Question 1080811


{{{x^2 + y^2 -12y = 0}}} to polar form:

Remember the conversions between Cartesian and polar.

{{{r^2 = x^2 + y^2}}}

{{{y = r*sin(theta)}}}
{{{x = r*cos(theta)}}}
 
Now let's convert.

{{{x^2 + y^2 -12y = 0}}}

{{{(r*cos(theta))^2 + (r*sin(theta))^2 -12r*sin(theta) = 0}}}

{{{(r^2*cos^2(theta) + r^2*sin^2(theta)) -12r*sin(theta) = 0}}}

{{{r^2(cos^2(theta) + sin^2(theta)) -12r*sin(theta) = 0}}}

{{{r^2(1) -12r*sin(theta) = 0}}}

{{{r^cross(2)  = 12cross(r)*sin(theta)}}}

{{{r  = 12*sin(theta)}}}