18.02 Problem Set 11 (due Friday, May 7, 1999)

Part I     (7 points)

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

Lecture 33 (Tues. May 4): Line integrals in space, $\mathop{curl}{\bf F}$, exactness, potentials.

Read SN: Sections 11, 12 and p. 15.1; Problems: SN p. 11.5 nos. 1, 2, 4, 5 (S.40)     SN p. 12.4 and 12.5 nos 1, 2, 3ab(ii) (both methods), 5 (S. 41-2)

Lecture 34 (Thurs. May 6): Stokes' Theorem.

Read: SN Section 13, EP Section 15.7. Problems: SP.7 nos. 5-B3, B4, B5, B7 (S. 43-4).

Lecture 35 (Fri. May 7): Stokes' Theorem, cont'd. Applications.

Part II     (11 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. 2 pts)

(a)

For what value(s) of the constant $a,b,c$ will the field

$${\bf F}= axyz{\bf i}+(bx^2z + z^2){\bf j}+ (x^2y + cyz + 2z){\bf k}$$

be conservative?

(b)

Using these values of the constants, find a potential function for the field, by either method described in the Notes. (Show systematic work.)

Solution:

(a)

${\bf F}$ will be conservative if ${\bf\nabla}\times{\bf F}=0.$ Since

$$\begin{multline*} {\bf\nabla}\times (P{\bf i}+Q{\bf j}+R {\bf k})=(R_y-Q_z){\bf ... ...k}\\ =(x^2+cz-bx^2-2z){\bf i}(axy-2xy){\bf j}+(2bxz-axz){\bf k} \end{multline*}$$

this vanishes (identically) exactly when $a=2,$ $b=1$ and $c=2.$

(b)

Method 1: Calculate the work integral of ${\bf F}$ along the curve which starts at the orgin, goes along the x-axis to $(x,0,0),$ then in the direction of $y$ to $(x,y,0)$ then in the z-direction to $(x,y,z).$

$$f(x,y,x)=\int_0^x0dx+\int_0^y0dy+\int_0^z(x^2y+2yz+2z)dz=x^2yz+yz^2+z^2.$$

Method 2: Solve $f_x=2xyz,$ giving $f=x^2yz+g(y,z).$ Then substitute in $f_y=x^2z+z^2$ giving $g_y=z^2$ so $g=z^2y+h(z)$ and $f=x^2yz+z^2y+h(z).$ Finally substitute this in $f_z=x^2y+2yz+2z$ giving $h'(z)=2z,$ so $h(z)=z^2+C$ and the general answer is $f=x^2yz+z^2y+z^2+C$ as before.

2.

(Thurs. 2 pts) Suppose that in $3$-space, ${\bf F}=\mathop{curl}{\bf G}$, where the components of ${\bf G}$ have continuous second partial derivatives. Prove that, if $S$ is a closed surface, then

$${\text{\LARGE O}}\kern-14pt\int\kern-9pt\int_S{\bf F}\cdot{\bf\hat n} \, dS=0$$

in two ways:

(a)

by the divergence theorem;

(b)

by drawing a closed curve $C$ dividing $S$ into two parts and applying Stokes' theorem to each.

Solution:

(a)

By the divergence theorem,

$${\text{\LARGE O}}\kern-14pt\int\kern-9pt\int_S{\bf F}\cdot{\bf\hat n} \, dS= \iiint_D \mathop{div}{\bf F}dV.$$

Since ${\bf F}=\mathop{curl}{\bf G},$

$$\begin{multline*} \mathop{div}{\bf F}=\mathop{div}(\mathop{curl}{\bf G}) =(R_y-Q... ...x)_y+(Q_x-P_y)_z\\ =R_{yx}-Q_{zx}+P_{zy}-R_{xy}+Q_{xz}-P_{yz}=0 \end{multline*}$$

by the equality of mixed second derivatives.

(b)

Choose a closed curve $C$ which divides $S$ into two parts (such as a little circular curve). Let $S^+$ be the half with outward normal which has the correct orientation for Stokes' theorem. Then the other half, $S^-$ with outward normal has the correct orientation for $C'$ which is $C$ run backwards. Applying Stokes' theorem twice

$$\oint_C {\bf G}\cdot d{\bf r}=\iint_{S^+}\mathop{curl}{\bf G}\cdot d{\bf S},$$

$$\oint_{C'} {\bf G}\cdot d{\bf r}=\iint_{S^-}\mathop{curl}{\bf G}\cdot d{\bf S}$$

with both $S^+$ and $S^-$ having the outward orientation of $S.$ Since

$$\oint_C{\bf G}\cdot d{\bf r}=-\oint_{C'}{\bf G}\cdot d{\bf r}$$

it follows that

$${\text{\LARGE O}}\kern-14pt\int\kern-9pt\int_S{\bf F}\cdot d{\bf ... ...hop{curl}{\bf G}\cdot d{\bf S}+\iint_{S^-}\mathop{curl}{\bf G}\cdot d{\bf S}=0.$$

3.

(Thurs, 3 pts) Which of the following differentials are exact? For each one which is, express it in the form $df$ for a suitable function $f(x,y,z)$.

(a)

$x^2 \, dx + y^2 \, dy + z^2 \, dz$

(b)

$y^2 z \, dx + 2xyz \, dy + xy^2 \, dz$

(c)

$y(6x^2 +z) \, dx + x(2x^2 +z) \, dy + xy \, dz$

Solution - in each case compute $P_y-Q_x,$ $P_z-R_x,$ $Q_z-R_y.$

(a)

Exact, it is $df$ for $f=\frac13(x^3+y^3+z^3).$

(b)

Exact, it is $df$ for $f=xy^2z.$

(c)

Exact, it is $df$ for $f=2x^3y+xyz.$

4.

(Thurs 1 pt) Find $\mathop{curl}{\bf F}$, if ${\bf F} = x^2 y {\bf i} + yz {\bf j} + xyz^2 {\bf k}$.

Solution: From the formula above,

$$\mathop{curl}{\bf F}=(xz^2-y){\bf i}-yz^2{\bf j}-x^2{\bf k}$$

5.

(Thurs, 2 pts) The fields ${\bf F}$ below are defined for all $x$, $y$, $z$. For each,

(a)

Show that curl ${\bf F} = \vec{0}$.

(b)

Find a potential function $f(x,y,z)$. Use either method, or inspection.

i.

$x {\bf i} + y{\bf j} + z {\bf k}$

ii.

$(2xy +z) {\bf i} + x^2 {\bf j} + x {\bf k}$

iii.

$yz {\bf i} + xz {\bf j} + xy {\bf k}$

Solution -

(a)

$P_y=0=Q_x,$ $P_z=0=R_x,$ $Q_z=0=R_y,$ so $\mathop{curl}{\bf F} = {\bf0}.$ ${\bf F}=\mathop{grad}f,$ $f=\frac12(x^2+y^2+z^2).$

(b)

$P_y=2x=Q_x,$ $P_z=1=R_x,$ $Q_z=0=R_y,$ so $\mathop{curl}{\bf F} = {\bf0}.$ ${\bf F}=\mathop{grad}f,$ $f=x^2y+xz.$

(c)

$P_y=z=Q_x,$ $P_z=y=R_x,$ $Q_z=x=R_y,$ so $\mathop{curl}{\bf F} = {\bf0}.$ ${\bf F}=\mathop{grad}f,$ $f=xyz.$