Find the area of $A = \{ \langle x,y\rangle \in \mathbb{R}^2 \mathrel| (x+y)^40 \}$? I can't really think of how to set the limits
 A: there are two ways to find the area


*

*use wolframalpha,you can see the picture at once.

*let $x=rcos(\theta), y=rsin(\theta)$, then you get $r \le \dfrac{a*cos^2(\theta)sin(\theta)}{(cos(\theta)+sin(\theta))^4}$ 
since $x>0 \to -\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$,I think you can go further this way.
Edit:$ay>0$, if $a>0 \to y>0 \to 0< \theta < \dfrac{\pi}{2}$
the shape is here
EDIT: I add the process also for reference:
$S=\dfrac{1}{2} \int_{0}^ {\frac{\pi}{2}} r^2 (\theta)d \theta$  and
$ \int r^2 (\theta)d \theta=\int \left[\dfrac{a*cos^2(\theta)sin(\theta)}{(cos(\theta)+sin(\theta))^4} \right]^2 d \theta=a^2*\dfrac{-280 sin(\theta)-210 sin(3\theta)+14 sin(5\theta)-385 cos(\theta)+147 cos(3 \theta)+105 cos(5 \theta)+13 cos(7 \theta)}{6720 (sin(\theta)+cos(\theta))^7}+C$
$S=a^2*\dfrac{-280+210+14-(-385+147+105+13 )}{2*6720}=\dfrac{a^2}{210}$
A: You can parameterize the equation such as
$$x=x(t)$$
$$y=t*x(t)$$
and by replacing to the original equation
$$\big(x(t)+t\,x(t)\big)^4=a\,x(t)^2\,t\,x(t)\Rightarrow x(t)^4(1+t)^4=a\,t\,x(t)^3\Rightarrow x(t)=\frac{a\,t}{(1+t)^4}$$
The limits are $0$ and $\infty$. By Greens therom the area enclosed by the parametric curve is given by
$$A=\frac 12\int_0^{\infty}\bigg(x\,y'-y\,x'\bigg)dt=\frac 12\int_0^{\infty}\bigg(x\,(t\,x)'-t\,x\,x'\bigg)dt=\frac 12\int_0^{\infty}\bigg(x\,(x+t\,x')-t\,x\,x'\bigg)dt$$
$$A=\frac 12\int_0^{\infty}x^2dt=\frac 12\int_0^{\infty}\bigg(\frac{a\,t}{(1+t)^4}\bigg)^2dt$$
This is an improper integral and
$$A=\lim_{b \to \infty}\frac 12\int_0^b\bigg(\frac{a\,t}{(1+t)^4}\bigg)^2dt=\frac 12\lim_{b \to \infty}\bigg(-a^2\frac{1+7b+21b^2}{105(1+b)^7}+\frac{a^2}{105}\bigg)=\frac{a^2}{210}$$
