Parametrization of $x^2+y^2-ay=0$

Question

I am to find the circulation of $$y^2 dx + x^2 dy$$ along the (counterclockwise) path $$\Gamma : x^2+y^2-ay = 0$$ both with and without using Green's theorem. Apparently, $\Gamma$ is supposed to describe a circle but It is confusing because the radius would depend on $y$ ...

So, how should I look at this ? How can I find a parametrization ?

Answer 1

If $a\ne0,$ $$x^2+\left(y-\dfrac a2\right)^2=\left(\dfrac a2\right)^2\implies\left(\dfrac x{\dfrac a2}\right)^2+\left(\dfrac{y-\dfrac a2}{\dfrac a2}\right)^2=1$$

Use $\cos^2\theta+\sin^2\theta=1$

Answer 2

Hint:

$$x^2 + y^2 - ay = x^2 + (y-\frac a2)^2 - \frac{a^2}{4}$$

Meaning that $\Gamma$ is the circlle defined by the equation $$x^2 + (y-\frac a2)^2 = \left(\frac{a}{2}\right)^2$$

Answer 3

The equation of your curve rewrites as follows:(just complete the square)

$$x^2+\left(y-\frac{a}{2}\right)^2=\frac{a^2}{4}$$

Here's a circle centered at $(0,\frac{a}{2})$ of radius $\frac{a}{2}$

A parametrisation of that circle is

$$\begin{cases} x=\frac{a}{2}\cos\theta\\y=\frac{a}{2}\left(1+\sin\theta\right)\end{cases}$$

Can you take it from here?