매개변수 곡면

Note that, since $x^2+y^2=1+2z^2$ on $S$, the condition $z\ge 1$ is equivalent to $x^2+y^2\ge 3$, $z\ge 0$. So the hyperboloid is $\big\{ \ (x,y,z)\ \big|\ x^2+y^2=1+2z^2,\ 3\le x^2+y^2\le 9, z\ge 0\ \big\}$.

(a) No. Under this parametrization, the condition $3\le x^2+y^2\le 9$ is $3\le u^2+v^2\le 9$, not $2\le u^2+v^2\le 9$.

(b) Yes. Under this parametrization, $x=u\sin v$, $y=-u\cos v$ and $z=\sqrt{\frac{u^2}{2}-\frac{1}{2}}$. So

• $x^2+y^2-2z^2 = u^2-2\left(\frac{u^2}{2}-\frac{1}{2}\right) = 1$, as desired.
• The condition $x^2+y^2\le 9$ is equivalent to $u\le 3$, since $u\ge 0$.
• The condition $x^2+y^2\ge 3$ is equivalent to $u\ge\sqrt{3}$, since $u\ge 0$.
• $z=\sqrt{\frac{u^2}{2}-\frac{1}{2}}\ge 0$

(c) Yes. Under this parametrization, $x=\sqrt{1+2v^2}\cos u$, $y=\sqrt{1+2v^2}\sin u$ and $z=v$. So

• $x^2+y^2-2z^2 = 1+2v^2 -2v^2 = 1$, as desired.
• The condition $x^2+y^2\le 9$ is equivalent to $1+2v^2\le 9$, which is equivalent to $v\le 2$, since $v\ge 0$.
• The condition $x^2+y^2\ge 3$ is equivalent to $1+2v^2\ge 3$, which is equivalent to $v\ge 1$, since $v\ge 0$.
• $z=v\ge 0$

(d) Yes. Under this parametrization, $x=\sqrt{1+u}\sin v$, $y=\sqrt{1+u}\cos v$ and $z=\sqrt{u/2}$. So

• $x^2+y^2-2z^2 = 1+u -2(u/2) = 1$, as desired.
• The condition $x^2+y^2\le 9$ is equivalent to $1+u\le 9$, which is equivalent to $u\le 8$.
• The condition $x^2+y^2\ge 3$ is equivalent to $1+u\ge 3$, which is equivalent to $u\ge 2$.
• $z=\sqrt{u/2}\ge 0$

(e) No. Under this parametrization, $x=\sqrt{u}\cos v$, $y=-\sqrt{u}\sin v$ and $z=\sqrt{(u+1)/2}$. So

• $x^2+y^2-2z^2 = u -2(u+1)/2 = -1$, not $+1$

(a) No. $z=\sin\phi\sin \theta$ is negative when $0 <\phi\le\frac{\pi}{4}$, $\pi < \theta <2\pi$.

(b) Yes. Note that $x^2+\big(-y\big)^2+\big(\sqrt{2-x^2-y^2}\big)^2=2$ and that, for $x^2+y^2\le 1$, we have both $x^2+(-y)^2\le 1$ and $\sqrt{2-x^2-y^2}\ge 0$.

(c) No. $(u\sin\theta)^2+(u\cos\theta)^2=u^2>1$ for $1<u\le 2$. Also $\sqrt{2-u^2}$ is not defined for $\sqrt{2}<u\le 2$.

(d) Yes. Note that

• $\big(\sqrt{2}\sin \phi\cos\theta\big)^2 +\big(\sqrt{2}\sin \phi\sin\theta\big)^2 +\big(\sqrt{2}\cos \phi\big)^2=2$
• For $0\le\phi\le\frac{\pi}{4}$, we have $z=\sqrt{2}\cos \phi>0$.
• As $\phi$ runs from $0$ to $\frac{\pi}{4}$, $r(\phi)=\sqrt{2}\sin\phi$ runs from $0$ to $1$, so that $\big(x=r(\phi)\cos\theta\,,\,y=r(\phi)\sin\theta\big)$ covers all of $x^2+y^2\le1$ as $\phi$ runs from $0$ to $\frac{\pi}{4}$ and $\theta$ runs from $0$ to $2\pi$.

(e) Yes. Note that

• $\big(-\sqrt{2-z^2}\sin \phi\big)^2 +\big(\sqrt{2-z^2}\cos \phi\big)^2 +\big(z\big)^2=2$
• For $1\le z\le\sqrt{2}$, we have obviously have $z>0$.
• As $z$ runs from $1$ to $\sqrt{2}$, $r(z)=\sqrt{2-z^2}$ runs from $1$ to $0$, so that $\big(x=-r(z)\sin\phi\,,\,y=r(z)\cos\phi\big)$ covers all of $x^2+y^2\le1$ as $z$ runs from $1$ to $\sqrt{2}$ and $\phi$ runs from $0$ to $2\pi$.

(a) No. When $u=v=0$, $z=4$ is not between $0$ and $1$.

(b) Yes. Note that when $x=\sqrt{4-u}\,\cos v$, $y=\sqrt{4-u}\,\sin v$ and $z= u$ with $0 \le u \le 1$, $0 \le v \le 2\pi$,

• $z+x^2+y^2=4$
• $0\le z=u\le 1$
• For each fixed $z=u$ between $0$ and $1$, $(x,y)$ runs once around the circle $x^2+y^2 =4-z =4-u$ as $v$ runs from $0$ to $2\pi$.

(c) Yes. Note that when $x=u\,\cos v$, $y=u\,\sin v$ and $z= 4-u^2$, with $\sqrt{3} \le u \le 2$, $0 \le v \le 2\pi$

• $z+x^2+y^2=4$
• $0\le z=4-u^2\le 1$
• For each fixed $z=4-u^2$ between $0$ and $1$, $(x,y)$ runs once around the circle $x^2+y^2 =4-z =u^2$ as $v$ runs from $0$ to $2\pi$.

A normal vector to the surface $x^2 + 9y^2 + 4z^2 = 17$ at the point $(x,y,z)$ is $( 2x\,,\, 18y\,,\,8z)$. A normal vector to the plane $x - 8z = 0$ is $( 1\,,\,0\,,\,-8)$. So we want $( 2x\,,\, 18y\,,\,8z)$ to be parallel to $( 1\,,\,0\,,\,-8)$, i.e. to be a nonzero constant times $( 1\,,\,0\,,\,-8)$. This is the case whenever $y=0$ and $z=-2x$ with $x\ne 0$. In addition, we want $(x,y,z)$ to lie on the surface $x^2 + 9y^2 + 4z^2 = 17$. So we want $y=0$, $z=-2x$ and

$$\begin{aligned} 17= x^2 + 9y^2 + 4z^2 =x^2 +4(-2x)^2=17x^2 \implies x=\pm 1 \end{aligned}$$

So the allowed points are $\pm(1,0,-2)$.

Let $(x_0,y_0,z_0)$ be a point on the hyperboloid $z^2=4x^2+y^2-1$ where the tangent plane is parallel to the plane $2x-y+z=0$. A normal vector to the plane $2x-y+z=0$ is $( 2,-1,1)$. Because the hyperboloid is $G(x,y,z)=4x^2+y^2-z^2-1$ and ${ \boldsymbol{\nabla} } G(x,y,z) = ( 8x,2y,-2z)$, a normal vector to the hyperboloid at $(x_0,y_0,z_0)$ is ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)=( 8x_0,2y_0,-2z_0)$. So $(x_0,y_0,z_0)$ satisfies the required conditions if and only if there is a nonzero $t$ obeying

$$\begin{aligned} &( 8x_0,2y_0,-2z_0) =t( 2,-1,1) \text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff \frac{t^2}{4}= \frac{t^2}{4}+ \frac{t^2}{4}-1\text{ and } x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\\ & \iff t=\pm 2\qquad (x_0,y_0,z_0)=\pm \big(\tfrac{1}{2},-1,-1\big) \end{aligned}$$