14. 스토크스 정리와 발산 정리Stokes' Theorem and the Divergence Theorem

Solution. The curl is defined by

Since $F_1=x^2-z,\ F_2=xe^z,\ F_3=xy$, computing each component gives

Therefore

Proof. We first show that for any function $f(x,y,z)$ of class $C^2$, $\nabla\times(\nabla f)=\mathbf 0$. Since $\nabla f=(f_x,f_y,f_z)$,

Since $f$ is of class $C^2$, Clairaut's theorem (equality of mixed partial derivatives) gives $f_{zy}=f_{yz},\ f_{xz}=f_{zx},\ f_{yx}=f_{xy}$. Hence $\nabla\times(\nabla f)=\mathbf 0$.

(1) $\mathbf F=2x\mathbf i+2y\mathbf j+2z\mathbf k=\nabla(x^2+y^2+z^2)$, and $x^2+y^2+z^2$ is a polynomial, hence of class $C^2$. So by the identity above, $\nabla\times\mathbf F=\mathbf 0$.

(2) $\mathbf F=\nabla(xy^2z^3)$, and $xy^2z^3$ is a polynomial, hence of class $C^2$. So by the identity above, $\nabla\times\mathbf F=\mathbf 0$.

Proof. The curl of $\mathbf F=y\mathbf i-x\mathbf j$ is

(1) Orient $S$ with $\mathbf n$ pointing outward from the origin (with a positive $\mathbf k$-component); by the right-hand rule, this induces a counterclockwise orientation on $C: x^2+y^2=9,\ z=0$ as viewed from above.

First compute the right side. Since $\nabla\times\mathbf F$ is a constant vector, replacing $S$ by the flat disk $S':x^2+y^2\leq 9,\ z=0$ (with orientation $\mathbf n=\mathbf k$), which has the same boundary $C$, gives the same value of the surface integral by Stokes' Theorem, so

Next compute the left side. Parametrizing $C$ by $\mathbf r(t)=(3\cos t,3\sin t,0),\ 0\leq t\leq 2\pi$, we get $d\mathbf r=(-3\sin t,3\cos t,0)\,dt$ and

so

Both sides equal $-18\pi$, confirming that Stokes' Theorem holds.

(2) Let $S$ be the disk of radius $3$ in the $xy$-plane; this is the same surface as $S'$ in (1). Hence

which agrees exactly with the value found in (1). That is, whether $S$ is taken to be the hemisphere or the flat disk, the surface integral of the curl over $S$ is the same as long as the boundary $C$ is the same — this illustrates that Stokes' Theorem depends only on the boundary, not on the choice of surface.

Proof. The curl of $\mathbf F=y\mathbf i-x\mathbf j+x^2\mathbf k$ is

Right side (surface integral). Parametrize $S$ by $z=g(x,y)=y^2-x^2\ (x^2+y^2\leq 1)$, oriented with a positive $\mathbf k$-component; then

so

and since $4xy$ is an odd function on $D:x^2+y^2\leq 1$, symmetry gives $\iint_D 4xy\,dA=0$. Hence

Left side (line integral). The boundary $C$ lies on the cylinder $x^2+y^2=1$, oriented counterclockwise as viewed from above, corresponding to the orientation of $S$. Parametrizing by $\mathbf r(t)=(\cos t,\sin t,\sin^2t-\cos^2t)=(\cos t,\sin t,-\cos 2t),\ 0\leq t\leq 2\pi$,

so

Since $2\cos^2t\sin 2t=4\sin t\cos^3t$ has antiderivative $-\cos^4t$, which is periodic, its definite integral over one full period is $0$. Therefore

Both sides equal $-2\pi$, confirming that Stokes' Theorem holds.

Solution. The curve of intersection of $z=2$ and $z=\sqrt{x^2+y^2}$ is the circle $\sqrt{x^2+y^2}=2$, i.e. $x^2+y^2=4,\ z=2$. Taking the flat disk $S:x^2+y^2\leq 4,\ z=2$ bounded by this circle, oriented with $\mathbf n=\mathbf k$ (upward), matches the counterclockwise orientation of $C$ as viewed from above.

By Stokes' Theorem,

The curl of $\mathbf F=(x^2-y,4z,x^2)$ is

so $(\nabla\times\mathbf F)\cdot\mathbf k=1$ is constant. Therefore

Solution. To find the curve of intersection of $x^2+y^2+z^2=1$ and $z=\sqrt{x^2+y^2}$, substitute $x^2+y^2=z^2$ into the sphere's equation: $z^2+z^2=1$, so $z=\dfrac{1}{\sqrt2}$ (taking the positive root). Then $x^2+y^2=z^2=\dfrac12$, so $C$ is a circle of radius $\dfrac1{\sqrt2}$ at height $z=\dfrac1{\sqrt2}$. Taking the flat disk $S:x^2+y^2\leq \dfrac12,\ z=\dfrac1{\sqrt2}$ bounded by this circle, oriented with $\mathbf n=\mathbf k$, matches the orientation of $C$.

The curl of $\mathbf F=(x^2+z,y^2+2x,z^2-y)$ is

so $(\nabla\times\mathbf F)\cdot\mathbf k=2$ is constant. By Stokes' Theorem,

Solution. Since $C$ is a closed curve, Stokes' Theorem applies. Parametrize the plane $2x+y+z=2\ (x,y,z\geq 0)$ by $z=g(x,y)=2-2x-y$, with domain the triangle $D:x\geq0,\ y\geq0,\ 2x+y\leq2$ in the $xy$-plane. Since $C$ runs counterclockwise as viewed from above, $S$ is oriented upward, and

The curl of $\mathbf F=(xz,xy,3xz)$ is

so

Substituting $z=2-2x-y$ gives $x-3(2-2x-y)+y=7x+4y-6$, so

Evaluating the inner integral,

and then the outer integral,

Therefore $\displaystyle\int_C\mathbf F\cdot d\mathbf r=-1$.

Solution. The boundary $C$ of $S$ is the ellipse $x^2+4y^2=1$ on the plane $z=1$; since $S$ is oriented with a positive $\mathbf k$-component, $C$ runs counterclockwise as viewed from above. The curl of $\mathbf F=(y,-xz,xz^2)$ is

Since $\nabla\times\mathbf F$ is polynomial and hence smooth everywhere, Stokes' Theorem lets us replace $S$ by the flat elliptical disk $S':x^2+4y^2\leq1,\ z=1$ (with orientation $\mathbf n=\mathbf k$), which has the same boundary $C$, without changing the value of the flux. On $S'$, $z=1$, so

is constant. The ellipse $x^2+4y^2\leq1$ has semi-axes $1$ and $\dfrac12$, so its area is $\pi\cdot1\cdot\dfrac12=\dfrac{\pi}2$. Therefore

Solution. Orient $S$ with $\mathbf n$ pointing outward from the origin. The curl of $\mathbf F=y\mathbf i$ is

The boundary of $S$ is $C:x^2+y^2=1,\ z=0$, and by the right-hand rule the outward orientation of the upper hemisphere induces a counterclockwise orientation on $C$ as viewed from above. Since $\nabla\times(y\mathbf i)$ is a constant vector, Stokes' Theorem lets us replace $S$ by the flat disk $S':x^2+y^2\leq1,\ z=0$ (with orientation $\mathbf n=\mathbf k$), which has the same boundary, without changing the value. Hence

(Check: since $\mathbf F\cdot d\mathbf r=y\,dx$, parametrizing $\mathbf r(t)=(\cos t,\sin t,0)$ gives $\displaystyle\oint_C y\,dx=\int_0^{2\pi}\sin t\cdot(-\sin t)\,dt=-\pi$, in agreement.)

Solution. We use the definition $\nabla\cdot\mathbf F=\dfrac{\partial F_1}{\partial x}+\dfrac{\partial F_2}{\partial y}+\dfrac{\partial F_3}{\partial z}$.

(1) $\mathbf F=x\mathbf i+y\mathbf j+z\mathbf k$: $\nabla\cdot\mathbf F=1+1+1=3$.

(2) $\mathbf F=-x\mathbf i-y\mathbf j-z\mathbf k$: $\nabla\cdot\mathbf F=-1-1-1=-3$.

(3) $\mathbf F=-y\mathbf i+x\mathbf j$: $\nabla\cdot\mathbf F=\dfrac{\partial(-y)}{\partial x}+\dfrac{\partial x}{\partial y}+0=0+0=0$.

(4) $\mathbf F=z\mathbf j$: $\nabla\cdot\mathbf F=0+\dfrac{\partial z}{\partial y}+0=0$.

Proof. (1) Let $\mathbf F=P\mathbf i+Q\mathbf j+R\mathbf k$. For the divergence product rule, since $g\mathbf F=(gP)\mathbf i+(gQ)\mathbf j+(gR)\mathbf k$,

and applying the product rule to each term,

For the curl product rule, the $\mathbf i$-component of $g\mathbf F$ is

which equals $g(\nabla\times\mathbf F)_{\mathbf i}+(\nabla g\times\mathbf F)_{\mathbf i}$. The $\mathbf j$- and $\mathbf k$-components follow the same way under the cyclic substitution $(x,P)\to(y,Q)\to(z,R)\to(x,P)$, so

(2) Let $\mathbf F_1=P_1\mathbf i+Q_1\mathbf j+R_1\mathbf k,\ \mathbf F_2=P_2\mathbf i+Q_2\mathbf j+R_2\mathbf k$. By linearity of partial differentiation,

Each component of the curl distributes in the same way by linearity of partial differentiation, so

Proof. The divergence of $\mathbf F=x\mathbf i+y\mathbf j+z\mathbf k$ is $\nabla\cdot\mathbf F=1+1+1=3$.

First compute the right side. Since $D$ is a ball of radius $a$, its volume is $\dfrac43\pi a^3$, so

Next compute the left side. At a point $(x,y,z)$ on the sphere $S$, the outward orientation is $\mathbf n=\dfrac{(x,y,z)}a$, so

is constant on $S$. Since the surface area of $S$ is $4\pi a^2$,

Both sides equal $4\pi a^3$, confirming that the Divergence Theorem holds.

Proof. The divergence of $\mathbf F=x^2\mathbf i+4xyz\mathbf j+ze^x\mathbf k$ is

First compute the right side.

Integrating over $x$ gives $\displaystyle\int_0^3 2x\,dx=9,\ \int_0^3 4xz\,dx=18z,\ \int_0^3 e^x\,dx=e^3-1$. Integrating over $y$ multiplies by $2$, since the integrand does not depend on $y$. Thus

and since $\displaystyle\int_0^1 18\,dz=18,\ \int_0^1 36z\,dz=18,\ \int_0^1(2e^3-2)\,dz=2e^3-2$,

Next compute the left side as the sum of the outward flux through the six faces. On the faces $x=0,\ y=0,\ z=0$ we get $\mathbf F\cdot\mathbf n=-F_1=-x^2=0,\ -F_2=-4xyz=0,\ -F_3=-ze^x=0$, respectively, so these contribute nothing. On the remaining three faces:

• $x=3$ ($\mathbf n=\mathbf i$): $\mathbf F\cdot\mathbf n=x^2=9$ (constant), area $2\times1=2$, so the flux is $9\cdot2=18$.
• $y=2$ ($\mathbf n=\mathbf j$): $\mathbf F\cdot\mathbf n=4xyz=8xz$, so the flux is $\displaystyle\int_0^1\int_0^3 8xz\,dx\,dz=8\cdot\frac92\cdot\frac12=18$.
• $z=1$ ($\mathbf n=\mathbf k$): $\mathbf F\cdot\mathbf n=ze^x\big|_{z=1}=e^x$, so the flux is $\displaystyle\int_0^2\int_0^3 e^x\,dx\,dy=2(e^3-1)=2e^3-2$.

Therefore

Both sides equal $34+2e^3$, confirming that the Divergence Theorem holds.

Solution. The divergence of $\mathbf F=xy\mathbf i+yz\mathbf j+zx\mathbf k$ is

Applying the Divergence Theorem to the cube $D=[0,1]^3$,

and by symmetry $\iiint_Dx\,dV=\iiint_Dy\,dV=\iiint_Dz\,dV$, so

Therefore the flux is $\dfrac32$.

Solution. (1) Let $\rho=\sqrt{x^2+y^2+z^2}$, so $\mathbf F=\dfrac{x\mathbf i+y\mathbf j+z\mathbf k}{\rho^3}$. Differentiating the $x$-component,

and the $y$- and $z$-components are computed the same way, so

On $D$ we have $\rho\geq b\gt0$, so $\mathbf F$ is smooth throughout $D$, and by the Divergence Theorem,

(2) Now find the flux across the sphere $S:\rho=a$ alone. Since the interior of $S$ contains the origin, where $\mathbf F$ is undefined, the Divergence Theorem cannot be applied directly, and the flux must be computed directly. On $S$, the outward orientation is $\mathbf n=\dfrac{(x,y,z)}a$, so

is constant on $S$. Since the surface area of $S$ is $4\pi a^2$,

(Remark: this value is $4\pi$ regardless of $a$, so the outward flux through the spheres of radii $a$ and $b$ are each $4\pi$; with the orientation induced by $D$ — outward on the outer sphere, toward the origin on the inner sphere — this is exactly consistent with the result $4\pi-4\pi=0$ from part (1).)

Solution. Since $\nabla f$ is smooth throughout $D$ (as $\rho\geq b\gt0$), the Divergence Theorem gives

Since $f=\ln(x^2+y^2+z^2)$, $\dfrac{\partial f}{\partial x}=\dfrac{2x}{\rho^2}$ (where $\rho^2=x^2+y^2+z^2$), so

and the $y$- and $z$-terms have the same form, so

In spherical coordinates $(\rho,\phi,\theta)$, $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$, so $\nabla^2f\,dV=2\sin\phi\,d\rho\,d\phi\,d\theta$. Since $D$ has $\rho:b\to a$, and $z\geq0$ corresponds to $\phi:0\to\dfrac{\pi}2$ while $x,y\geq0$ (the first octant) corresponds to $\theta:0\to\dfrac{\pi}2$,