Solution (i):
Given, x = 6 , y = 6 , z = 8 x=6,\ y=6, z=8 x = 6 , y = 6 , z = 8
In rectangular coordinates,
x = r cos θ , y = r sin θ , z = z x=r\cos\theta,\ y=r\sin\theta,\ z=z x = r cos θ , y = r sin θ , z = z
So, 6 = r cos θ , 6 = r sin θ , z = 8 6=r\cos\theta,\ 6=r\sin\theta,\ z=8 6 = r cos θ , 6 = r sin θ , z = 8
Also, r 2 = x 2 + y 2 = 6 2 + 6 2 = 36 + 36 = 72 r^2=x^2+y^2=6^2+6^2=36+36=72 r 2 = x 2 + y 2 = 6 2 + 6 2 = 36 + 36 = 72
⇒ r = 72 = 6 2 \Rightarrow r=\sqrt{72}=6\sqrt2 ⇒ r = 72 = 6 2 (Taking positive value only)
And, tan θ = y x = 6 6 = 1 \tan\theta=\dfrac yx=\dfrac 66=1 tan θ = x y = 6 6 = 1
⇒ θ = tan − 1 1 = π 4 \Rightarrow \theta=\tan^{-1}1=\dfrac{\pi}4 ⇒ θ = tan − 1 1 = 4 π
So, cylindrical coordinates are ( r , θ , z ) = ( 6 2 , π 4 , 8 ) (r,\theta,z)=(6\sqrt2,\dfrac {\pi}4,8) ( r , θ , z ) = ( 6 2 , 4 π , 8 )
Solution (ii):
Given, x = 2 , y = 1 , z = 1 x=\sqrt2,\ y=1, z=1 x = 2 , y = 1 , z = 1
In rectangular coordinates,
x = r cos θ , y = r sin θ , z = z x=r\cos\theta,\ y=r\sin\theta,\ z=z x = r cos θ , y = r sin θ , z = z
So, 2 = r cos θ , 1 = r sin θ , z = 1 \sqrt2=r\cos\theta,\ 1=r\sin\theta,\ z=1 2 = r cos θ , 1 = r sin θ , z = 1
Also, r 2 = x 2 + y 2 = ( 2 ) 2 + 1 2 = 2 + 1 = 3 r^2=x^2+y^2=(\sqrt2)^2+1^2=2+1=3 r 2 = x 2 + y 2 = ( 2 ) 2 + 1 2 = 2 + 1 = 3
⇒ r = 3 \Rightarrow r=\sqrt3 ⇒ r = 3 (Taking positive value only)
And, tan θ = y x = 1 2 \tan\theta=\dfrac yx=\dfrac 1{\sqrt2} tan θ = x y = 2 1
⇒ θ = tan − 1 ( 1 2 ) = 0.6154 \Rightarrow \theta=\tan^{-1}(\dfrac 1{\sqrt2})=0.6154 ⇒ θ = tan − 1 ( 2 1 ) = 0.6154 rad
So, cylindrical coordinates are ( r , θ , z ) = ( 3 , 0.6154 , 1 ) (r,\theta,z)=(\sqrt3,0.6154,1) ( r , θ , z ) = ( 3 , 0.6154 , 1 )
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