13. 면적분Surface Integrals

Solution. Use cylindrical coordinates. Setting $x=r\cos\theta$, $y=r\sin\theta$ $(r\geq0,\ 0\leq\theta\leq2\pi)$, a point on the cone satisfies $z=\sqrt{x^2+y^2}=\sqrt{r^2}=r$, so $0\leq z\leq1$ corresponds to $0\leq r\leq1$. Hence the parametrization is

$$\mathbf{r}(r,\theta)=(r\cos\theta)\mathbf{i}+(r\sin\theta)\mathbf{j}+r\mathbf{k}\qquad(0\leq r\leq1,\ 0\leq\theta\leq2\pi).$$

Solution.

(1) Use spherical coordinates. Let $\phi$ $(0\leq\phi\leq\pi)$ be the polar angle measured from the center and $\theta$ $(0\leq\theta\leq2\pi)$ the azimuthal angle. Then the sphere of radius $a$ is parametrized by

$$\mathbf{r}(\phi,\theta)=(a\sin\phi\cos\theta)\mathbf{i}+(a\sin\phi\sin\theta)\mathbf{j}+(a\cos\phi)\mathbf{k}\qquad(0\leq\phi\leq\pi,\ 0\leq\theta\leq2\pi).$$

(2) The surface is the cylinder obtained by extending, in the $z$-direction over $0\leq z\leq5$, the circle in the $xy$-plane with center $(0,3)$ and radius $3$. Setting $x=3\cos\theta$, $y=3+3\sin\theta$ gives

$$\mathbf{r}(\theta,z)=(3\cos\theta)\mathbf{i}+(3+3\sin\theta)\mathbf{j}+z\mathbf{k}\qquad(0\leq\theta\leq2\pi,\ 0\leq z\leq5).$$

Solution. Solving the plane's equation for $y$ gives $y=x+2z-2$. Since the surface lies inside the cylinder $x^2+z^2=3$, write $x,z$ in polar form as $x=r\cos\theta$, $z=r\sin\theta$ $(0\leq r\leq\sqrt3,\ 0\leq\theta\leq2\pi)$, so that $x^2+z^2=r^2\leq3$ matches the "inside the cylinder" condition. Substituting into $y=x+2z-2$ gives the parametrization

$$\mathbf{r}(r,\theta)=(r\cos\theta)\mathbf{i}+(r\cos\theta+2r\sin\theta-2)\mathbf{j}+(r\sin\theta)\mathbf{k}\qquad(0\leq r\leq\sqrt3,\ 0\leq\theta\leq2\pi).$$

Indeed $x-y+2z=r\cos\theta-(r\cos\theta+2r\sin\theta-2)+2r\sin\theta=2$, so this surface lies on the given plane.

Solution. Revolving the point $(f(u),g(u),0)$ on $C$ about the $x$-axis fixes the $x$-coordinate at $f(u)$ while tracing a circle of radius $g(u)$ in the plane perpendicular to the $x$-axis. Letting $v$ $(0\leq v\leq2\pi)$ be the angle of revolution, a point on the surface of revolution is parametrized by

$$\mathbf{r}(u,v)=f(u)\mathbf{i}+\big(g(u)\cos v\big)\mathbf{j}+\big(g(u)\sin v\big)\mathbf{k}$$

with parameter range $a\leq u\leq b$, $0\leq v\leq2\pi$. (Since $g(u) \gt 0$, a circle of radius $g(u)$ is indeed obtained for each $u$.)

Solution.

(1) Parametrize by scaling each component of spherical coordinates by the corresponding semi-axis. For $\phi$ $(0\leq\phi\leq\pi)$, $\theta$ $(0\leq\theta\leq2\pi)$, set

$$\mathbf{r}(\phi,\theta)=(a\sin\phi\cos\theta)\mathbf{i}+(b\sin\phi\sin\theta)\mathbf{j}+(c\cos\phi)\mathbf{k}.$$

Then $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\sin^2\phi\cos^2\theta+\sin^2\phi\sin^2\theta+\cos^2\phi=\sin^2\phi+\cos^2\phi=1$.

(2) Use the hyperbolic identity $\cosh^2u-\sinh^2u=1$. For $u\in\mathbb{R}$, $\theta$ $(0\leq\theta\leq2\pi)$, set

$$\mathbf{r}(u,\theta)=(a\cosh u\cos\theta)\mathbf{i}+(b\cosh u\sin\theta)\mathbf{j}+(c\sinh u)\mathbf{k}.$$

Then $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=\cosh^2u\cos^2\theta+\cosh^2u\sin^2\theta-\sinh^2u=\cosh^2u-\sinh^2u=1$.

Solution. The surface is $z=x^2+y^2$, and $z\leq4$ corresponds to the projected region $R:x^2+y^2\leq4$ (a disk of radius $2$).

(1) Area formula for a parametrized surface. Parametrize by $\mathbf{r}(x,y)=x\mathbf{i}+y\mathbf{j}+(x^2+y^2)\mathbf{k}$. Then

$$\mathbf{r}_x=\mathbf{i}+2x\mathbf{k},\qquad \mathbf{r}_y=\mathbf{j}+2y\mathbf{k},\qquad \mathbf{r}_x\times\mathbf{r}_y=-2x\mathbf{i}-2y\mathbf{j}+\mathbf{k},$$

so $d\sigma=|\mathbf{r}_x\times\mathbf{r}_y|\,dx\,dy=\sqrt{4x^2+4y^2+1}\,dx\,dy$. Hence

$$A=\iint_R\sqrt{4x^2+4y^2+1}\,dA=\int_0^{2\pi}\int_0^2\sqrt{4r^2+1}\,r\,dr\,d\theta.$$

Substituting $u=4r^2+1$, $du=8r\,dr$, with $r:0\to2$ giving $u:1\to17$,

$$A=2\pi\int_1^{17}\frac{\sqrt u}{8}\,du=2\pi\left[\frac{u^{3/2}}{12}\right]_1^{17}=\frac{\pi(17\sqrt{17}-1)}{6}.$$

(2) Area formula for an implicitly defined surface. Let $F(x,y,z)=x^2+y^2-z$, so $\nabla F=(2x,2y,-1)$. Projecting onto the $xy$-plane with $\mathbf{p}=\mathbf{k}$, $d\sigma=\dfrac{|\nabla F|}{|\nabla F\cdot\mathbf{p}|}\,dA$. Since $|\nabla F|=\sqrt{4x^2+4y^2+1}$ and $|\nabla F\cdot\mathbf{k}|=1$, we again get $d\sigma=\sqrt{4x^2+4y^2+1}\,dA$, the same integrand as in (1). Hence $A=\dfrac{\pi(17\sqrt{17}-1)}{6}$ by the same computation.

(3) Area formula for an explicitly defined surface. With $z=g(x,y)=x^2+y^2$, $g_x=2x$, $g_y=2y$, and $d\sigma=\sqrt{1+g_x^2+g_y^2}\,dA=\sqrt{1+4x^2+4y^2}\,dA$, again the same integrand as in (1) and (2). Hence $A=\dfrac{\pi(17\sqrt{17}-1)}{6}$.

All three methods reduce to the same integrand $\sqrt{4x^2+4y^2+1}$, so they agree on the value $A=\dfrac{\pi(17\sqrt{17}-1)}{6}$.

Solution.

(1) Use the same parametrization as in Example 01, $\mathbf{r}(r,\theta)=(r\cos\theta,r\sin\theta,r)$ $(0\leq r\leq4,\ 0\leq\theta\leq2\pi)$.

$$\mathbf{r}_r=(\cos\theta,\sin\theta,1),\quad \mathbf{r}_\theta=(-r\sin\theta,r\cos\theta,0),\quad \mathbf{r}_r\times\mathbf{r}_\theta=(-r\cos\theta,-r\sin\theta,r),$$

so $d\sigma=|\mathbf{r}_r\times\mathbf{r}_\theta|\,dr\,d\theta=r\sqrt2\,dr\,d\theta$. Hence

$$A=\int_0^{2\pi}\int_0^4 r\sqrt2\,dr\,d\theta=\sqrt2\cdot2\pi\cdot\left[\frac{r^2}{2}\right]_0^4=16\sqrt2\,\pi.$$

(2) The area of a sphere of radius $a$ is the well-known $4\pi a^2$.

(3) Revolving $x=\cos z$, $y=0$ about the $z$-axis, with parameter $u=z$ $(-\pi/2\leq u\leq\pi/2)$, gives a surface of revolution with radius $\cos u\ (\geq0)$. The arc-length element of the generating curve is $ds=\sqrt{(dx/du)^2+1}\,du=\sqrt{\sin^2u+1}\,du$, so the area of the surface of revolution is

$$A=\int_{-\pi/2}^{\pi/2}2\pi(\cos u)\sqrt{1+\sin^2u}\,du.$$

Substituting $s=\sin u$, $ds=\cos u\,du$, with $u:-\pi/2\to\pi/2$ giving $s:-1\to1$,

$$A=2\pi\int_{-1}^1\sqrt{1+s^2}\,ds=2\pi\left[\frac{s}{2}\sqrt{1+s^2}+\frac12\ln\!\left(s+\sqrt{1+s^2}\right)\right]_{-1}^1=2\pi\big(\sqrt2+\ln(1+\sqrt2)\big),$$

where the last equality uses that the integrand is even.

Solution. The circle in the $xz$-plane is parametrized by $x=R+r\cos\phi$, $z=r\sin\phi$ $(0\leq\phi\leq2\pi)$. Revolving it about the $z$-axis by an angle $\theta$ $(0\leq\theta\leq2\pi)$ gives the torus

$$\mathbf{r}(\phi,\theta)=\big((R+r\cos\phi)\cos\theta\big)\mathbf{i}+\big((R+r\cos\phi)\sin\theta\big)\mathbf{j}+(r\sin\phi)\mathbf{k}.$$

Then

$$\mathbf{r}_\phi=(-r\sin\phi\cos\theta,\,-r\sin\phi\sin\theta,\,r\cos\phi),\qquad \mathbf{r}_\theta=\big(-(R+r\cos\phi)\sin\theta,\,(R+r\cos\phi)\cos\theta,\,0\big),$$

and a direct computation gives $\mathbf{r}_\phi\times\mathbf{r}_\theta=-r(R+r\cos\phi)\big(\cos\phi\cos\theta,\cos\phi\sin\theta,\sin\phi\big)$, so

$$d\sigma=|\mathbf{r}_\phi\times\mathbf{r}_\theta|\,d\phi\,d\theta=r(R+r\cos\phi)\,d\phi\,d\theta$$

(here $R+r\cos\phi \gt 0$ since $0 \lt r \lt R$). Hence the area is

$$A=\int_0^{2\pi}\int_0^{2\pi} r(R+r\cos\phi)\,d\phi\,d\theta=2\pi\cdot r\left[R\phi+r\sin\phi\right]_0^{2\pi}=2\pi\cdot r\cdot2\pi R=4\pi^2Rr.$$

Solution. The sphere has radius $\rho=\sqrt2$. Using spherical coordinates $z=\rho\cos\phi$, $\sqrt{x^2+y^2}=\rho\sin\phi$ $(0\leq\phi\leq\pi)$, the condition that the surface lies above the cone $z=\sqrt{x^2+y^2}$ becomes $\rho\cos\phi\geq\rho\sin\phi$, i.e., $\phi\leq\pi/4$. So the required area is that of the spherical cap on $\rho=\sqrt2$ with $0\leq\phi\leq\pi/4$.

For the sphere $\mathbf{r}(\phi,\theta)=(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi)$, we have $d\sigma=\rho^2\sin\phi\,d\phi\,d\theta$, so

$$A=\int_0^{2\pi}\int_0^{\pi/4}\rho^2\sin\phi\,d\phi\,d\theta=2\pi\rho^2\big[-\cos\phi\big]_0^{\pi/4}=2\pi\rho^2\left(1-\frac{\sqrt2}{2}\right).$$

Substituting $\rho^2=2$,

$$A=2\pi\cdot2\left(1-\frac{\sqrt2}{2}\right)=4\pi-2\sqrt2\,\pi=2\pi(2-\sqrt2).$$

Solution. Use the parametrization of Example 01, $\mathbf{r}(r,\theta)=(r\cos\theta,r\sin\theta,r)$ $(0\leq r\leq1,\ 0\leq\theta\leq2\pi)$. As computed in 유제 02-1(1), $d\sigma=r\sqrt2\,dr\,d\theta$. Since $G=x^2=r^2\cos^2\theta$,

$$\iint_S G\,d\sigma=\int_0^{2\pi}\int_0^1 (r^2\cos^2\theta)(r\sqrt2)\,dr\,d\theta=\sqrt2\int_0^{2\pi}\cos^2\theta\,d\theta\int_0^1r^3\,dr.$$

Since $\displaystyle\int_0^{2\pi}\cos^2\theta\,d\theta=\pi$ and $\displaystyle\int_0^1r^3\,dr=\frac14$,

$$\iint_S G\,d\sigma=\sqrt2\cdot\pi\cdot\frac14=\frac{\sqrt2\,\pi}{4}.$$

Solution. The cube is $[0,1]\times[0,1]\times[0,1]$, and its boundary consists of the three coordinate-plane faces ($x=0$, $y=0$, $z=0$) together with the three opposite faces ($x=1$, $y=1$, $z=1$). Since $G=xyz$, $G=0$ on each coordinate-plane face, so those contribute nothing. On each of the remaining three faces, $d\sigma$ is the area element in the two free variables.

On $x=1$ ($0\leq y\leq1,\ 0\leq z\leq1$): $G=yz$, so $\displaystyle\iint yz\,dy\,dz=\left(\int_0^1y\,dy\right)\left(\int_0^1z\,dz\right)=\frac12\cdot\frac12=\frac14$.

On $y=1$ ($0\leq x\leq1,\ 0\leq z\leq1$): $G=xz$, and likewise $\displaystyle\iint xz\,dx\,dz=\frac14$.

On $z=1$ ($0\leq x\leq1,\ 0\leq y\leq1$): $G=xy$, and likewise $\displaystyle\iint xy\,dx\,dy=\frac14$.

Hence

$$\iint_S G\,d\sigma=\frac14+\frac14+\frac14=\frac34.$$

Solution. Use the parametrization from 유제 02-1(3), $\mathbf{r}(u,\theta)=(\cos u\cos\theta,\ \cos u\sin\theta,\ u)$ $(-\pi/2\leq u\leq\pi/2,\ 0\leq\theta\leq2\pi)$. Then

$$\mathbf{r}_u=(-\sin u\cos\theta,-\sin u\sin\theta,1),\qquad \mathbf{r}_\theta=(-\cos u\sin\theta,\cos u\cos\theta,0),$$

and a direct computation gives $\mathbf{r}_u\times\mathbf{r}_\theta=(-\cos u\cos\theta,-\cos u\sin\theta,-\sin u\cos u)$, so

$$|\mathbf{r}_u\times\mathbf{r}_\theta|=\sqrt{\cos^2u+\sin^2u\cos^2u}=\cos u\sqrt{1+\sin^2u}$$

(the last equality uses $\cos u\geq0$ for $-\pi/2\leq u\leq\pi/2$). Hence $d\sigma=\cos u\sqrt{1+\sin^2u}\,du\,d\theta$.

On the surface, $x^2+y^2=\cos^2u$, so $G=\sqrt{1-x^2-y^2}=\sqrt{1-\cos^2u}=|\sin u|$. Thus

$$\iint_S G\,d\sigma=\int_0^{2\pi}\int_{-\pi/2}^{\pi/2}|\sin u|\cos u\sqrt{1+\sin^2u}\,du\,d\theta.$$

Since the integrand $|\sin u|\cos u\sqrt{1+\sin^2u}$ is even in $u$,

$$\int_{-\pi/2}^{\pi/2}|\sin u|\cos u\sqrt{1+\sin^2u}\,du=2\int_0^{\pi/2}\sin u\cos u\sqrt{1+\sin^2u}\,du.$$

Substituting $w=\sin u$, $dw=\cos u\,du$, with $u:0\to\pi/2$ giving $w:0\to1$,

$$2\int_0^1 w\sqrt{1+w^2}\,dw=2\left[\frac{(1+w^2)^{3/2}}{3}\right]_0^1=\frac23\big(2\sqrt2-1\big).$$

Hence

$$\iint_S G\,d\sigma=2\pi\cdot\frac23(2\sqrt2-1)=\frac{4\pi(2\sqrt2-1)}{3}.$$

Solution. $S$ is the graph of the explicit function $z=g(x,y)=y^2/2$. Since $g_x=0$, $g_y=y$,

$$d\sigma=\sqrt{1+g_x^2+g_y^2}\,dA=\sqrt{1+y^2}\,dA.$$

On the surface, $z=y^2/2$, so the integrand is

$$\sqrt{x(1+2z)}=\sqrt{x(1+y^2)}=\sqrt{x}\,\sqrt{1+y^2},$$

giving

$$\sqrt{x(1+2z)}\,d\sigma=\sqrt{x}\,\sqrt{1+y^2}\cdot\sqrt{1+y^2}\,dA=\sqrt{x}\,(1+y^2)\,dA.$$

Hence

$$\iint_S\sqrt{x(1+2z)}\,d\sigma=\iint_R\sqrt{x}\,(1+y^2)\,dx\,dy=\int_0^1(1+y^2)\left[\int_0^{1-y}\sqrt x\,dx\right]dy=\int_0^1(1+y^2)\cdot\frac23(1-y)^{3/2}\,dy.$$

Substituting $t=1-y$, $y=1-t$, $dy=-dt$, with $y:0\to1$ giving $t:1\to0$,

$$\frac23\int_0^1(1+(1-t)^2)\,t^{3/2}\,dt=\frac23\int_0^1\big(2t^{3/2}-2t^{5/2}+t^{7/2}\big)\,dt.$$

Integrating each term,

$$\int_0^1 2t^{3/2}\,dt=\frac45,\qquad \int_0^1 2t^{5/2}\,dt=\frac47,\qquad \int_0^1 t^{7/2}\,dt=\frac29,$$

so

$$\frac23\left(\frac45-\frac47+\frac29\right)=\frac23\cdot\frac{252-180+70}{315}=\frac23\cdot\frac{142}{315}=\frac{284}{945}.$$

Hence $\displaystyle\iint_S\sqrt{x(1+2z)}\,d\sigma=\frac{284}{945}$.

Solution. Parametrize $S$ by $x,z$: $\mathbf{r}(x,z)=x\mathbf{i}+x^2\mathbf{j}+z\mathbf{k}$ $(0\leq x\leq1,\ 0\leq z\leq4)$. Then

$$\mathbf{r}_x=(1,2x,0),\qquad \mathbf{r}_z=(0,0,1),\qquad \mathbf{r}_x\times\mathbf{r}_z=(2x,-1,0).$$

The $\mathbf{i}$-component of this normal is $2x\geq0$ for $0\leq x\leq1$, so it satisfies $\mathbf{n}\cdot\mathbf{i}\geq0$. Hence

$$\mathbf{n}\,d\sigma=(2x,-1,0)\,dx\,dz.$$

On the surface $y=x^2$, so $\mathbf{F}=yz\mathbf{i}+x\mathbf{j}-z^2\mathbf{k}=x^2z\mathbf{i}+x\mathbf{j}-z^2\mathbf{k}$, and

$$\mathbf{F}\cdot\mathbf{n}\,d\sigma=(x^2z)(2x)+(x)(-1)+(-z^2)(0)=2x^3z-x.$$

Hence the flux is

$$\iint_S\mathbf{F}\cdot\mathbf{n}\,d\sigma=\int_0^1\int_0^4(2x^3z-x)\,dz\,dx=\int_0^1\Big[x^3z^2-xz\Big]_0^4dx=\int_0^1(16x^3-4x)\,dx=\Big[4x^4-2x^2\Big]_0^1=2.$$

Solution. Parametrize $S$ by $\mathbf{r}(x,\theta)=x\mathbf{i}+(\cos\theta)\mathbf{j}+(\sin\theta)\mathbf{k}$ $(0\leq x\leq1,\ 0\leq\theta\leq\pi)$ (with $y=\cos\theta$, $z=\sin\theta\geq0$ requiring $0\leq\theta\leq\pi$). Then

$$\mathbf{r}_x=(1,0,0),\qquad \mathbf{r}_\theta=(0,-\sin\theta,\cos\theta),\qquad \mathbf{r}_\theta\times\mathbf{r}_x=(0,\cos\theta,\sin\theta).$$

The $\mathbf{k}$-component of this normal is $\sin\theta\geq0$ for $0\leq\theta\leq\pi$, so the orientation satisfying $\mathbf{n}\cdot\mathbf{k}\geq0$ is $\mathbf{n}\,d\sigma=(0,\cos\theta,\sin\theta)\,dx\,d\theta$.

On the surface, $y=\cos\theta$, $z=\sin\theta$, so $\mathbf{F}=yz\mathbf{i}+z^2\mathbf{k}=(\cos\theta\sin\theta,\,0,\,\sin^2\theta)$, and

$$\mathbf{F}\cdot\mathbf{n}\,d\sigma=(\cos\theta\sin\theta)(0)+0\cdot\cos\theta+\sin^2\theta\cdot\sin\theta=\sin^3\theta.$$

Hence the flux is

$$\iint_S\mathbf{F}\cdot\mathbf{n}\,d\sigma=\int_0^1\int_0^\pi\sin^3\theta\,d\theta\,dx=\int_0^\pi\sin^3\theta\,d\theta.$$

Since $\displaystyle\int_0^\pi\sin^3\theta\,d\theta=\int_0^\pi\sin\theta(1-\cos^2\theta)\,d\theta=\left[-\cos\theta+\frac{\cos^3\theta}{3}\right]_0^\pi=\left(1-\frac13\right)-\left(-1+\frac13\right)=\frac43$,

$$\iint_S\mathbf{F}\cdot\mathbf{n}\,d\sigma=\frac43.$$

Check. Viewing $S$ as the graph $z=\sqrt{1-y^2}$ (independent of $x$), $\mathbf{n}\,d\sigma=(-g_x,-g_y,1)\,dx\,dy=\left(0,\dfrac{y}{\sqrt{1-y^2}},1\right)dx\,dy$ (already satisfies $\mathbf{n}\cdot\mathbf{k}\geq0$ since the $\mathbf{k}$-component is $1>0$), and $\mathbf{F}\cdot\mathbf{n}\,d\sigma=z^2=1-y^2$, so

$$\int_0^1\int_{-1}^1(1-y^2)\,dy\,dx=\left[y-\frac{y^3}3\right]_{-1}^1=\frac43,$$

the same value.

Solution. View $S$ as the graph of $z=x^2+y^2$ ($x^2+y^2\leq1$) and parametrize it by $x,y$: $\mathbf{r}(x,y)=x\mathbf{i}+y\mathbf{j}+(x^2+y^2)\mathbf{k}$, with $(x,y)$ ranging over the disk $x^2+y^2\leq1$. Then

$$\mathbf{r}_x=(1,0,2x),\qquad \mathbf{r}_y=(0,1,2y),\qquad \mathbf{r}_x\times\mathbf{r}_y=(-2x,-2y,1).$$

The horizontal part $(-2x,-2y)$ of this normal points toward the origin (i.e., toward the $z$-axis), opposite to the orientation pointing away from the $z$-axis. So we must take the opposite normal,

$$\mathbf{n}\,d\sigma=(2x,2y,-1)\,dx\,dy,$$

whose horizontal part points radially outward from the axis, as required. On the surface $\mathbf{F}=4x\mathbf{i}+4y\mathbf{j}+2\mathbf{k}$, so

$$\mathbf{F}\cdot\mathbf{n}\,d\sigma=(4x)(2x)+(4y)(2y)+(2)(-1)=8x^2+8y^2-2.$$

Using polar coordinates $x=r\cos\theta$, $y=r\sin\theta$, $dx\,dy=r\,dr\,d\theta$, the flux is

$$\iint_S\mathbf{F}\cdot\mathbf{n}\,d\sigma=\int_0^{2\pi}\int_0^1(8r^2-2)r\,dr\,d\theta=2\pi\Big[2r^4-r^2\Big]_0^1=2\pi(2-1)=2\pi.$$

Check. Verify via the Divergence Theorem. $\operatorname{div}\mathbf{F}=4+4+0=8$. Capping $S$ with the disk $D=\{x^2+y^2\leq1,\ z=1\}$ oriented by $\mathbf{k}$, $S\cup D$ bounds the solid $V=\{(x,y,z): x^2+y^2\leq z\leq1\}$, and the outward orientation of $S$ for this solid agrees with $\mathbf{n}\,d\sigma=(2x,2y,-1)\,dx\,dy$ found above. By the Divergence Theorem,

$$\iint_{S\cup D}\mathbf{F}\cdot\mathbf{n}\,d\sigma=\iiint_V 8\,dV=8\int_0^{2\pi}\int_0^1(1-r^2)r\,dr\,d\theta=8\cdot2\pi\left[\frac{r^2}2-\frac{r^4}4\right]_0^1=8\cdot2\pi\cdot\frac14=4\pi.$$

Since $\mathbf{F}\cdot\mathbf{k}=2$ on $D$, $\displaystyle\iint_D\mathbf{F}\cdot\mathbf{n}\,d\sigma=2\cdot\pi(1)^2=2\pi$, so

$$\iint_S\mathbf{F}\cdot\mathbf{n}\,d\sigma=4\pi-2\pi=2\pi,$$

the same value.