극좌표 이중적분

The left hand sketch below contains the points, $(x_1,y_1)$, $(x_3,y_3)$, $(x_5,y_5)$, that are on the axes. The right hand sketch below contains the points, $(x_2,y_2)$, $(x_4,y_4)$, that are not on the axes.

Recall that the polar coordinates $r$, $\theta$ are related to the cartesian coordinates $x$, $y$, by $x=r\cos\theta$, $y=r\sin\theta$. So $r=\sqrt{x^2+y^2}$ and $\tan\theta=\frac{y}{x}$ (assuming that $x\ne 0$ and $r>0$) and

$$\begin{aligned} (x_1,y_1) &= (3,0) &&\implies &r_1=3,\ \tan\theta_1=0 &&\implies \theta_1&=0 \text{ as $(x_1,y_1)$ is on the positive $x$-axis} \\ (x_2,y_2) &= (1,1) &&\implies &r_2=\sqrt{2},\ \tan\theta_2=1 &&\implies \theta_2&=\frac{\pi}{4} \text{ as $(x_2,y_2)$ is in the first octant} \\ (x_3,y_3) &= (0,1) &&\implies &r_3=1,\ \cos\theta_3=0 &&\implies \theta_3&=\frac{\pi}{2} \text{ as $(x_3,y_3)$ is on the positive $y$-axis} \\ (x_4,y_4) &= (-1,1) &&\implies &r_4=\sqrt{2},\ \tan\theta_4=-1 &&\implies \theta_4&=\frac{3\pi}{4} \text{ as $(x_4,y_4)$ is in the third octant} \\ (x_5,y_5) &= (-2,0) &&\implies &r_5=2,\ \tan\theta_5=0 &&\implies \theta_5&=\pi \text{ as $(x_5,y_5)$ is on the negative $x$-axis} \end{aligned}$$

In this solution, we'll supress the subscripts. That is, we'll write $r$ in place of $r_i$ and $\theta$ in place of $\theta_i$. Note that the distance from the point $\big(r\cos\theta\,,\,r\sin\theta\big)$ to the origin is

$$\sqrt{r^2\cos^2\theta + r^2\sin^2\theta} =\sqrt{r^2} =|r|$$

Thus $r$ can be either the distance to the origin or minus the distance to the origin.

(a) The distance from $(-2,0)$ to the origin is $2$. So either $r=2$ or $r=-2$.

• If $r=2$, then $\theta$ must obey

$$\begin{aligned} (-2,0) = \big(2\cos\theta\,,\,2\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=-1 \\ &\iff \theta= n\pi,\ n\text{ integer },\ \cos\theta=-1 \\ &\iff \theta= n\pi,\ n\text{ odd integer } \end{aligned}$$

• If $r=-2$, then $\theta$ must obey

$$\begin{aligned} (-2,0) = \big(-2\cos\theta\,,\,-2\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=1 \\ &\iff \theta= n\pi,\ n\text{ integer },\ \cos\theta=1 \\ &\iff \theta= n\pi,\ n\text{ even integer } \end{aligned}$$

In particular, $\big(r=-2\,,\,\theta= 0\big)$ has $r<0$ and $0\le\theta<2\pi$.

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\pi$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=\pi$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=0$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=0$, $r<0$.

(b) The distance from $(1,1)$ to the origin is $\sqrt{2}$. So either $r=\sqrt{2}$ or $r=-\sqrt{2}$.

• If $r=\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (1,1) = \big(\sqrt{2}\,\cos\theta\,,\,\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta={1}/{\sqrt{2}} \\ &\iff \theta= {\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

• If $r=-\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (1,1) = \big(-\sqrt{2}\,\cos\theta\,,\,-\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=-{1}/{\sqrt{2}} \\ &\iff \theta= {5\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

In particular, $\big(r=-\sqrt{2}\,,\, \theta= {5\pi}/{4}\big)$ has $r<0$ and $0\le\theta<2\pi$.

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r<0$.

(c) The distance from $(-1,-1)$ to the origin is $\sqrt{2}$. So either $r=\sqrt{2}$ or $r=-\sqrt{2}$.

• If $r=\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (-1,-1) = \big(\sqrt{2}\,\cos\theta\,,\,\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=-{1}/{\sqrt{2}} \\ &\iff \theta= {5\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

• If $r=-\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (-1,-1) = \big(-\sqrt{2}\,\cos\theta\,,\,-\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta={1}/{\sqrt{2}} \\ &\iff \theta= {\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

In particular, $\big(r=-\sqrt{2}\,,\, \theta= {\pi}/{4}\big)$ has $r<0$ and $0\le\theta<2\pi$.

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r<0$.

(d) The distance from $(3,0)$ to the origin is $3$. So either $r=3$ or $r=-3$.

• If $r=3$, then $\theta$ must obey

$$\begin{aligned} (3,0) = \big(3\,\cos\theta\,,\,3\,\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=1 \\ &\iff \theta= 0 + 2n\pi,\ n\text{ integer } \end{aligned}$$

• If $r=-3$, then $\theta$ must obey

$$\begin{aligned} (3,0) = \big(-3\,\cos\theta\,,\,-3\,\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=-1 \\ &\iff \theta= \pi + 2n\pi,\ n\text{ integer } \end{aligned}$$

In particular, $\big(r=-3\,,\, \theta= \pi\big)$ has $r<0$ and $0\le\theta<2\pi$.

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\pi$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=\pi$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=0$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=0$, $r<0$.

(e) The distance from $(0,1)$ to the origin is $1$. So either $r=1$ or $r=-1$.

• If $r=1$, then $\theta$ must obey

$$\begin{aligned} (0,1) = \big(\cos\theta\,,\,\sin\theta\big) &\iff \cos\theta=0,\ \sin\theta=1 \\ &\iff \theta= {\pi}/{2} + 2n\pi,\ n\text{ integer } \end{aligned}$$

• If $r=-1$, then $\theta$ must obey

$$\begin{aligned} (0,1) = \big(-\cos\theta\,,\,-\sin\theta\big) &\iff \cos\theta=0,\ \sin\theta=-1 \\ &\iff \theta= {3\pi}/{2} + 2n\pi,\ n\text{ integer } \end{aligned}$$

In particular, $\big(r=-1\,,\, \theta= {3\pi}/{2}\big)$ has $r<0$ and $0\le\theta<2\pi$.

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{3\pi}{2}$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{3\pi}{2}$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{2}$, $r>0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{2}$, $r<0$.

(a) The lengths are

$$\begin{aligned} |\hat{\mathbf{e}}_r(\theta)| &= \sqrt{\cos^2\theta+\sin^2\theta} = 1 \\ |\hat{\mathbf{e}}_\theta(\theta)| &= \sqrt{(-\sin\theta)^2+\cos^2\theta} = 1 \end{aligned}$$

As

$$\hat{\mathbf{e}}_r(\theta) \cdot \hat{\mathbf{e}}_\theta(\theta) = (\cos\theta)(-\sin\theta) +(\sin\theta)(\cos\theta)=0$$

the two vectors are perpendicular and the angle between them is $\frac{\pi}{2}$. The cross product is

$$\begin{aligned} \hat{\mathbf{e}}_r(\theta) \times \hat{\mathbf{e}}_\theta(\theta) &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \end{matrix}\right] =\hat{\mathbf{k}} \end{aligned}$$

(b) Note that for $\theta$ determined by $x=r\cos\theta$, $y=r\sin\theta$,

• the vector $\hat{\mathbf{e}}_r(\theta)$ is a unit vector in the same direction as the vector from $(0,0)$ to $(x,y)$ and
• the vector $\hat{\mathbf{e}}_\theta(\theta)$ is a unit vector that is perpendicular to $\hat{\mathbf{e}}_r(\theta)$.
• The $y$-component of $\hat{\mathbf{e}}_\theta(\theta)$ has the same sign as the $x$-component of $\hat{\mathbf{e}}_r(\theta)$. The $x$-component of $\hat{\mathbf{e}}_\theta(\theta)$ has opposite sign to that of the $y$-component of $\hat{\mathbf{e}}_r(\theta)$.