Question

# Express the plane z=x in cylindrical and spherical coordinates. a) cykindrical z=r\cos(\theta) b) spherical coordinates \theta=\arcsin(\cot(\phi))

Express the plane $$z=x$$ in cylindrical and spherical coordinates.
a) cykindrical
$$z=r\cos(\theta)$$
b) spherical coordinates
$$\theta=\arcsin(\cot(\phi))$$

2021-05-28
Step 1
The plane $$z=x$$ goes through the line intersection of the planes
$$x=0$$ and $$z=0$$ and makes a $$\frac{\pi}{4}$$ angle with those planes.
So, a point $$(x,y,z)=(x,y,x)=(x,y)$$ is on the plane.
a) Cylindrical coordinates:
Let $$x=r\cos\theta,\ y=r\sin\theta$$ with $$r\geq0$$ and $$0\leq\theta\leq2\pi$$
Then. $$z=x$$
$$\Rightarrow z=r\cos\theta$$
So, a point on the plane takes the form $$(x,y,z)=(r\cos\theta,\ r\sin\theta,\ r\cos\theta)$$
Step 2
b) Spherical coordinates:
Let $$x=\rho\sin\phi\cos\theta,\ y=\rho\sin\phi\theta,\ z=\rho\cos\phi$$ with $$\rho\geq0,\ 0\leq\theta\leq2\pi$$ and $$0\leq\phi\leq\pi$$
Then, $$z=x$$
$$\Rightarrow\rho\cos\phi=\rho\sin\phi\cos\theta$$
$$\Rightarrow\frac{\rho\cos\phi}{\rho\sin\phi}=\cos\theta$$
$$\Rightarrow\cot\phi=\cos\theta$$
$$\Rightarrow\theta=\arccos(\cot\phi)$$
So, a point on the plane takes the form
$$(x,y,z)=(\rho\sin\phi\cos\theta,\ \rho\sin\phi\sin\theta,\ \rho\cos\phi)$$
$$=(\rho\sin\phi\cos(\arccos(\cot\phi)),\ \rho\sin\phi\sin(\arccos(\cot\phi)),\ \rho\cos\phi)$$
$$=(\rho\sin\phi\cos(\arccos(\cot\phi)),\ \rho\sin\phi\sin(\arcsin(\sqrt{1-\cot^{2}\phi})),\ \rho\cos\phi)$$
$$=(\rho\sin\phi\cot\phi,\ \rho\sin\phi\sqrt{1-\cot^{2}\phi},\ \rho\cos\phi)$$
$$=(\rho\cos\phi,\ \rho\sin\phi\sqrt{1-\cot^{2}\phi},\ \rho\cos\phi)$$