§16.7 면적분Surface Integrals

Solution. Parametrize the unit sphere spherically,

As already computed in §16.6, this parametrization satisfies $\lvert\mathbf r_\phi\times\mathbf r_\theta\rvert=\sin\phi$. On $S$, $x=\sin\phi\cos\theta$, so $x^{2}=\sin^{2}\phi\cos^{2}\theta$, and Equation (2) gives

The half-angle identity $\cos^{2}\theta=\tfrac12(1+\cos2\theta)$ gives the first factor as $\pi$. For the second, write $\sin^{3}\phi=\sin\phi\,(1-\cos^{2}\phi)$ and substitute $w=\cos\phi$:

Multiplying the two factors, $\displaystyle\iint_S x^{2}\,dS=\pi\cdot\frac43=\frac{4\pi}{3}$.

Solution. The surface is the graph $z=g(x,y)=x+y^{2}$ over $D=[0,1]\times[0,2]$. Since $g_x=1$ and $g_y=2y$, the area factor of Equation (4) is

independent of $x$. Hence

Substituting $u=2+4y^{2}$, $du=8y\,dy$, so that $y=0,2$ correspond to $u=2,18$,

Since $18^{3/2}=18\cdot3\sqrt2=54\sqrt2$ and $2^{3/2}=2\sqrt2$, the numerator is $52\sqrt2$, so $\displaystyle\iint_S y\,dS=\frac{52\sqrt2}{12}=\frac{13\sqrt2}{3}$.

Solution. By the definition of an integral over a piecewise-smooth surface, evaluate the three integrals over $S_1$, $S_2$, $S_3$ separately and add them.

The lateral piece $S_1$ is parametrized by $\mathbf r(\theta,z)=\langle\cos\theta,\sin\theta,z\rangle$, $0\le\theta\le2\pi$, $0\le z\le1+\cos\theta$, and $\lvert\mathbf r_\theta\times\mathbf r_z\rvert=1$. Hence

Expanding the integrand as $1+2\cos\theta+\cos^{2}\theta$ gives $\int_0^{2\pi}1\,d\theta=2\pi$, $\int_0^{2\pi}2\cos\theta\,d\theta=0$, and $\int_0^{2\pi}\cos^{2}\theta\,d\theta=\pi$, so the sum is $3\pi$ and $\iint_{S_1} z\,dS=\dfrac{3\pi}{2}$.

$S_2$ lies in the plane $z=0$, so the integrand $z$ vanishes identically on it, and $\iint_{S_2} z\,dS=0$.

$S_3$ is the graph $z=1+x$ over the unit disk $D:x^{2}+y^{2}\le1$, and since $g_x=1,g_y=0$ the area factor of Equation (4) is the constant $\sqrt2$. Therefore

Here $\iint_D 1\,dA=\pi$, and $x$ is an odd function on the disk about the origin, so $\iint_D x\,dA=0$. Thus $\iint_{S_3} z\,dS=\sqrt2\,\pi$.

Adding the three values, $\displaystyle\iint_S z\,dS=\frac{3\pi}{2}+0+\sqrt2\,\pi=\left(\frac32+\sqrt2\right)\pi$.

Solution. Use the same spherical parametrization as in Example 1, $\mathbf r(\phi,\theta)=\langle\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi\rangle$. As was checked earlier when the sphere was oriented, the vector $\mathbf r_\phi\times\mathbf r_\theta$ arising from this ordering of the parameters is a positive multiple of the position vector, hence outward, so no sign correction is needed to match the orientation.

and on $S$, $\mathbf F=\langle z,y,x\rangle=\langle\cos\phi,\ \sin\phi\sin\theta,\ \sin\phi\cos\theta\rangle$, so computing the dot product component by component gives

By Equation (9),

The first term vanishes entirely because $\int_0^{2\pi}\cos\theta\,d\theta=0$. In the second, the $\theta$-integral is $\int_0^{2\pi}\sin^{2}\theta\,d\theta=\pi$ and the $\phi$-integral is the value $\int_0^{\pi}\sin^{3}\phi\,d\phi=\dfrac43$ already found in Example 1, so

Solution. View $S=S_1\cup S_2$ as a piecewise-smooth closed surface with the positive (outward) orientation. The upper piece $S_1:z=1-x^{2}-y^{2}$ is the top of the solid, so its outward orientation is the upward one; the lower piece $S_2:z=0$ is the bottom, so its outward orientation is the downward one, $\mathbf n=-\mathbf k$.

On $S_1$, $g=1-x^{2}-y^{2}$ gives $g_x=-2x$, $g_y=-2y$, and $\mathbf F=\langle y,x,z\rangle$ has components $P=y,Q=x,R=z$. Substituting into Equation (10),

so switching to polar coordinates $x=r\cos\theta,y=r\sin\theta$ turns $4xy$ into $2r^{2}\sin2\theta$, which vanishes upon integrating over $\theta$, leaving

On $S_2$ the orientation is $\mathbf n=-\mathbf k$, so $\mathbf F\cdot\mathbf n=-z$, which is identically $0$ since $z=0$ throughout $S_2$. Hence $\iint_{S_2}\mathbf F\cdot d\mathbf S=0$, and

Solution. Put the center of the ball at the origin. Since the temperature is proportional to the squared distance from the center, write the proportionality constant as $C$, so $u(x,y,z)=C(x^{2}+y^{2}+z^{2})$ and the heat flow field is

($K$ the conductivity). The outward unit normal on the sphere $S$ (center the origin, radius $a$) is the scaled position vector $\mathbf n=\dfrac1a\langle x,y,z\rangle$, so

Every point of $S$ satisfies $x^{2}+y^{2}+z^{2}=a^{2}$, so this dot product is constant on $S$:

Since the integrand is constant, the surface integral is that constant times the area $A(S)=4\pi a^{2}$, so

The negative sign means heat is flowing net against the chosen (outward) orientation, that is, inward toward the center.