(1) Given spherical coordinates: ( 5 , π 2 , π 2 ) (5,\frac{\pi }{2},\frac{\pi }{2}) ( 5 , 2 π , 2 π )
Require to find the rectangular coordinates.
Now ( ρ , θ , φ ) = ( 5 , π 2 , π 2 ) (\rho ,\theta ,\varphi )=(5,\frac{\pi }{2},\frac{\pi }{2}) ( ρ , θ , φ ) = ( 5 , 2 π , 2 π )
⇒ ρ = 5 , θ = π 2 , φ = π 2 \Rightarrow \rho=5,\theta=\frac{\pi }{2},\varphi =\frac{\pi }{2} ⇒ ρ = 5 , θ = 2 π , φ = 2 π
The relation between the rctangular coordinates and spherical coordinates are given by the equaitons
x = ρ s i n φ c o s θ , y = ρ s i n φ s i n θ , z = ρ c o s φ x=\rho sin\varphi cos\theta ,y=\rho sin\varphi sin\theta,z=\rho cos\varphi x = ρ s in φ cos θ , y = ρ s in φ s in θ , z = ρ cos φ
Subtituting the given values, we get
x = 5 s i n ( π 2 ) c o s ( π 2 ) = 5 ( 1 ) ( 0 ) = 0 x=5sin(\frac{\pi }{2})cos(\frac{\pi }{2})=5(1)(0)=0 x = 5 s in ( 2 π ) cos ( 2 π ) = 5 ( 1 ) ( 0 ) = 0
y = 5 s i n ( π 2 ) s i n ( π 2 ) = 5 ( 1 ) ( 1 ) = 5 y=5sin(\frac{\pi }{2})sin(\frac{\pi }{2})=5(1)(1)=5 y = 5 s in ( 2 π ) s in ( 2 π ) = 5 ( 1 ) ( 1 ) = 5
z = 5 c o s ( π 2 ) = 5 ( 0 ) = 0 z=5cos(\frac{\pi }{2})=5(0)=0 z = 5 cos ( 2 π ) = 5 ( 0 ) = 0
Therefore, ( x , y , z ) = ( 0 , 5 , 0 ) (x,y,z)=(0,5,0) ( x , y , z ) = ( 0 , 5 , 0 )
(2) Gievn: ( 4 , π 3 , 2 p i 3 ) (4,\frac{\pi }{3},\frac{2pi }{3}) ( 4 , 3 π , 3 2 p i )
x = 4 s i n ( 2 p i 3 ) c o s ( p i 3 ) = 4 ( 3 2 ) ( 1 2 ) = 3 x=4sin(\frac{2pi }{3})cos(\frac{pi }{3})=4(\frac{\sqrt{3}}{2})(\frac{1}{2})=\sqrt{3} x = 4 s in ( 3 2 p i ) cos ( 3 p i ) = 4 ( 2 3 ) ( 2 1 ) = 3
y = 4 s i n ( 2 p i 3 ) s i n ( p i 3 ) = 4 ( 3 2 ) ( 3 2 ) = 3 y=4sin(\frac{2pi }{3})sin(\frac{pi }{3})=4(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})=3 y = 4 s in ( 3 2 p i ) s in ( 3 p i ) = 4 ( 2 3 ) ( 2 3 ) = 3
z = 4 c o s ( 2 p i 3 ) = 4 ( − 1 2 ) = − 2 z=4cos(\frac{2pi }{3})=4(-\frac{1}{2})=-2 z = 4 cos ( 3 2 p i ) = 4 ( − 2 1 ) = − 2
Therefore, ( x , y , z ) = ( 3 , 3 , − 2 ) (x,y,z)=(\sqrt{3},3,-2) ( x , y , z ) = ( 3 , 3 , − 2 )
(3)Given: ( 0 , π 11 , π 5 ) (0,\frac{\pi }{11},\frac{\pi }{5}) ( 0 , 11 π , 5 π )
⇒ ρ = 0 , θ = π 11 , φ = π 5 \Rightarrow \rho=0,\theta=\frac{\pi }{11},\varphi =\frac{\pi }{5} ⇒ ρ = 0 , θ = 11 π , φ = 5 π
x = 0 s i n ( p i 5 ) c o s ( p i 11 ) = 0 x=0sin(\frac{pi }{5})cos(\frac{pi }{11})=0 x = 0 s in ( 5 p i ) cos ( 11 p i ) = 0
y = 0 s i n ( p i 5 ) s i n ( p i 11 ) = 0 y=0sin(\frac{pi }{5})sin(\frac{pi }{11})=0 y = 0 s in ( 5 p i ) s in ( 11 p i ) = 0
z = 0 c o s ( p i 5 ) = 0 z=0cos(\frac{pi }{5})=0 z = 0 cos ( 5 p i ) = 0
Therefore, ( x , y , z ) = ( 0 , 0 , 0 ) (x,y,z)=(0,0,0) ( x , y , z ) = ( 0 , 0 , 0 )
(4) Given: ( 2 , 5 p i 3 , 3 p i 4 ) (2,\frac{5pi }{3},\frac{3pi }{4}) ( 2 , 3 5 p i , 4 3 p i )
⇒ ρ = 2 , θ = 5 p i 3 , φ = 3 p i 4 \Rightarrow \rho=2,\theta=\frac{5pi }{3},\varphi =\frac{3pi }{4} ⇒ ρ = 2 , θ = 3 5 p i , φ = 4 3 p i
x = 2 s i n ( 3 p i 4 ) c o s ( 5 p i 3 ) = 2 ( 1 2 ) ( 1 2 ) = 1 2 x=2sin(\frac{3pi }{4})cos(\frac{5pi }{3})=2(\frac{1}{\sqrt2})(\frac{1}{2})=\frac{1}{\sqrt2} x = 2 s in ( 4 3 p i ) cos ( 3 5 p i ) = 2 ( 2 1 ) ( 2 1 ) = 2 1
y = 2 s i n ( 3 p i 4 ) s i n ( 5 p i 3 ) = 2 ( 1 2 ) ( − 3 2 ) = − 3 2 y=2sin(\frac{3pi }{4})sin(\frac{5pi }{3})=2(\frac{1}{\sqrt2})(\frac{-\sqrt3}{2})=-\frac{\sqrt3}{\sqrt2} y = 2 s in ( 4 3 p i ) s in ( 3 5 p i ) = 2 ( 2 1 ) ( 2 − 3 ) = − 2 3
z = 2 c o s ( 3 p i 4 ) = 2 ( − 1 2 ) = − 2 z=2cos(\frac{3pi }{4})=2(-\frac{1}{\sqrt2})=-\sqrt2 z = 2 cos ( 4 3 p i ) = 2 ( − 2 1 ) = − 2
Therefore, ( x , y , z ) = ( 1 2 , − 3 2 , − 2 ) (x,y,z)=(\frac{1}{\sqrt2},-\frac{\sqrt3}{\sqrt2},-\sqrt2) ( x , y , z ) = ( 2 1 , − 2 3 , − 2 )
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