원기둥·구면좌표 삼중적분

(a), (b) Since the cylindrical coordinate $r(x,y,z)$ of a point $(x,y,z)$ is the distance, $\sqrt{x^2+y^2}$, from $(x,y,z)$ to the $z$-axis, the sets

$$\begin{aligned} \big\{ \ (x,y,z)\ \big|\ r(x,y,z)=0\ \big\} &=\big\{ \ (x,y,z)\ \big|\ x^2+y^2=0\ \big\} =\big\{ \ (x,y,z)\ \big|\ x=y=0\ \big\} \\ &=\text{the $z$-axis} \\ \big\{ \ (x,y,z)\ \big|\ r(x,y,z)=1\ \big\} &=\big\{ \ (x,y,z)\ \big|\ x^2+y^2=1\ \big\} \\ &=\text{the cylinder of radius $1$ centred on the $z$-axis} \end{aligned}$$

(c), (d) Since the cylindrical coordinate $\theta(x,y,z)$ of a point $(x,y,z)$ is the angle between the positive $x$-axis and the line from $(0,0,0)$ to $(x,y,0)$, the sets

$$\begin{aligned} \big\{ \ (x,y,z)\ \big|\ \theta(x,y,z)=0\ \big\} &=\text{the half of the $xz$-plane with $x>0$} \\ \big\{ \ (x,y,z)\ \big|\ \theta(x,y,z)={\pi}/{4}\ \big\} &=\text{the half of the plane $y=x$ with $x>0$} \end{aligned}$$

The sketch is below. To help build up this sketch, it is useful to recall the following facts.

• The cylindrical coordinate $r$ is the distance of the point from the $z$-axis. In particular all points with $r=0$ lie on the $z$-axis (for all values of $\theta$).
• The cylindrical coordinate $z$ is the distance of the point from the $xy$-plane. In particular all points with $z=0$ lie on the $xy$-plane.

(a) When $\theta=0$, $\sin\theta=0$ and $\cos\theta=1$, so that the polar coordinates $r=1$, $\theta=0$, $z=0$ correspond to the Cartesian coordinates

$$(x,y,z) = (r\cos\theta,r\sin\theta,z) = (1\times\cos 0,\ 1\times\sin 0,\ 0) =(1,0,0)$$

(b) When $\theta=\frac{\pi}{4}$, $\sin\theta=\cos\theta=\frac{1}{\sqrt{2}}$, so that the polar coordinates $r=1$, $\theta=\frac{\pi}{4}$, $z=0$ correspond to the Cartesian coordinates

$$(x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(1\times\cos \frac{\pi}{4},\ 1\times\sin \frac{\pi}{4},\ 0\right) = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)$$

(c) When $\theta=\frac{\pi}{2}$, $\sin\theta=1$ and $\cos\theta=0$, so that the polar coordinates $r=1$, $\theta=\frac{\pi}{2}$, $z=0$ correspond to the Cartesian coordinates

$$(x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(1\times\cos \frac{\pi}{2},\ 1\times\sin \frac{\pi}{2},\ 0\right) = (0,1,0)$$

(d) When $\theta=\pi$, $\sin\theta=0$ and $\cos\theta=-1$, so that the polar coordinates $r=0$, $\theta=\pi$, $z=1$ correspond to the Cartesian coordinates

$$(x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(0\times\cos\pi,\ 0\times\sin\pi,\ 1\right) = (0,0,1)$$

(e) When $\theta=\frac{\pi}{4}$, $\sin\theta=\cos\theta=\frac{1}{\sqrt{2}}$, so that the polar coordinates $r=1$, $\theta=\frac{\pi}{4}$, $z=1$ correspond to the Cartesian coordinates

$$(x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(1\times\cos \frac{\pi}{4},\ 1\times\sin \frac{\pi}{4},\ 1\right) = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1\right)$$

(a) The cylindrical coordinates must obey

$$1=x=r\cos\theta\qquad 1=y=r\sin\theta\qquad 2=z$$

So $z=2$, $r=\sqrt{1^2+1^2}=\sqrt{2}$ and $\tan\theta=\frac{y}{x} =\frac{1}{1}=1$. Recall that $\tan\left(\frac{\pi}{4}+k\pi\right)=1$ for all integers $k$. As $(x,y)=(1,1)$ lies in the first quadrant, $0\le\theta\le\frac{\pi}{2}$. So $\theta=\frac{\pi}{4}$ (plus possibly any integer multiple of $2\pi$).

(b) The cylindrical coordinates must obey

$$-1=x=r\cos\theta\qquad -1=y=r\sin\theta\qquad 2=z$$

So $z=2$, $r=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}$ and $\tan\theta=\frac{y}{x} =\frac{-1}{-1}=1$. Recall that $\tan\left(\frac{\pi}{4}+k\pi\right)=1$ for all integers $k$. As $(x,y)=(-1,-1)$ lies in the third quadrant, $\pi\le\theta\le\frac{3\pi}{2}$. So $\theta=\frac{5\pi}{4}$ (plus possibly any integer multiple of $2\pi$).

(c) The cylindrical coordinates must obey

$$-1=x=r\cos\theta\qquad \sqrt{3}=y=r\sin\theta\qquad 0=z$$

So $z=0$, $r=\sqrt{(-1)^2+\big(\sqrt{3}\big)^2}=2$ and $\tan\theta=\frac{y}{x} =\frac{\sqrt{3}}{-1}=-\sqrt{3}$. Recall that $\tan\left(\frac{2\pi}{3}+k\pi\right)=-\sqrt{3}$ for all integers $k$. As $(x,y)=(-1,\sqrt{3})$ lies in the second quadrant, $\frac{\pi}{2}\le\theta\le\pi$. So $\theta=\frac{2\pi}{3}$ (plus possibly any integer multiple of $2\pi$).

(d) The cylindrical coordinates must obey

$$0=x=r\cos\theta\qquad 0=y=r\sin\theta\qquad 1=z$$

So $z=1$, $r=\sqrt{0^2+0^2}=0$ and $\theta$ is completely arbitrary.

(a) As $x=r\cos\theta$ and $y=r\sin\theta$,

$$z=2xy \iff z=2r^2\cos\theta\sin\theta = r^2\sin(2\theta)$$

(b) As $x=r\cos\theta$ and $y=r\sin\theta$,

$$x^2+y^2+z^2=1 \iff r^2\cos^2\theta+r^2\sin^2\theta + z^2 =1 \iff r^2+z^2=1$$

(c) As $x=r\cos\theta$ and $y=r\sin\theta$,

$$\begin{aligned} (x-1)^2 + y^2 =1 &\iff (r\cos\theta-1)^2 + (r\sin\theta)^2 =1 \\ &\iff r^2\cos^2\theta -2r\cos\theta +1 + r^2\sin^2\theta = 1 \\ &\iff r^2=2r\cos\theta \iff r=2\cos\theta\text{ or }r=0 \\ &\iff r=2\cos\theta \end{aligned}$$

Note that the solution $r=0$ is included in $r=2\cos\theta$ — just choose $\theta=\frac{\pi}{2}$.

Since the spherical coordinate $\varphi(x,y,z)$ of a point $(x,y,z)$ is the angle between the positive $z$-axis and the radius vector from $(0,0,0)$ to $(x,y,z)$, the sets

$$\begin{aligned} \big\{ \ (x,y,z)\ \big|\ \varphi(x,y,z)=0\ \big\} &=\text{the positive $z$-axis} \\ \big\{ \ (x,y,z)\ \big|\ \varphi(x,y,z)={\pi}/{2}\ \big\} &=\text{the $xy$-plane} \\ \big\{ \ (x,y,z)\ \big|\ \varphi(x,y,z)=\pi\ \big\} &=\text{the negative $z$-axis} \end{aligned}$$

Alternatively, $\tan\varphi(x,y,z)=\frac{z}{\sqrt{x^2+y^2}}$, so that, for any $0<\Phi<\pi$,

$$\begin{aligned} \big\{ \ (x,y,z)\ \big|\ \varphi(x,y,z)=\Phi\ \big\} &=\big\{ \ (x,y,z)\ \big|\ z=\tan\Phi\sqrt{x^2+y^2}\ \big\} \\ &=\text{the cone that makes the angle $\Phi$ with the positive $z$-axis} \end{aligned}$$

The sketch is below. To help build up this sketch, it is useful to recall the following facts.

• The spherical coordinate $\rho$ is the distance of the point from the origin $(0,0,0)$. In particular if $\rho=0$, then the point is the origin (regardless of the values of $\theta$ and $\varphi$). If $\rho=1$ then the point lies on the sphere of radius $1$ centred on the origin.
• The spherical coordinate $\varphi$ is the angle between the positive $z$-axis and the radial line segment from the origin to $(x,y,z)$. In particular, all points with $\varphi=0$ lie on the positive $z$–axis (regardless of the value of $\theta$). All points with $\varphi=\frac{\pi}{2}$ lie in the $xy$-plane.

(a) The point $(-2,0,0)$

• lies in the $xy$-plane (i.e. has $z=\rho\cos\varphi=0$) and so has $\varphi=\frac{\pi}{2}$ and
• lies on the negative $x$-axis and so has $\theta=\pi$ and
• is a distance $2$ from the origin and so has $\rho=2$.

(b) The point $(0,3,0)$

• lies in the $xy$-plane (i.e. has $z=\rho\cos\varphi=0$) and so has $\varphi=\frac{\pi}{2}$ and
• lies on the positive $y$-axis and so has $\theta=\frac{\pi}{2}$ and
• is a distance $3$ from the origin and so has $\rho=3$.

(c) The point $(0,0,-4)$

• lies on the negative $z$-axis and so has $\varphi=\pi$ and $\theta$ arbitrary and
• is a distance $4$ from the origin and so has $\rho=4$.

(d) The point $\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\sqrt{3}\right)$

• has $\rho=\sqrt{x^2+y^2+z^2} =\sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 +\left(\frac{1}{\sqrt{2}}\right)^2 +\left(\sqrt{3}\right)^2}=\sqrt{4}=2$ and
• has $\sqrt{3}=z=\rho\cos\varphi = 2\cos\varphi$ so that $\cos\varphi = \frac{\sqrt{3}}{2}$ and $\varphi=\frac{\pi}{6}$ and
• has $-\frac{1}{\sqrt{2}}=x=\rho\sin\varphi\cos\theta = 2\big(\frac{1}{2}\big)\cos\theta$ so that $\cos\theta = -\frac{1}{\sqrt{2}}$. As $\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ is in the second quadrant, we have $\frac{\pi}{2}\le\theta\le\pi$ and so $\theta=\frac{3\pi}{4}$.