Solution: We know that "The cylindrical coordinates are denoted by ( r , θ , z ) (r,\theta,z) ( r , θ , z ) and rectangular coordinates are denoted by ( x , y , z ) (x,y,z) ( x , y , z ) "
To convert from cylindrical coordinates to rectangular coordinates we use the equations
x = r c o s θ x=r~cos~\theta x = r cos θ
y = r s i n θ y=r~sin~\theta y = r s in θ and
z = z z=z z = z
(1) Given cylindrical coordinates are ( 5 , π 6 , 3 ) (5, \frac{\pi}{6},3) ( 5 , 6 π , 3 )
∴ ( r , θ , z ) = ( 5 , π 6 , 3 ) \therefore (r,\theta,z)=(5, \frac{\pi}{6},3) ∴ ( r , θ , z ) = ( 5 , 6 π , 3 )
Now, to find rectangular coordinates, we have
x = r c o s θ = 5 c o s ( π 6 ) = 5 ( 3 2 ) = 5 3 2 x=r~cos~\theta=5~cos~(\frac{\pi}{6})=5(\frac{\sqrt{3}}{2})=\frac{5\sqrt{3}}{2} x = r cos θ = 5 cos ( 6 π ) = 5 ( 2 3 ) = 2 5 3
y = r s i n θ = 5 s i n ( π 6 ) = 5 ( 1 2 ) = 5 2 y=r~sin~\theta=5~sin~(\frac{\pi}{6})=5(\frac{1}{2})=\frac{5}{2} y = r s in θ = 5 s in ( 6 π ) = 5 ( 2 1 ) = 2 5
z = z = 3 z=z=3 z = z = 3
Therefore rectangular coordinates ( x , y , z ) = ( 5 3 2 , 5 2 , 3 ) (x,y,z)=(\frac{5\sqrt{3}}{2},\frac{5}{2},3) ( x , y , z ) = ( 2 5 3 , 2 5 , 3 )
(2) Given cylindrical coordinates are ( 6 , π 3 , − 5 ) (6, \frac{\pi}{3},-5) ( 6 , 3 π , − 5 )
∴ ( r , θ , z ) = ( 6 , π 3 , − 5 ) \therefore (r,\theta,z)=(6, \frac{\pi}{3},-5) ∴ ( r , θ , z ) = ( 6 , 3 π , − 5 )
Now, to find rectangular coordinates, we have
x = r c o s θ = 6 c o s ( π 3 ) = 6 ( 1 2 ) = 6 2 = 3 x=r~cos~\theta=6~cos~(\frac{\pi}{3})=6(\frac{1}{2})=\frac{6}{2}=3 x = r cos θ = 6 cos ( 3 π ) = 6 ( 2 1 ) = 2 6 = 3
y = r s i n θ = 6 s i n ( π 3 ) = 6 ( 3 2 ) = 6 3 2 = 3 3 y=r~sin~\theta=6~sin~(\frac{\pi}{3})=6(\frac{\sqrt{3}}{2})=\frac{6\sqrt{3}}{2}=3\sqrt{3} y = r s in θ = 6 s in ( 3 π ) = 6 ( 2 3 ) = 2 6 3 = 3 3
z = z = − 5 z=z=-5 z = z = − 5
Therefore rectangular coordinates ( x , y , z ) = ( 3 , 3 3 , − 5 ) (x,y,z)=(3,3\sqrt{3},-5) ( x , y , z ) = ( 3 , 3 3 , − 5 )
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