What the is the polar form of #2 = x^2y-x/y^2 +xy^2 #?

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Answer 1 · 2018-07-28T12:47:31

# r^4sin theta cos theta ( cos theta+sin theta )#
#- 2 r - cos theta csc^2theta= 0, r ne 0.#

I have realized that my previous answer was wrong. Thank you

Sonnhard. Now, I do it again.

Using

#0 <= sqrt ( x^2 + y^2 ) = r and ( x, y ) = r ( cos theta, sin theta )#,

#2 = x^2y -x/y^2 + xy^2# converts to

# r^4sin theta cos theta ( cos theta+sin theta )#

#- 2 r - cos theta csc^2theta= 0, r ne 0.#

Graph, from the given Cartesian equation:
graph{ x^2y -x/y^2 + xy^2 - 2 = 0}

As #x and y to 0, parabolic x/y^2 to - 2#.

In polars, as #r to 0, cos theta/sin^2theta to 0 rArr cos theta to 0#

#rArr abs theta to pi/2#