part I problems.

Solution. Flux through $S:$

$$\iint_S \vec{F} \cdot \vec{n} dS = \iint_S z \, dS$$

$$= \iint _{x^2+y^2 \leq 1/2} \, dx \, dy/ \sqrt{2} = \frac{\pi}{2 \sqrt{2}} \, .$$

Since $z= 1/ \sqrt{2}$ on the disk.

Flux through $T:$

$$\iint_T \vec{F} \cdot \vec{n} \, dS = \iint_T z^2 \sin \varphi \, d \varphi \, d \theta$$

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

$$= \frac{2 \pi}{3} - \frac{\pi}{3 \sqrt{2}} \, .$$

Flux through $U:$ On $U$

$$\begin{multline*} z= \sqrt{x^2 + y^2} = f(x,y) \, , {\bf\hat{n}} \, dS = \left... ...t{2}}_0 r \cdot r \, dr \, d \theta = \frac{\pi}{3 \sqrt{2}}. \end{multline*}$$

2.

(Thurs. 3 pts.) A pattern of heat generation and absorption produces the temperature distribution in space ($u$ is the temperature at the point $(x,y,z))$:

$$u=e^{-x^2-y^2-z^2} \, .$$

Find the heat flow across a sphere of radius $a$ centered at the origin (cf. EP  p. 1000; call the heat conductivity $k$).

For what radius $a$ will the heat flow across the sphere be greatest?

Solution: The heat flow across a surface is the flux of the vector field $-k\nabla u$ where $u$ is the temerature distribution. In this case,

$$\nabla u=e^{-x^2-y^2-z^2}(-2x{\bf i}-2y{\bf j}-2z{\bf k}.$$

Thus the flux across the sphere of radius $a$ (outwards) is

$$\begin{multline*} \iint_S{\bf F}\cdot{\bf\hat n}dS\\ =(-k)\iint (-2)e^{-x^2-y^2... ...dot\frac1a(x{\bf i}+y{\bf j}+z{\bf k})dS\\ =8\pi a^3 ke^{-a^2}. \end{multline*}$$

The maximum value for this will occur when the derivative vanishes, that is when $8\pi k a^2e^{-a^2}(3-2a^2)=0$ so $a=\sqrt\frac32.$

3.

(Fri. 4 pts: 2,2) In problem 1 above, you calculated directly the flux of the vector field $F=z{\bf k}$ over three surfaces shown in cross-section at the right. The unit with the spherical cap $T$, the cone $U$ given by $z=r$, and the disc $S$ bounded by the circle in which $T$ and $U$ intersect. In each case, the normal vector is the one with a positive $k$-component.

The correct value for the flux over the disc $S$ is $\pi \sqrt{2}/4$. Using this:

(a)

Use the divergence theorem (and the formula for the volume of a solid cone) to get the flux over the conical surface $U$. (Watch for orientations!)

(b)

Similarly, use the divergence theorem to get the flux over $T$. (get the volume of the whole ice cream cone and subtract the cone volume to get the volume of the spherical cap on top.)

Solution:

(a)

By the divergence theorem,

$$\iint_{S}{\bf F}\cdot{\bf\hat n}dS+\iint_{U'}{\bf F}\cdot{\bf\hat n}dS=\iiint_{C}(\nabla\cdot{\bf F})dV= (C)=\frac{\sqrt2\pi}{12}$$

where $U'$ is $U$ with the opposite normal and $C$ is the solid cone. Therefore,

$$\iint{U}{\bf F}\cdot{\bf\hat n}dS=-\iint_{U'}{\bf F}\cdot{\bf\hat n}dS =\frac{\sqrt2\pi}6.$$

(b)

Volume of solid is

$$\int_0^{2\pi}\int_0^{\pi/4}\int_0^1\rho ^2\sin\varphi d\rho \,d\varphi \,d\theta =\frac{2\pi}3-\frac{\sqrt2\pi}3.$$

So the volume of the cap is $\frac{2\pi}3-\frac{5\sqrt2\pi}{12}.$ By the divergence theorem

$$\iint_{S'}{\bf F}\cdot{\bf\hat n}dS+\iint_{T}{\bf F}\cdot{\bf\hat n}dS=$$

so

$$\iint_{T}{\bf F}\cdot{\bf\hat n}dS=\frac{2\pi}3-\frac{\sqrt2\pi}6.$$

4.

(Fri. 2 pts.) Let $S$ be a smooth closed surface. Show that the field $x{\bf i}+y{\bf j}+z{\bf k}$ cannot be tangent to $S$ at every point $(x,y,z)$ of $S$.

Solution: By the divergence theorem $\iint_S{\bf F}\cdot{\bf\hat n}dS=\iiint_R\nabla\cdot{\bf F}dV.$ If ${\bf F}=x{\bf i}+y{\bf j}+z{\bf k}$ then $\nabla\cdot{\bf F}=3$ and the volume integral is $3$Vol$(R)\not=0.$ Thus the flux of ${\bf F}$ through $S$ cannot be zero, so ${\bf F}\cdot{\bf\hat n}=0$ at some point on $S$ and therefore $x{\bf i}+y{\bf j}+z{\bf k}$ cannot be tangent to the surface everywhere.

5.

(Fri. 3 pts.) Prove that if $f(x,y,z)$ satisfies Laplace's equation (see Notes P), then the flux of its gradient field $\triangledown f$ across any smooth closed surface is 0.

Solution: If ${\bf F}=\nabla f=f_x{\bf i}+f_y{\bf j}+f_z{\bf k}$ then $\nabla\cdot{\bf F}=f_{xx}+f_{yy}+f_{zz}=0,$ precisely the given condition that $f$ satisfy Laplace's equation. The by the divergence theorem the flux of $\nabla f$ through any closed surface is $\iint_S{\nabla a}\cdot{\bf\hat n} dS=0.$