Find the area: $(\frac xa+\frac yb)^2 = \frac xa-\frac yb , y=0 , b>a$ Find the area: $$\left(\frac xa+\frac yb\right)^2 = \frac xa-\frac yb,$$ $ y=0 , b>a$
I work in spherical coordinates. 
$x=a\cdot r\cdot \cos(\phi)\;\;,$
$y=b\cdot r\cdot \cos(\phi)$
Then I get the equation and don't know to do with, cause "a" and "b" are dissapearing.
For what are the conditions: $y=0, b>a?$..How to define the limits of integration?

Your shape for $a=1$, $b=2$ is as below

It is much easier if you would use line parametrization rather than polar coordinates. Let $y=m\,x$ then
$$\bigg(\frac xa+\frac{mx}b\bigg)^2=\frac xa-\frac {mx}b\Rightarrow x(m)=\frac{1/a-m/b}{\big(1/a+m/b\big)^2}$$
and
$$y(m)=m\,x(m)=m\frac{1/a-m/b}{\big(1/a+m/b\big)^2}$$
To find the limits you must determine where $y(m)$ becomes zero due to boundary of $y=0$
$$y(m)=m\frac{1/a-m/b}{\big(1/a+m/b\big)^2}\Rightarrow m=0\text{ and }m=b/a$$
Since $x(m)$ is zero as $m=b/a$ the integration will be from $b/a$ to $0$. Therefore
$$A=\int_{b/a}^0ydx=\int_{b/a}^0y(m)\frac{dx}{dm}dm=\int_{b/a}^0m\frac{1/a-m/b}{\big(1/a+m/b\big)^2}\frac{a^2b(a\,m-3b)}{(am+b)^3}dm=\frac{ab}{12}$$
