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?
 A: 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
$$A=\int_{b/a}^0ydx=\int_{b/a}^0y(m)\frac{dx}{dm}dm=\int_{b/a}^0m\frac{1/a-m/b}{\big(1/a+m/b\big)^2}\frac{a^2b(a\,m-3b)}{(am+b)^3}dm=\frac{ab}{12}$$
A: $X=\frac{x}{a}=r\cos \theta\\ Y=\frac{y}{b}=r\sin \theta$
The curve becomes $(X+Y)^2=X-Y$ or $R^2(cos^2\theta+sin^2\theta+2\cos \theta \sin \theta)=R(\cos \theta -\sin \theta)$ which becomes $R=\displaystyle{\frac{\cos \theta -\sin \theta}{1+2\cos \theta \sin \theta}}$.
area$\displaystyle{=\int_0^{\frac{\tau}{2}=\pi}\int_0^R}ab \ \mathrm{d}r\mathrm{d}\theta= ab \int_0^{\frac{\tau}{2}}R \ \mathrm{d}\theta= ab \left[ \frac{-1}{\sin \theta + \cos \theta }\right]_0^{\frac{\tau}{2}}=2ab$
