Problems were equations in rectangular form are converted to polar form, using the relationship between polar and rectangular coordinates, are presented along with detailed solutions.
In what follows the polar coordinates of a point are (R , t) where R is the radial coordinate and t is the angular coordinate.
The relationships between the rectangualr (x,y) and polar (R,t) coordinates of a points are given by
R^{ 2} = x^{ 2} + y^{ 2} y = R sin t x = R cos t
Problems on Converting Rectangualar Equations to polar form Problem 1
Convert the equation
2x^{ 2} + 2y^{ 2}  x + y = 0
to polar form.
Solution to Problem 1

Let us rewrite the equations as follows:
2 ( x^{ 2} + y^{ 2} )  x + y = 0

We now use the formulas giving the relationship between polar and rectangular coordinates: R^{ 2} = x^{ 2} + y^{ 2}, y = R sin t and x = R cos t:
2 ( R^{ 2} )  R cos t + R sin t = 0

Factor out R
R ( 2 R  cos t + sin t ) = 0

The above equation gives:
R = 0
or
2 R  cos t + sin t = 0

The equation R = 0 is the pole. But the pole is included in the graph of the second equation 2 R  cos t + sin t = 0 (check that for t = π / 4 , R = 0). We therefore can keep only the second equation.
2 R  cos t + sin t = 0
or
R = (1 / 2)(cos t  sin t)
Problem 2
Convert the equation
x + y = 0
to polar form.
Solution to Problem 2

Use y = R sin t and x = R cos t into the given equation:
x + y = 0
R cos t + R sin t = 0

Factor out R
R ( cos t + sin t ) = 0

The above equation gives:
R = 0
or
cos t + sin t = 0

The equation R = 0 is the pole. But the pole is included in the graph of the second equation cos t + sin t = 0 since this equation is independent of R. We therefore keep only the second equation.
cos t + sin t = 0

The above equation may be written as.
tan t =  1

Solve for t to obtain
t = 3 π / 4

All points of the form (R , 3 π / 4) are on the graph of the above equation. It is the equation of a line in polar form.
More References and Links to Polar Coordinates and TrigonometryPolar Coordinates. 