converting cylindrical, spherical, and rectangular coordinates...?

Question

having trouble with converting between those coordinate systems. i know how to do it for points, but not for equations...

given in cylindrical coordinates

r=4sin(theta)

express the equation in rectangular coordinates...

<<< answer is x^2+(y-2)^2=4 >>>

um, how do i do that?


also...

given in spherical coordinates...

rho sin (theta) = 2 cos(theta)

<<< answer is (x-1)^2+y^2=1 >>>

express equation in rectangular coordinates...

um, same thing, i don't know how to do this... the book doesn't provide any examples for this and the professor didn't go over it, only went over converting points. it seems like it shouldn't be that hard of a problem, but i don't know how to start it. if anyone could, please help explain how to convert these equations... thanx in advance.

Answer 1

We have x=rcos(θ), and y=rsin(θ).


Therefore x=4sin(θ)cos(θ) and y=4sin(θ)sin(θ).
And x^2+y^2= 16sin(θ)^2(cos(θ)^2+sin(θ)^2) =16sin(θ)=4y.
Thus x^2+y^2-4y+4-4 =x^2+(y-2)^2-4=0 and x^2+(y-2)^2=4.

Are you sure the second one is given by:
ρsin(θ)=2cos(θ) and not ρsin(φ)=cos(θ)?

if it is ρsin(φ)=2cos(θ), then first notice that φ≠kπ for any integer k.
Then we have ρ=2cos(θ)/sin(φ)
Next we have x=ρcos(θ)sin(φ)=2cos(θ)^2, and y=ρsin(θ)sin(φ)=2cos(θ)sin(θ).

Thus x^2+y^2= 4cos(θ)^2(cos(θ)^2+sin(θ)^2) =2x,
and x^2-2x+1-1+y^2= (x-1)^2-1+y^2=0,
so (x-1)^2+y^2=1

Answer 2

spherical coord. x= rsin phi cosT y=rsin phi sinT z=rcos phi phi=pi/3= 60º , Let sin phi=sin60º=K = (1/2)sqrt3 x=r K cosT y=rK sinT z= r cos60 = r (1/2) , isolate r , r=2z x^2 +y^2 = (rK)^2 (cos^2 T+sin^2T) x^2+y^2= (rK)^2 x^2+y^2 = K^2 r^2 , but r=2z x^2+y^2 = (1/4) (3) 4z^2 x^2 +y^2- 3z^2=0 ( Cartesian ) cylindrical x=rcosT y=rsinT z=z plug , r^2 -3z^2=0 z= r / sqrt3 ( Cylindrical) it is an inverted cone at 60º with respect Z axis .-