Vector Calculus

(a) Evaluate and draw. $\mathbf{F}(1,0)=⟨1,0⟩$, $\mathbf{F}(-1,0)=⟨-1,0⟩$, $\mathbf{F}(0,1)=⟨0,-1⟩$, $\mathbf{F}(0,-1)=⟨0,1⟩$, $\mathbf{F}(1,1)=⟨1,-1⟩$, $\mathbf{F}(1,-1)=⟨1,1⟩$, $\mathbf{F}(-1,1)=⟨-1,-1⟩$, $\mathbf{F}(-1,-1)=⟨-1,1⟩$.

Drawn at their base points: along the $x$-axis the arrows point away from the origin, along the $y$-axis they point toward it, and at the four diagonal points they point away from the $y$-axis and toward the $x$-axis. The picture is a saddle — fluid arriving along the $y$-axis is turned and sent out along the $x$-axis. Every arrow has length $|\mathbf{F}|=\sqrt{x^2+y^2}$, the distance from the origin, and $\mathbf{F}(0,0)=0$ is the one stagnation point.

(b) Find the potential. We need $f_x=x$ and $f_y=-y$. Integrating the first in $x$ gives $f=\frac{1}{2}{x}^2+g(y)$\; then $f_y=g^{\prime }(y)$ must equal $-y$, so $g(y)=-\frac{1}{2}{y}^2$ and

$$f(x,y)=\frac{1}{2}\left(x^2-y^2\right).$$

Check: $\nabla f=⟨x,-y⟩=\mathbf{F}$, so $\mathbf{F}$ is a gradient field and $f$ is a potential.

The level curve $f=0$ is $x^2=y^2$: the pair of lines $y=x$ and $y=-x$. The level curve $f=\frac{1}{2}$ is $x^2-y^2=1$: a hyperbola opening left and right, vertices $(\pm 1,0)$, asymptotes $y=\pm x$.

At $(1,0)$ the hyperbola has a vertical tangent and the arrow $⟨1,0⟩$ is horizontal: perpendicular. At $(1,1)$ the level curve is the line $y=x$, direction $⟨1,1⟩$, and $⟨1,-1⟩\cdot ⟨1,1⟩=0$: perpendicular again, as a gradient must be to the level curves of its potential.

(c) Tangency to the hyperbolas. The curve $xy=k$ is a level curve of $g(x,y)=xy$, so $\nabla g=⟨y,x⟩$ is normal to it at each of its points, and $\nabla g\ne 0$ there because $k\ne 0$ forces $x\ne 0$ and $y\ne 0$. Now

$$\mathbf{F}\cdot \nabla g=⟨x,-y⟩\cdot ⟨y,x⟩=xy-xy=0,$$

so $\mathbf{F}$ is perpendicular to the normal, that is, tangent to the hyperbola, at every one of its points. The branch of $xy=1$ through $(1,1)$ lies in the first quadrant with the two axes as asymptotes, and it crosses the line $y=x$ at right angles at $(1,1)$ — consistent with (b), since the arrow $⟨1,-1⟩$ there is tangent to one curve and normal to the other.

A particle at $(1,1)$ has velocity $⟨1,-1⟩$: right and down. Since the velocity is always tangent to $xy=1$, the particle slides along that branch\; its $x$-velocity is $x>0$, so $x$ keeps increasing, and its $y$-velocity is $-y<0$, so $y$ decreases toward $0$. It sweeps down the hyperbola toward the positive $x$-axis, never reaching it, and speeds up as $|\mathbf{F}|$ grows with distance from the origin. The whole flow is fluid arriving from above and below along the $y$-axis, slowing to a halt at the origin, and leaving left and right along the $x$-axis.

(a) Parametrise by $x$ itself. Take $\mathbf{r}(t)=⟨t,t^2⟩$, $0\le t\le 2$. Then ${\mathbf{r}}^{\prime }(t)=⟨1,2t⟩$ and $ds=|{\mathbf{r}}^{\prime }(t)|dt=\sqrt{1+4t^2}dt$, so

$${\int }_{C}xds={\int }_0^{2}t\sqrt{1+4t^2}dt.$$

Substitute $u=1+4t^2$, $du=8tdt$, so $tdt=\frac{1}{8}du$, with limits $u=1$ to $u=17$:

$$\frac{1}{8}{\int }_1^{17}{u}^{1/2}du=\frac{1}{8}\cdot \frac{2}{3}[u^{3/2}{]}_1^{17}=\frac{17^{3/2}-1}{12}=\frac{17\sqrt{17}-1}{12}.$$

Sanity check: $17\sqrt{17}\approx 70.1$, so the value is about $5.8$. The arc is a little under $4.7$ long and $x$ averages somewhat more than $1$ along it, so this is the right size.

(b) Same curve, different element. With the same parametrisation $dy=y^{\prime }(t)dt=2tdt$, so

$${\int }_{C}xdy={\int }_0^{2}t\cdot 2tdt=[\frac{2}{3}{t}^3{]}_0^2=\frac{16}{3}.$$

Along $-C$: ${\int }_{-C}xds=\frac{17\sqrt{17}-1}{12}$, unchanged, while ${\int }_{-C}xdy=-\frac{16}{3}$.

The reason sits in the two Riemann sums. The arc-length integral weights each sample value by $\Delta s_i$, the length of a subarc, which is positive whichever way the arc is walked, so reversing the direction changes nothing. The $dy$ integral weights by $\Delta y_i=y_i-y_{i-1}$, the change in $y$ across the subarc\; walking the arc backwards swaps the roles of $y_i$ and $y_{i-1}$ and negates every term, hence the sum and its limit.

(c) Mass and centre of mass. Parametrise $\mathbf{r}(t)=⟨2\cos t,2\sin t⟩$, $0\le t\le \pi$. Then $|{\mathbf{r}}^{\prime }(t)|=\sqrt{4{\sin }^{2}t+4{\cos }^{2}t}=2$, so $ds=2dt$ — not $dt$\; the speed is $2$. The mass is

$$m={\int }_{C}yds={\int }_0^{\pi }(2\sin t)2dt=4[-\cos t{]}_0^{\pi }=8.$$

By the symmetry of both the arc and the density under $x\mapsto -x$, $\bar{x}=0$\; explicitly, ${\int }_{C}xyds={\int }_0^{\pi }(2\cos t)(2\sin t)2dt=8[\frac{1}{2}{\sin }^{2}t{]}_0^{\pi }=0$. For $\bar{y}$,

$$\bar{y}=\frac{1}{m}{\int }_{C}y\cdot yds=\frac{1}{8}{\int }_0^{\pi }(2\sin t{)}^{2}2dt={\int }_0^{\pi }{\sin }^{2}tdt={\int }_0^{\pi }\frac{1-\cos 2t}{2}dt=\frac{\pi }{2}.$$

The centre of mass is $(0,\pi /2)$. It is reasonable: $\pi /2\approx 1.57$ lies inside the semicircle, off the wire, as the centre of mass of a curved wire does\; and it is higher than the $4/\pi \approx 1.27$ of a uniform semicircular wire of radius $2$, because the density $y$ puts more of the mass near the top of the arc.

(a) The segment. $\mathbf{r}(t)=⟨t,t⟩$, $0\le t\le 1$, so ${\mathbf{r}}^{\prime }(t)=⟨1,1⟩$ and on the curve $\mathbf{F}=⟨t^2,t-t⟩=⟨t^2,0⟩$. Then $\mathbf{F}\cdot {\mathbf{r}}^{\prime }=t^2$ and

$${\int }_{C_1}\mathbf{F}\cdot d\mathbf{r}={\int }_0^1{t}^{2}dt=\frac{1}{3}.$$

(b) The parabola. $\mathbf{r}(t)=⟨t,t^2⟩$, $0\le t\le 1$, so ${\mathbf{r}}^{\prime }(t)=⟨1,2t⟩$ and $\mathbf{F}=⟨t\cdot t^2,t-t^2⟩=⟨t^3,t-t^2⟩$. Then

$$\mathbf{F}\cdot {\mathbf{r}}^{\prime }=t^3+2t\left(t-t^2\right)=2t^2-t^3,{\int }_{C_2}\mathbf{F}\cdot d\mathbf{r}={\int }_0^1\left(2t^2-t^3\right)dt=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}.$$

(c) Two legs. On the first leg $\mathbf{r}(t)=⟨t,0⟩$, $0\le t\le 1$, so ${\mathbf{r}}^{\prime }=⟨1,0⟩$ and $\mathbf{F}=⟨0,t⟩$\; the dot product is $0$ and the leg contributes nothing — the field is vertical on the $x$-axis while the motion is horizontal. On the second leg $\mathbf{r}(t)=⟨1,t⟩$, $0\le t\le 1$, so ${\mathbf{r}}^{\prime }=⟨0,1⟩$ and $\mathbf{F}=⟨t,1-t⟩$\; the dot product is $1-t$ and

$${\int }_{C_3}\mathbf{F}\cdot d\mathbf{r}=0+{\int }_0^1(1-t)dt=\frac{1}{2}.$$

(d) Path dependence. Three routes between the same two points gave three different values, $\frac{1}{3}$, $\frac{5}{12}$ and $\frac{1}{2}$, so the line integral of $\mathbf{F}$ is not independent of path. By the Fundamental Theorem for line integrals a conservative field gives the same value on every path between two points, so $\mathbf{F}$ is not conservative: no function $f$ has $\nabla f=\mathbf{F}$. Equivalently, the closed loop out along $C_1$ and back along $-C_3$ has circulation $\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}\ne 0$.

The derivative check: with $P=xy$ and $Q=x-y$, $P_y=x$ while $Q_x=1$. For a conservative field with continuous partials the cross partials must agree everywhere (this is Clairaut's theorem applied to $f_{xy}=f_{yx}$), and here they disagree at every point off the line $x=1$. Either argument alone is conclusive.

(a) The three conditions. With $P=2xy+z^2$, $Q=x^2+2yz$, $R=y^2+2xz$:

$P_y=2x$ and $Q_x=2x$\; $P_z=2z$ and $R_x=2z$\; $Q_z=2y$ and $R_y=2y$.

All three pairs agree, which says exactly that $\mathrm{curl}\mathbf{F}=0$. The components are polynomials, so their partials are continuous, and $\mathbf{F}$ is defined on all of ${\mathbb{R}}^3$, which is simply connected. Under those hypotheses the curl test is a two-way test: zero curl on a simply connected region implies conservative. So $\mathbf{F}$ is conservative and a potential must exist.

(b) Build the potential one variable at a time. Integrate $P$ with respect to $x$, holding $y$ and $z$ fixed:

$$f(x,y,z)=x^{2}y+x{z}^2+g(y,z),$$

where the constant of integration may depend on both remaining variables. Differentiate in $y$: $f_y=x^2+g_y(y,z)$, and this must equal $Q=x^2+2yz$, so $g_y=2yz$ and $g(y,z)=y^{2}z+h(z)$. Now $f=x^{2}y+x{z}^2+y^{2}z+h(z)$\; differentiate in $z$: $f_z=2xz+y^2+h^{\prime }(z)$, which must equal $R=y^2+2xz$, so $h^{\prime }(z)=0$ and $h$ is a constant, which we take to be $0$. A potential is

$$f(x,y,z)=x^{2}y+x{z}^2+y^{2}z.$$

Check: $\nabla f=⟨2xy+z^2,x^2+2yz,2xz+y^2⟩=\mathbf{F}$. Note the pleasant cyclic symmetry $x\to y\to z\to x$ of $f$, which the field itself has.

(c) Endpoints only. By the Fundamental Theorem for line integrals, since $\mathbf{F}=\nabla f$ with $\nabla f$ continuous everywhere, the integral equals $f(\mathbf{r}(1))-f(\mathbf{r}(0))$, whatever the path does in between. The endpoints are $\mathbf{r}(0)=⟨\cos 0,0,\sin 0⟩=⟨1,0,0⟩$ and $\mathbf{r}(1)=⟨\cos \pi ,1\cdot e^0,\sin (\pi /2)⟩=⟨-1,1,1⟩$. Hence

$${\int }_C\mathbf{F}\cdot d\mathbf{r}=f(-1,1,1)-f(1,0,0)=[(1)(1)+(-1)(1)+(1)(1)]-[0+0+0]=1.$$

Parametrising directly would have meant integrating products of $\cos (\pi t)$, $t{e}^{t-1}$ and $\sin (\pi t/2)$ against their derivatives — possible, but pointless.

(d) A closed curve. The value is $0$: the circle is a closed path, so its initial and terminal points coincide and the Fundamental Theorem gives $f(\text{end})-f(\text{start})=0$ for the conservative field $\mathbf{F}$.

(a) Green's theorem on a disc. Neither $e^{x^2}$ nor $\sin (y^2)$ has an elementary antiderivative, so a direct parametrisation is hopeless. Take $P=e^{x^2}-y^3$ and $Q=x^3+\sin (y^2)$. Both have continuous partials on the whole plane, $C$ is a simple closed piecewise-smooth curve, and it is positively oriented, so Green's theorem applies with $D$ the disc $x^2+y^2\le 4$:

$$Q_x-P_y=3x^2-\left(-3y^2\right)=3\left(x^2+y^2\right).$$

The awkward terms vanished because each depends only on the variable it is not differentiated in. In polar coordinates,

$${\oint }_{C}Pdx+Qdy={\iint }_{D}3\left(x^2+y^2\right)dA={\int }_0^{2\pi }{\int }_0^{2}3r^2\cdot rdrd\theta =2\pi \cdot 3{\left[\frac{r^4}{4}\right]}_0^2=2\pi \cdot 12=24\pi .$$

Sanity check on the sign: $Q_x-P_y\ge 0$ throughout $D$, so a counterclockwise circuit must give a positive answer.

(b) Area from the boundary. Choosing $P=-\frac{1}{2}y$, $Q=\frac{1}{2}x$ gives $Q_x-P_y=1$, so Green's theorem reads $A=\frac{1}{2}{\oint }_{C}xdy-ydx$ for any positively oriented simple closed curve. Orientation: at $t=0$ the point is $(1,0)$, at $t=\pi /2$ it is $(0,1)$, and in between $x$ decreases while $y$ increases, so the curve moves from the positive $x$-axis to the positive $y$-axis through the first quadrant — counterclockwise. (The astroid has four cusps, but it is piecewise smooth, which is all the theorem needs.)

With $dx=-3{\cos }^{2}t\sin tdt$ and $dy=3{\sin }^{2}t\cos tdt$,

$$xdy-ydx=3{\cos }^{4}t{\sin }^{2}tdt+3{\sin }^{4}t{\cos }^{2}tdt=3{\sin }^{2}t{\cos }^{2}t\left({\cos }^{2}t+{\sin }^{2}t\right)dt=\frac{3}{4}{\sin }^{2}2tdt,$$

using $\sin t\cos t=\frac{1}{2}\sin 2t$. Over $[0,2\pi ]$ the function ${\sin }^{2}2t$ runs through two full periods and averages $\frac{1}{2}$, so ${\int }_0^{2\pi }{\sin }^{2}2tdt=\pi$ and

$$A=\frac{1}{2}\cdot \frac{3}{4}\cdot \pi =\frac{3\pi }{8}.$$

Check: on the astroid, with $a={\cos }^{2}t$ and $b={\sin }^{2}t$ (so $a+b=1$), we have $|x|+|y|=a^{3/2}+b^{3/2}\le a+b=1$, so the astroid lies inside the diamond $|x|+|y|\le 1$ of area $2$. And $3\pi /8\approx 1.18<2$, as it must be.

(c) A region with a hole. The positive orientation of a boundary keeps the region on the walker's left. On the outer circle $x^2+y^2=4$ that means counterclockwise\; on the inner circle $x^2+y^2=1$ it means clockwise, since the annulus lies outside that circle. With $D$ now the annulus, on which $P$ and $Q$ still have continuous partials, Green's theorem for a region with a hole gives

$${\oint }_{C}Pdx+Qdy={\iint }_{D}3\left(x^2+y^2\right)dA={\int }_0^{2\pi }{\int }_1^{2}3r^{3}drd\theta =2\pi \cdot 3{\left[\frac{r^4}{4}\right]}_1^2=2\pi \cdot \frac{45}{4}=\frac{45\pi }{2}.$$

Cross-check: the inner circle run counterclockwise gives, by the computation in (a) with radius $1$, the value $2\pi \cdot 3\cdot \frac{1}{4}=\frac{3\pi }{2}$\; run clockwise it gives $-\frac{3\pi }{2}$, and $24\pi -\frac{3\pi }{2}=\frac{45\pi }{2}$, in agreement.

(a) The divergence product rule. By definition $\mathrm{div}(f\mathbf{F})=\frac{\partial }{\partial x}(fP)+\frac{\partial }{\partial y}(fQ)+\frac{\partial }{\partial z}(fR)$. Apply the one-variable product rule to each term:

$$\mathrm{div}(f\mathbf{F})=\left(f_{x}P+f{P}_x\right)+\left(f_{y}Q+f{Q}_y\right)+\left(f_{z}R+f{R}_z\right).$$

Collect the three terms with $f$ undifferentiated and the three with $f$ differentiated:

$$=f\left(P_x+Q_y+R_z\right)+\left(f_{x}P+f_{y}Q+f_{z}R\right)=f\mathrm{div}\mathbf{F}+\nabla f\cdot \mathbf{F}.■$$

(b) The curl product rule. By definition the first component of $\mathrm{curl}(f\mathbf{F})$ is

$$\frac{\partial }{\partial y}(fR)-\frac{\partial }{\partial z}(fQ)=f_{y}R+f{R}_y-f_{z}Q-f{Q}_z=f\left(R_y-Q_z\right)+\left(f_{y}R-f_{z}Q\right).$$

The first bracket is $f$ times the first component of $\mathrm{curl}\mathbf{F}$. The second is the first component of $\nabla f\times \mathbf{F}=⟨f_x,f_y,f_z⟩\times ⟨P,Q,R⟩$, which by the cross-product formula is $f_{y}R-f_{z}Q$. So the first components agree.

The second component: $\frac{\partial }{\partial z}(fP)-\frac{\partial }{\partial x}(fR)=f_{z}P+f{P}_z-f_{x}R-f{R}_x=f\left(P_z-R_x\right)+\left(f_{z}P-f_{x}R\right)$, and the second component of $\nabla f\times \mathbf{F}$ is $f_{z}P-f_{x}R$. The third: $\frac{\partial }{\partial x}(fQ)-\frac{\partial }{\partial y}(fP)=f\left(Q_x-P_y\right)+\left(f_{x}Q-f_{y}P\right)$, and the third component of $\nabla f\times \mathbf{F}$ is $f_{x}Q-f_{y}P$. (Each line is the previous one with $x\to y\to z\to x$ and $P\to Q\to R\to P$, which is how both the curl and the cross product are built.) All three components agree, so

$$\mathrm{curl}(f\mathbf{F})=f\mathrm{curl}\mathbf{F}+\nabla f\times \mathbf{F}.■$$

(c) Two consequences. Apply (b) with $\mathbf{F}=\nabla g$, which has continuous partials because $g$ has continuous second partials:

$$\mathrm{curl}(f\nabla g)=f\mathrm{curl}(\nabla g)+\nabla f\times \nabla g=f0+\nabla f\times \nabla g=\nabla f\times \nabla g,$$

since the curl of a gradient is zero (the components of $\mathrm{curl}\nabla g$ are differences of mixed second partials of $g$, equal by Clairaut's theorem).

Apply (a) with $\mathbf{F}=\nabla f$:

$$\mathrm{div}(f\nabla f)=f\mathrm{div}(\nabla f)+\nabla f\cdot \nabla f=f{\nabla }^{2}f+|\nabla f{|}^2,$$

because $\mathrm{div}(\nabla f)=f_{xx}+f_{yy}+f_{zz}={\nabla }^{2}f$ is the Laplacian and $\nabla f\cdot \nabla f=|\nabla f{|}^2$. If $f$ is harmonic then ${\nabla }^{2}f=0$ and the first term drops out, leaving $\mathrm{div}(f\nabla f)=|\nabla f{|}^2$. $■$

(a) Step 3 is the illegal one. Steps 1 and 2 are correct wherever $P$ and $Q$ are defined. Green's theorem, however, requires $P$ and $Q$ to have continuous partial derivatives on an open region containing the whole of $D$, the disc $x^2+y^2\le 4$ — not merely on the curve. Both functions are undefined at the origin, which lies inside $D$, so the hypothesis fails and the theorem says nothing at all about this integral. The derivatives in Step 2 are valid only on the punctured plane.

A theorem whose hypotheses fail does not assert the opposite of its conclusion\; it is simply silent. The true value could be anything, and here it happens to be $0$. An argument that reaches a true statement by an invalid step has proved nothing, which is why the number must be recovered honestly.

(b) Two honest evaluations. Parametrise. Take $x=2\cos t$, $y=2\sin t$, $0\le t\le 2\pi$, so $dx=-2\sin tdt$, $dy=2\cos tdt$ and $x^2+y^2=4$ on $C$. Then

$$xdx+ydy=\left(2\cos t\right)\left(-2\sin t\right)dt+\left(2\sin t\right)\left(2\cos t\right)dt=0,$$

so the integrand vanishes identically along $C$ and the integral is $0$. (Geometrically: $⟨x,y⟩$ is radial and the circle's tangent is perpendicular to it.)

Find a potential. Let $f(x,y)=\frac{1}{2}\ln \left(x^2+y^2\right)$ on the punctured plane ${\mathbb{R}}^2\setminus \{(0,0)\}$. Then

$$f_x=\frac{1}{2}\cdot \frac{2x}{x^2+y^2}=\frac{x}{x^2+y^2}=P,f_y=\frac{y}{x^2+y^2}=Q,$$

so the field is $\nabla f$, with $\nabla f$ continuous on the punctured plane, which contains $C$. By the Fundamental Theorem for line integrals, ${\oint }_C\nabla f\cdot d\mathbf{r}=f(\text{end})-f(\text{start})=0$, since $C$ is closed. Note the moral: the field is conservative on the punctured plane, even though the cross-partial test cannot certify that on a region with a hole — the test being inconclusive is not the same as the field failing.

(c) The same steps, a different field. Now $P=\frac{-y}{x^2+y^2}$ and $Q=\frac{x}{x^2+y^2}$, and by the quotient rule

$$Q_x=\frac{\left(x^2+y^2\right)-2x^2}{{\left(x^2+y^2\right)}^2}=\frac{y^2-x^2}{{\left(x^2+y^2\right)}^2},P_y=\frac{-\left(x^2+y^2\right)+2y^2}{{\left(x^2+y^2\right)}^2}=\frac{y^2-x^2}{{\left(x^2+y^2\right)}^2},$$

so again $Q_x-P_y=0$ and the student's Step 3 would give $0$. The truth, by the same parametrisation:

$$-ydx+xdy=\left(-2\sin t\right)\left(-2\sin t\right)dt+\left(2\cos t\right)\left(2\cos t\right)dt=4dt,$$

and dividing by $x^2+y^2=4$ leaves an integrand of $1$, so the integral is ${\int }_0^{2\pi }1dt=2\pi \ne 0$.

Why the difference? For either field, Green's theorem does apply on the annulus between $C$ and a small circle $C_{\epsilon }$ about the origin, where the hypotheses hold and $Q_x-P_y=0$\; it shows ${\oint }_C={\oint }_{C_{\epsilon }}$. So the whole value of the integral is decided by what the field does arbitrarily close to its singular point. For the radial field that small-circle contribution is $0$\; for the rotational field it is $2\pi$, no matter how small $\epsilon$ is. The student's argument silently replaces this contribution by $0$ without ever computing it, which is harmless exactly when it happens to be $0$. Equivalently: the first field has a potential on the whole punctured plane, and the second cannot, because a gradient field integrates to zero around every closed curve and this one gives $2\pi$.

(a) Identify the surface. From the first two components, $x^2+y^2=u^2{\cos }^{2}v+u^2{\sin }^{2}v=u^2$, and the third gives $z=u$. So every point satisfies $x^2+y^2=z^2$ with $z=u\ge 0$: $S$ is the cone $z=\sqrt{x^2+y^2}$, apex at the origin, opening upward, cut off at height $z=2$ where its rim is the circle $x^2+y^2=4$, $z=2$. Its sides make an angle of $45^{\circ }$ with the axis.

Grid curves with $u=u_0$ fixed: $⟨u_0\cos v,u_0\sin v,u_0⟩$ is the horizontal circle of radius $u_0$ at height $u_0$ — the circles of latitude of the cone. Grid curves with $v=v_0$ fixed: $u⟨\cos v_0,\sin v_0,1⟩$ is a straight ray from the apex — the rulings.

The sketch shows the cone with apex at the origin and rim circle at height $2$\; the circle $u=1$ at height $1$ with radius $1$\; the ruling $v=0$, which is the segment from $(0,0,0)$ to $(2,0,2)$ in the $xz$-plane\; and the point $\mathbf{r}(1,\pi /2)=(0,1,1)$, on the circle $u=1$ directly above the positive $y$-axis.

(b) Normal vector and smoothness. ${\mathbf{r}}_u=⟨\cos v,\sin v,1⟩$ and ${\mathbf{r}}_v=⟨-u\sin v,u\cos v,0⟩$. Expanding the determinant,

$${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨\sin v\cdot 0-1\cdot u\cos v,1\cdot (-u\sin v)-\cos v\cdot 0,\cos v\cdot u\cos v-\sin v\cdot (-u\sin v)⟩=⟨-u\cos v,-u\sin v,u⟩.$$

This vanishes exactly when $u=0$, that is, at the single point $\mathbf{r}(0,v)=(0,0,0)$ for every $v$: the apex is the only non-smooth point, and indeed a cone has no tangent plane at its tip. Everywhere else the vector is $u⟨-\cos v,-\sin v,1⟩$: its horizontal part $⟨-\cos v,-\sin v⟩$ points from the base point toward the $z$-axis and its vertical part is upward, so the normal points into the solid cone above $S$ (up and toward the axis), not outward.

At $(u,v)=(1,\pi /2)$ the point is $(0,1,1)$ and the normal is $⟨0,-1,1⟩$. The tangent plane is $0\cdot (x-0)-1\cdot (y-1)+1\cdot (z-1)=0$, that is,

$$z=y.$$

Sanity check: this plane contains the whole ruling $v=\pi /2$, the points $(0,u,u)$, as the tangent plane to a cone along a ruling must\; on the sketch it is the plane through the $x$-axis tilted at $45^{\circ }$, touching the cone along that ruling.

(c) Area. $|{\mathbf{r}}_u\times {\mathbf{r}}_v|=\sqrt{u^2{\cos }^{2}v+u^2{\sin }^{2}v+u^2}=\sqrt{2u^2}=u\sqrt{2}$, positive for $u>0$. The parameter rectangle covers $S$ exactly once (each point has a unique $u=z$ and, off the apex, a unique angle $v$), so

$$A(S)={\int }_0^{2\pi }{\int }_0^{2}u\sqrt{2}dudv=\sqrt{2}\cdot 2\pi {\left[\frac{u^2}{2}\right]}_0^2=4\sqrt{2}\pi .$$

Check: the lateral area of a cone of base radius $r$ and slant height $\ell$ is $\pi r\ell$. Here $r=2$ and $\ell =\sqrt{2^2+2^2}=2\sqrt{2}$, giving $\pi \cdot 2\cdot 2\sqrt{2}=4\sqrt{2}\pi$. A second check: the shadow of $S$ is the disc of area $4\pi$, and a surface tilted at $45^{\circ }$ to the plane everywhere has $\sqrt{2}$ times the area of its shadow.

(a) A scalar surface integral. With $\mathbf{r}(u,v)=⟨u\cos v,u\sin v,u⟩$ on $0\le u\le 2$, $0\le v\le 2\pi$, one computes ${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨-u\cos v,-u\sin v,u⟩$, of length $u\sqrt{2}$, so $dS=u\sqrt{2}dudv$. On $S$ the integrand is $z=u$, hence

$${\iint }_{S}zdS={\int }_0^{2\pi }{\int }_0^{2}u\cdot u\sqrt{2}dudv=\sqrt{2}\cdot 2\pi {\left[\frac{u^3}{3}\right]}_0^2=\frac{16\sqrt{2}\pi }{3}.$$

The graph description gives the same thing: $g(x,y)=\sqrt{x^2+y^2}$ has $g_x=x/\sqrt{x^2+y^2}$ and $g_y=y/\sqrt{x^2+y^2}$, so $\sqrt{1+g_x^2+g_y^2}=\sqrt{1+1}=\sqrt{2}$ and the integral is $\sqrt{2}{\iint }_D\sqrt{x^2+y^2}dA=\sqrt{2}{\int }_0^{2\pi }{\int }_0^{2}r\cdot rdrd\theta$, identical after renaming.

Sanity check: the area of $S$ is $4\sqrt{2}\pi$ and the average height of the cone, weighted by area (which grows like $u$), is $\frac{{\int }_0^2{u}^{2}du}{{\int }_0^{2}udu}=\frac{8/3}{2}=\frac{4}{3}$\; and $4\sqrt{2}\pi \cdot \frac{4}{3}=\frac{16\sqrt{2}\pi }{3}$.

(b) Flux through the cone. The parametrisation's normal $⟨-u\cos v,-u\sin v,u⟩$ points toward the axis and upward, into the solid cone. The required normal is the opposite one,

$$\mathbf{N}=⟨u\cos v,u\sin v,-u⟩,$$

which points away from the axis (and downward, as the outward normal on the slanted side of the solid must). On $S$, $\mathbf{F}=⟨u\cos v,u\sin v,2u⟩$, so

$$\mathbf{F}\cdot \mathbf{N}=u^2{\cos }^{2}v+u^2{\sin }^{2}v-2u^2=-u^2,$$

and

$${\iint }_S\mathbf{F}\cdot d\mathbf{S}={\int }_0^{2\pi }{\int }_0^2\left(-u^2\right)dudv=-2\pi \cdot \frac{8}{3}=-\frac{16\pi }{3}.$$

The sign is right. The outward unit normal on the cone is $\frac{1}{\sqrt{2}}⟨x/z,y/z,-1⟩$, and $\mathbf{F}\cdot \mathbf{n}=\frac{1}{\sqrt{2}}\left(\frac{x^2+y^2}{z}-2z\right)=\frac{1}{\sqrt{2}}(z-2z)=-\frac{z}{\sqrt{2}}<0$ on $S$: the radial part $⟨x,y,0⟩$ of the field pushes outward through the sloping side, but the vertical part $2z\mathbf{k}$ is twice as strong and pushes upward, which through a normal tilted downward counts as inward. Net flow is into the solid through its side. (Via the graph formula: the upward normal is $⟨-g_x,-g_y,1⟩$, the required one is its negative $⟨g_x,g_y,-1⟩$, and $\mathbf{F}\cdot ⟨g_x,g_y,-1⟩=\frac{x^2+y^2}{\sqrt{x^2+y^2}}-2\sqrt{x^2+y^2}=-r$, whose polar integral $-{\int }_0^{2\pi }{\int }_0^2{r}^{2}drd\theta$ is again $-\frac{16\pi }{3}$.)

(c) The lid, and the closed surface. On the disc $z=2$ the upward unit normal is $\mathbf{k}$ and $dS=dA$, so $\mathbf{F}\cdot \mathbf{k}=2z=4$ is constant there and the flux is $4\cdot \text{area}=4\cdot 4\pi =16\pi$. Upward is outward for the solid cone at its top, so the total outward flux through the closed surface is

$$16\pi +\left(-\frac{16\pi }{3}\right)=\frac{32\pi }{3}.$$

Sanity check by the Divergence theorem, which applies because the closed surface bounds the solid cone and $\mathbf{F}$ is smooth everywhere: $\mathrm{div}\mathbf{F}=1+1+2=4$ and the cone has volume $\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \cdot 4\cdot 2=\frac{8\pi }{3}$, so the outward flux should be $4\cdot \frac{8\pi }{3}=\frac{32\pi }{3}$. It is.

(a) Curl of $\mathbf{a}\times \mathbf{r}$. By the cross-product formula,

$$\mathbf{a}\times \mathbf{r}=⟨a_{2}z-a_{3}y,a_{3}x-a_{1}z,a_{1}y-a_{2}x⟩=⟨P,Q,R⟩.$$

Every component is linear, so the partials are constants:

$$R_y-Q_z=a_1-(-a_1)=2a_1,P_z-R_x=a_2-(-a_2)=2a_2,Q_x-P_y=a_3-(-a_3)=2a_3.$$

Hence $\mathrm{curl}(\mathbf{a}\times \mathbf{r})=⟨2a_1,2a_2,2a_3⟩=2\mathbf{a}$. $■$ (For $\mathbf{a}=\mathbf{k}$ this is the familiar fact that the rotation field $⟨-y,x,0⟩$ has curl $2\mathbf{k}$.)

(b) Area as a line integral. Put $\mathbf{F}=\mathbf{n}\times \mathbf{r}$, with $\mathbf{n}$ constant. Its components are linear polynomials, so they have continuous partials on all of ${\mathbb{R}}^3$\; $S$ is a smooth oriented surface and $C$ its positively oriented boundary. Stokes' theorem therefore applies:

$${\oint }_C(\mathbf{n}\times \mathbf{r})\cdot d\mathbf{r}={\iint }_S\mathrm{curl}(\mathbf{n}\times \mathbf{r})\cdot \mathbf{n}dS={\iint }_{S}2\mathbf{n}\cdot \mathbf{n}dS={\iint }_{S}2dS=2A,$$

using (a) with $\mathbf{a}=\mathbf{n}$, then $\mathbf{n}\cdot \mathbf{n}=|\mathbf{n}{|}^2=1$, then ${\iint }_{S}1dS=A$. Dividing by $2$ gives the formula. $■$

For $\mathbf{n}=\mathbf{k}$: $\mathbf{k}\times ⟨x,y,z⟩=⟨0\cdot z-1\cdot y,1\cdot x-0\cdot z,0⟩=⟨-y,x,0⟩$, so

$$\frac{1}{2}{\oint }_C(\mathbf{k}\times \mathbf{r})\cdot d\mathbf{r}=\frac{1}{2}{\oint }_C-ydx+xdy,$$

which is the plane formula\; and positive orientation with respect to $\mathbf{k}$ is counterclockwise seen from above, exactly the orientation Green's theorem requires.

(c) A tilted ellipse. The plane is $-2y+z=0$, with unit normal $\mathbf{n}=\frac{1}{\sqrt{5}}⟨0,-2,1⟩$ (either sign would do\; this one has positive $\mathbf{k}$-component, so the positive boundary orientation is counterclockwise seen from above). The boundary curve is the intersection of the plane with the cylinder, parametrised by

$$\mathbf{r}(t)=⟨\cos t,\sin t,2\sin t⟩,0\le t\le 2\pi ,$$

which runs counterclockwise seen from above, as required. Then ${\mathbf{r}}^{\prime }(t)=⟨-\sin t,\cos t,2\cos t⟩$ and, computing the cross product with $⟨0,-2,1⟩$ first,

$$⟨0,-2,1⟩\times ⟨\cos t,\sin t,2\sin t⟩=⟨(-2)(2\sin t)-(1)(\sin t),(1)(\cos t)-0,0-(-2)(\cos t)⟩=⟨-5\sin t,\cos t,2\cos t⟩.$$

Dotting with ${\mathbf{r}}^{\prime }(t)$: $5{\sin }^{2}t+{\cos }^{2}t+4{\cos }^{2}t=5{\sin }^{2}t+5{\cos }^{2}t=5$, a constant. Therefore