§16.6 매개곡면과 그 넓이Parametric Surfaces and Their Areas

Solution. Reading off components, $x=2\cos u$, $y=v$, $z=2\sin u$. Here $y$ is just the parameter $v$ carried over unchanged, so it is unconstrained, while $x,z$ depend on $u$ alone. Whatever $u$ is,

holds identically, so every point of $S$ satisfies $x^2+z^2=4$. Conversely, any point with $x^2+z^2=4$ has some angle $u_0$ with $x=2\cos u_0$, $z=2\sin u_0$ (a point on the unit circle scaled by $2$), and taking $v=y$ realizes that point as $\mathbf r(u_0,y)$, so it does lie on $S$. Thus $S$ coincides exactly with the set of points satisfying $x^2+z^2=4$.

Since $x,z$ stay on a circle of radius $2$ no matter what value the free coordinate $y$ takes, $S$ is a circular cylinder of radius $2$ whose axis is the $y$-axis. The parameter $u$ plays the role of the angle around that circle, and $v$ the role of displacement along the axis.

Solution. Examine the two families of grid curves separately.

Fix $v=v_0$. Then $x=(2+\sin v_0)\cos u$ and $y=(2+\sin v_0)\sin u$, so writing $\rho:=2+\sin v_0$ for the constant,

holds identically — as $u$ varies, $x,y$ always lie on the circle of radius $\rho$. Meanwhile $z=u+\cos v_0$ increases in direct proportion to $u$, the angle swept around that circle. As $u$ runs from $0$ to $4\pi$ the curve winds twice around the circle while its height climbs steadily, so this curve is a coil (a spiral) of radius $\rho$. This is the grid curve with $v$ held constant.

Fix $u=u_0$. Then $x=(2+\sin v)\cos u_0$, $y=(2+\sin v)\sin u_0$, and since $\cos u_0,\sin u_0$ are constants, the point $(x,y)$ always lies in the fixed vertical half-plane through the origin in direction $(\cos u_0,\sin u_0)$. Inside that half-plane, set $X=2+\sin v$ (horizontal distance from the axis) and $Z=u_0+\cos v$; then

so as $v$ runs through one full period, the curve traces a circle of radius $1$ centered at $(2,u_0)$ inside that half-plane. This is the grid curve with $u$ held constant, and it appears as a closed circle lying in a fixed vertical plane.

In summary, the grid curves with $v$ constant are the spirals winding along the tube, while the grid curves with $u$ constant are the circular cross-sections cutting straight across that spiral tube.

Solution. Let $P$ be an arbitrary point of the plane through $P_0$ that contains $\mathbf a,\mathbf b$. Since $\mathbf a,\mathbf b$ are not parallel, together they form a basis spanning that plane, so $\overrightarrow{P_0P}$ is written uniquely as

for some scalars $u,v$. The position vector of $P$ is $\mathbf r=\mathbf r_0+\overrightarrow{P_0P}$, so

Conversely, for any $u,v\in\mathbb R$ the point this formula gives lies in the plane, because $\overrightarrow{P_0P}=u\mathbf a+v\mathbf b$ is a linear combination of $\mathbf a,\mathbf b$. So this vector function sweeps out exactly the whole plane, with no restriction on the parameters.

Writing $\mathbf r_0=\langle x_0,y_0,z_0\rangle$, $\mathbf a=\langle a_1,a_2,a_3\rangle$, $\mathbf b=\langle b_1,b_2,b_3\rangle$ in components, the parametric equations read

Solution. Recall spherical coordinates $(\rho,\theta,\phi)$: a point of the sphere is exactly one whose distance from the origin is fixed at $\rho=a$. Taking the remaining two degrees of freedom — the azimuthal angle $\theta$ (measured from the positive $x$-axis) and the polar angle $\phi$ (measured from the positive $z$-axis) — as parameters, and substituting $\rho=a$ into the conversion formulas between spherical and rectangular coordinates, gives

As a vector function,

Substituting these components confirms that the parametrization does lie on the sphere:

an identity. To cover the whole sphere without duplication (aside from the two poles and the meridian at longitude $0$), it suffices to let $\phi$ range over $[0,\pi]$ and $\theta$ over $[0,2\pi]$, so the parameter domain is

The grid curves with $\phi$ constant are circles of constant latitude (the equator among them), and those with $\theta$ constant are meridians — semicircles of constant longitude joining the north and south poles.

Solution. Points with $x^2+y^2=4$ are exactly the points whose distance from the $z$-axis is fixed at $2$, so plane polar coordinates $x=2\cos\theta$, $y=2\sin\theta$ sweep out that circle. The remaining degree of freedom of the cylinder is height, and $z$ itself already serves as that parameter, so

that is,

Indeed $x^2+y^2=4\cos^2\theta+4\sin^2\theta=4$ holds identically regardless of $\theta$, so this parametrization does lie on the cylindrical surface. Covering the full cylinder once requires $\theta$ to range over $[0,2\pi]$, and the condition $0\le z\le1$ directly fixes the range of $z$, so the parameter domain is

Solution. This surface is already presented as a graph, with $z$ solved for as a function of $x,y$, so the recipe of Card 5 (every graph is a parametric surface) applies directly: take $x,y$ themselves as parameters and substitute the formula for $z$.

that is,

With this choice the third component automatically satisfies $z=x^2+2y^2$, so as $(x,y)$ ranges over the whole $xy$-plane (that is, $D=\mathbb R^2$), the image of this vector function coincides exactly with the given elliptic paraboloid.

Solution. Parametrization 1 — use $x,y$ themselves. Since $z=2\sqrt{x^2+y^2}$ is a graph, Card 5's recipe takes $x,y$ themselves as parameters:

This always works, but the square root left sitting in the third component is awkward once an integral over this surface is required.

Parametrization 2 — use polar coordinates. Since $x^2+y^2=r^2$, polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ ($r\ge0$) turn $\sqrt{x^2+y^2}$ into simply $r$, so the square root disappears:

Indeed $z^2=4r^2=4(x^2+y^2)$ holds, so the surface equation is satisfied. Choosing parameters adapted to the cone's own symmetry this way makes later computations much shorter, as Card 6 (parametrizations are not unique) points out: to cut out the piece with $z\le1$, Parametrization 2 needs only a change of parameter domain to $0\le r\le\tfrac12$, while Parametrization 1 must first solve the square-root condition $2\sqrt{x^2+y^2}\le1$ to reach the disk $x^2+y^2\le\tfrac14$.

Solution. Set $f(x)=\sin x$ and apply Equations (3) directly. The parametric equations of the surface of revolution about the $x$-axis are

Indeed $y^2+z^2=\sin^2x\cos^2\theta+\sin^2x\sin^2\theta=\sin^2x$ holds regardless of $\theta$, so the cross-section over the abscissa $x$ is always a circle of radius $|\sin x|$, matching what it means to revolve about the $x$-axis.

The hypothesis $f\ge0$ of Equations (3) holds only for $0\le x\le\pi$, however; on $\pi\lt x\lt2\pi$, $\sin x\lt 0$. The parametrization still works without trouble even so: with a negative radius $\sin x$, letting $\theta$ run through a full period $0$ to $2\pi$ still sweeps out exactly the same circle as it would with the angle shifted by $\pi$ (a negative radius merely reverses the phase, not the locus traced). The result is a radius that shrinks to $0$ at $x=\pi$ and then opens back out, producing two symmetric spindle-shaped lobes that meet at a single point (a vertex) at $x=\pi$.

Solution. First find the parameter pair $(u,v)$ corresponding to $(1,1,3)$. Since $u^2=1$, $u=\pm1$, and since $v^2=1$, $v=\pm1$; imposing the remaining condition $u+2v=3$ singles out $u=1,v=1$ (the other three combinations give $-1,1,-3$ respectively, none of which match). So the parameters for this point are $(u_0,v_0)=(1,1)$.

The vector equation is $\mathbf r(u,v)=u^2\,\mathbf i+v^2\,\mathbf j+(u+2v)\,\mathbf k$, so the tangent vectors are

By Card 9, $\mathbf r_u\times\mathbf r_v$ is normal to the tangent plane, so compute the cross product:

Substituting $(u,v)=(1,1)$ gives the normal vector

With normal vector $\langle a,b,c\rangle$ and the plane through $(x_0,y_0,z_0)$ given by $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$, substitution gives

Expanding gives $-2x-4y+4z-6=0$, and dividing both sides by $-2$,

Hence the tangent plane has equation $x+2y-2z+3=0$.

Solution. Start from the parametrization of Example 4, $\mathbf r(\phi,\theta)=a\sin\phi\cos\theta\,\mathbf i+a\sin\phi\sin\theta\,\mathbf j+a\cos\phi\,\mathbf k$, $D=[0,\pi]\times[0,2\pi]$. The tangent vectors are

and their cross product is

The square of its length is

(the second equality uses $\cos^2\theta+\sin^2\theta=1$; the third factors out $\sin^2\phi$ and uses $\sin^2\phi+\cos^2\phi=1$), so, since $\sin\phi\ge0$ on $0\le\phi\le\pi$,

Applying Definition 6,

The integrand does not involve $\theta$, so the outer integral contributes a factor of $2\pi$, and the inner integral is

so

Hence a sphere of radius $a$ has area $4\pi a^2$, agreeing with the familiar formula.

Solution. Since $z=x^2+y^2$ is a graph, Equation (9) applies directly. With $\partial z/\partial x=2x$ and $\partial z/\partial y=2y$, the integrand is

First pin down $D$: the plane $z=9$ meets the paraboloid along $x^2+y^2=9$, $z=9$, so the piece of the paraboloid below the plane is exactly the part whose projection is the disk $D=\{(x,y)\mid x^2+y^2\le9\}$ (radius $3$). Hence

Since the integrand depends only on $x^2+y^2$ and $D$ is a disk, switching to polar coordinates $x=r\cos\theta,\ y=r\sin\theta,\ dA=r\,dr\,d\theta$ gives

The integrand does not involve $\theta$, so the outer integral contributes a factor of $2\pi$. For the inner integral, substitute $w=1+4r^2$, so $dw=8r\,dr$, and $r=0,3$ correspond to $w=1,37$:

Multiplying by $2\pi$,

Hence the required area is $A(S)=\dfrac{\pi}{6}\bigl(37\sqrt{37}-1\bigr)\approx117.319$.