Answer to Question #151520 in Differential Geometry | Topology for Dolly

Question

Find the rectangular coordinates of the point whose spherical coordinates are
1) (5, π/2,π/2)
2) (4, π/3, 2π/3)
3) (0,π/11,π/5)
4) (2, 5π/3, 3π/4)

Answer 1

(1) Given spherical coordinates: $(5,\frac{\pi }{2},\frac{\pi }{2})$

Require to find the rectangular coordinates.

Now $(\rho ,\theta ,\varphi )=(5,\frac{\pi }{2},\frac{\pi }{2})$

$\Rightarrow \rho=5,\theta=\frac{\pi }{2},\varphi =\frac{\pi }{2}$

The relation between the rctangular coordinates and spherical coordinates are given by the equaitons

$x=\rho sin\varphi cos\theta ,y=\rho sin\varphi sin\theta,z=\rho cos\varphi$

Subtituting the given values, we get

$x=5sin(\frac{\pi }{2})cos(\frac{\pi }{2})=5(1)(0)=0$

$y=5sin(\frac{\pi }{2})sin(\frac{\pi }{2})=5(1)(1)=5$

$z=5cos(\frac{\pi }{2})=5(0)=0$

Therefore, $(x,y,z)=(0,5,0)$

(2) Gievn: $(4,\frac{\pi }{3},\frac{2pi }{3})$

$x=4sin(\frac{2pi }{3})cos(\frac{pi }{3})=4(\frac{\sqrt{3}}{2})(\frac{1}{2})=\sqrt{3}$

$y=4sin(\frac{2pi }{3})sin(\frac{pi }{3})=4(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})=3$

$z=4cos(\frac{2pi }{3})=4(-\frac{1}{2})=-2$

Therefore, $(x,y,z)=(\sqrt{3},3,-2)$

(3)Given: $(0,\frac{\pi }{11},\frac{\pi }{5})$

$\Rightarrow \rho=0,\theta=\frac{\pi }{11},\varphi =\frac{\pi }{5}$

$x=0sin(\frac{pi }{5})cos(\frac{pi }{11})=0$

$y=0sin(\frac{pi }{5})sin(\frac{pi }{11})=0$

$z=0cos(\frac{pi }{5})=0$

Therefore, $(x,y,z)=(0,0,0)$

(4) Given: $(2,\frac{5pi }{3},\frac{3pi }{4})$

$\Rightarrow \rho=2,\theta=\frac{5pi }{3},\varphi =\frac{3pi }{4}$

$x=2sin(\frac{3pi }{4})cos(\frac{5pi }{3})=2(\frac{1}{\sqrt2})(\frac{1}{2})=\frac{1}{\sqrt2}$

$y=2sin(\frac{3pi }{4})sin(\frac{5pi }{3})=2(\frac{1}{\sqrt2})(\frac{-\sqrt3}{2})=-\frac{\sqrt3}{\sqrt2}$

$z=2cos(\frac{3pi }{4})=2(-\frac{1}{\sqrt2})=-\sqrt2$

Therefore, $(x,y,z)=(\frac{1}{\sqrt2},-\frac{\sqrt3}{\sqrt2},-\sqrt2)$