Exercises — Surface integrals — scalar and vector (flux)

WHAT we do: differentiate the map in each parameter direction. ${\mathbf{r}}_u=(1,0,2),{\mathbf{r}}_v=(0,1,3).$ WHY: nudging $u$ moves us along $(1,0,2)$; nudging $v$ moves us along $(0,1,3)$. These are the two edges of the tiny parallelogram — the yellow and pink arrows in the figure above. Cross product (determinant of $\hat{\mathbf{i}},\hat{\mathbf{j}},\hat{\mathbf{k}}$ over the two rows): ${\mathbf{r}}_u\times {\mathbf{r}}_v=(0\cdot 3-2\cdot 1,2\cdot 0-1\cdot 3,1\cdot 1-0\cdot 0)=(-2,-3,1).$ Its length: $\Vert {\mathbf{r}}_u\times {\mathbf{r}}_v\Vert =\sqrt{(-2{)}^2+(-3{)}^2+1^2}=\sqrt{14}.$ So $dS=\sqrt{14}dudv$. Since the plane is flat, this stretch is the same everywhere — a constant.

The $z$-component is $+1>0$, so the normal points upward. To orient the plane downward instead, use ${\mathbf{r}}_v\times {\mathbf{r}}_u=(2,3,-1)$, which flips every sign.

Set $f=1$ in the scalar engine. From L1.1, $dS=\sqrt{14}dudv$. $\text{Area}={\iint }_{D}1\cdot \sqrt{14}dudv=\sqrt{14}\cdot (\text{area of unit square})=\sqrt{14}\approx \mathrm{3.742.}$ Sanity: the flat base has area $1$; the plane is tilted, so the true patch is bigger by exactly the stretch $\sqrt{14}$. ✔ (This is just a Double integrals of a constant.)

WHAT: use the graph formula $dS=\sqrt{1+g_x^2+g_y^2}dxdy$. Here $g=x+y\Rightarrow g_x=1,g_y=1$, so $\sqrt{1+1^2+1^2}=\sqrt{3}.$ On the surface $z=g=x+y$, so ${\iint }_{S}zdS={\iint }_D(x+y)\sqrt{3}dxdy=\sqrt{3}{\iint }_D(x+y)dxdy.$ The base integral over the triangle $x+y\le 1$: ${\iint }_D(x+y)dxdy={\int }_0^1{\int }_0^{1-x}(x+y)dydx.$ Inner: ${\int }_0^{1-x}(x+y)dy=x(1-x)+\frac{1}{2}(1-x{)}^2$. Expand: $x-x^2+\frac{1}{2}(1-2x+x^2)=\frac{1}{2}-\frac{1}{2}{x}^2$. Outer: ${\int }_0^1(\frac{1}{2}-\frac{1}{2}{x}^2)dx=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}.$ $\boxed{{\iint }_{S}zdS=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\approx \mathrm{0.5774.}}$

Use the flux-through-a-graph shortcut ${\iint }_D(-P{g}_x-Q{g}_y+R)dxdy$ with upward normal $(-g_x,-g_y,1)$. (Recall $\mathbf{F}=(P,Q,R)$; dotting $(P,Q,R)$ into $(-g_x,-g_y,1)$ gives exactly $-P{g}_x-Q{g}_y+R$.) Here $P=0,Q=0,R=z=g=1-x-y$, and $g_x=g_y=-1$ (but $P=Q=0$ so those terms vanish). $\text{Flux}={\iint }_D(1-x-y)dxdy.$ By symmetry with L2.2's method: ${\int }_0^1{\int }_0^{1-x}(1-x-y)dydx$. Inner $=\frac{1}{2}(1-x{)}^2$. Outer $={\int }_0^1\frac{1}{2}(1-x{)}^{2}dx=\frac{1}{6}$. $\boxed{\text{Flux}=\frac{1}{6}\approx \mathrm{0.1667.}}$

Downward normal is $(g_x,g_y,-1)$ — every sign flipped. So the flux flips sign: ${\text{Flux}}_{\text{down}}=-\frac{1}{6}\approx -\mathrm{0.1667.}$ Why: flux measures net flow through; reverse which side you call "out" and the same flow now counts as entering, i.e. negative. Scalar surface integrals would be unchanged (length has no sign).

Tangents: ${\mathbf{r}}_{\theta }=(-\sin \theta ,\cos \theta ,0),{\mathbf{r}}_z=(0,0,1).$ Cross product (this is our vector area element): ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(\cos \theta \cdot 1-0,0-(-\sin \theta )\cdot 1,0)=(\cos \theta ,\sin \theta ,0).$ Check orientation: at $\theta =0$ this is $(1,0,0)$, pointing straight away from the axis — outward. ✔ Dot with the field ($\mathbf{F}=(x,y,0)=(\cos \theta ,\sin \theta ,0)$ on the wall): $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)={\cos }^2\theta +{\sin }^2\theta =1.$ $\text{Flux}={\int }_0^2{\int }_0^{2\pi }1d\theta dz=2\pi \cdot 2=4\pi \approx \mathrm{12.566.}$ Meaning: the field pushes straight out through every point of the wall with strength $1$; total $=$ (wall area) $=2\pi rh=4\pi$. ✔

Direct — three pieces:

• Side wall: from L3.2, flux $=4\pi$.
• Top disk ($z=2$, outward normal $=(0,0,1)$): $\mathbf{F}\cdot (0,0,1)=0$ since $\mathbf{F}$ has no $z$-component. Flux $=0$.
• Bottom disk ($z=0$, outward normal $=(0,0,-1)$): $\mathbf{F}\cdot (0,0,-1)=0$. Flux $=0$. Total $=4\pi +0+0=4\pi$.

Via Divergence Theorem: ${\iint }_{\partial V}\mathbf{F}\cdot d\mathbf{S}={\iiint }_V(\nabla \cdot \mathbf{F})dV$. $\nabla \cdot \mathbf{F}=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial 0}{\partial z}=1+1+0=2.$ Volume of cylinder $=\pi r^{2}h=\pi \cdot 1\cdot 2=2\pi$. ${\iiint }_{V}2dV=2\cdot 2\pi =4\pi .$ Both give $4\pi$. ✔ The caps contribute nothing because the flow is purely horizontal.

Graph formula: $g=\sqrt{x^2+y^2}$, so $g_x=\frac{x}{\sqrt{x^2+y^2}},g_y=\frac{y}{\sqrt{x^2+y^2}},$ $1+g_x^2+g_y^2=1+\frac{x^2+y^2}{x^2+y^2}=1+1=2(\text{away from the tip}).$ So $dS=\sqrt{2}dxdy$ and on the surface $z=\sqrt{x^2+y^2}$: $\text{Mass}={\iint }_D\sqrt{x^2+y^2}\cdot \sqrt{2}dxdy.$ Switch to polar (Jacobian and change of variables: $dxdy=\rho d\rho d\theta$, and $\sqrt{x^2+y^2}=\rho$): $=\sqrt{2}{\int }_0^{2\pi }{\int }_0^1\rho \cdot \rho d\rho d\theta =\sqrt{2}\cdot 2\pi \cdot \frac{1}{3}=\frac{2\sqrt{2}\pi }{3}\approx \mathrm{2.962.}$ Note the tip: $g_x,g_y$ blow up at the origin, but the integrand $\rho \cdot \rho$ vanishes there fast enough — the integral is finite. ✔

Route chosen — sphere parametrization (so the outward normal is clean): $\mathbf{r}(\varphi ,\theta )=(\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi ),\varphi \in [0,\frac{\pi }{2}],\theta \in [0,2\pi ].$ Why ${\mathbf{r}}_{\varphi }\times {\mathbf{r}}_{\theta }=\sin \varphi \mathbf{r}$ here — the local justification: differentiate the map, ${\mathbf{r}}_{\varphi }=(\cos \varphi \cos \theta ,\cos \varphi \sin \theta ,-\sin \varphi ),{\mathbf{r}}_{\theta }=(-\sin \varphi \sin \theta ,\sin \varphi \cos \theta ,0).$ Taking the cross product component by component:

• $x$-comp: $\cos \varphi \sin \theta \cdot 0-(-\sin \varphi )\cdot \sin \varphi \cos \theta ={\sin }^2\varphi \cos \theta$,
• $y$-comp: $(-\sin \varphi )(-\sin \varphi \sin \theta )-\cos \varphi \cos \theta \cdot 0={\sin }^2\varphi \sin \theta$,
• $z$-comp: $\cos \varphi \cos \theta \cdot \sin \varphi \cos \theta -\cos \varphi \sin \theta \cdot (-\sin \varphi \sin \theta )=\sin \varphi \cos \varphi$.

So ${\mathbf{r}}_{\varphi }\times {\mathbf{r}}_{\theta }=\sin \varphi (\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi )=\sin \varphi \mathbf{r}$. It points along $\mathbf{r}$ (radially outward) with length $\sin \varphi$ — exactly the spherical area element. On the surface $z=\cos \varphi$, so $\mathbf{F}=(0,0,\cos \varphi )$, and $\mathbf{r}=(\dots ,\dots ,\cos \varphi )$: $\mathbf{F}\cdot (\sin \varphi \mathbf{r})=\sin \varphi (0\cdot r_x+0\cdot r_y+\cos \varphi \cdot \cos \varphi )=\sin \varphi {\cos }^2\varphi .$ $\text{Flux}={\int }_0^{2\pi }{\int }_0^{\pi /2}\sin \varphi {\cos }^2\varphi d\varphi d\theta .$ Inner: let $c=\cos \varphi$, $dc=-\sin \varphi d\varphi$; ${\int }_0^{\pi /2}\sin \varphi {\cos }^2\varphi d\varphi ={\int }_0^1{c}^{2}dc=\frac{1}{3}$. $\text{Flux}=2\pi \cdot \frac{1}{3}=\frac{2\pi }{3}\approx \mathrm{2.094.}$ Confirm via a closed surface + Divergence Theorem: close the hemisphere with the flat disk $z=0$ (outward normal there is $(0,0,-1)$, and $\mathbf{F}=(0,0,0)$ on $z=0$, so that cap contributes $0$). Then over the solid half-ball $\nabla \cdot \mathbf{F}=1$, volume $=\frac{1}{2}\cdot \frac{4}{3}\pi =\frac{2\pi }{3}$. Total closed flux $=\frac{2\pi }{3}$; subtract the $0$ cap → hemisphere flux $=\frac{2\pi }{3}$. ✔

By Stokes' Theorem, ${\iint }_S(\mathrm{curl}\mathbf{F})\cdot d\mathbf{S}={\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}$, which depends only on the boundary curve, not on the surface. So every $S$ with that same boundary gives the same number. Compute the boundary line integral (Line integrals) on $\mathbf{r}(t)=(\cos t,\sin t,0)$, $t\in [0,2\pi ]$, ${\mathbf{r}}^{\prime }(t)=(-\sin t,\cos t,0)$: $\mathbf{F}=(-\sin t,\cos t,0),\mathbf{F}\cdot {\mathbf{r}}^{\prime }={\sin }^{2}t+{\cos }^{2}t=1.$ ${\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}={\int }_0^{2\pi }1dt=2\pi .$ Direct check of the curl (for the flat disk): $\mathrm{curl}\mathbf{F}=(0,0,{\partial }_{x}x-{\partial }_y(-y))=(0,0,2)$, and through the flat disk (normal $(0,0,1)$, area $\pi$): $2\cdot \pi =2\pi$. ✔ $\boxed{\text{Flux of }\mathrm{curl}\mathbf{F}=2\pi \text{for any such }S.}$

Stretch factor of $z=2u+3v$? ::: $\sqrt{14}$ (constant, plane). Flux of $(0,0,z)$ up through $z=1-x-y$ over the triangle? ::: $\frac{1}{6}$. Same flux, oriented downward? ::: $-\frac{1}{6}$. Outward flux of $(x,y,0)$ through the unit cylinder wall, $0\le z\le 2$? ::: $4\pi$. Why do the caps add nothing there? ::: $\mathbf{F}$ has no $z$-component, so $\mathbf{F}\cdot (0,0,\pm 1)=0$. Mass of the cone $z=\sqrt{x^2+y^2}$, $z\le 1$, density $z$? ::: $\frac{2\sqrt{2}\pi }{3}$. Flux of $(0,0,z)$ out through the upper unit hemisphere? ::: $\frac{2\pi }{3}$. Flux of $\mathrm{curl}(-y,x,0)$ through any cap on the unit circle? ::: $2\pi$.