원기둥·구면좌표 삼중적분

(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 $\theta=\frac{3\pi}{4}$.