Need a Intuitive understanding of solving double integral where integrand has max function I'll try my best to keep the question brief. So I came upon this question:
$$\int \int _R^{ }\cos \left(\max \left\{x^3,\ y^{\frac{3}{2}}\right\}\right)dA$$ where R region is $[0,1]\times \ [0,1]$. This particular example has already been asked like 3 times. But none of them answered me completely. The way how its solved is like this:
you have max function = $max({x^3,y^{\frac{3}{2}})}$, which means either $x^3$ is biggest or $y^{\frac{3}{2}}$ is the biggest, which can be represented as $x^3>y^{\frac{3}{2}}$ $\Rightarrow $$x^2>y$ OR
$y^{\frac{3}{2}}>x^3$ $\Rightarrow $ $\sqrt{\ y}>x\ $
from this we can say that $y=x^2$ ; Now our region is already given by $[0,1]\times \ [0,1]$ so we divide the region R like this:
My problem is that our limit is clearly given as $y=0, y=1, x=0, x=1$ first of all why is there a need to divide our region R into 2? and 2nd, the Max function is inside our integrand, how are we taking that from integrand and using it in the limits? I mean the integrand $f(x,y) =z$ usually means its the top surface of the volume, so how is that hindering the limits on x-y plane? I want to understand it intuitively and pictorially/graphically instead of just memorizing this method of solving. Any help is greatly appreciated.

The $\max$ function can't be directly integrated. The integrand has different definitions above and below the line $y=x^2$ so you will have to split the integral into two integrals, one over the region under $y=x^2$ and the other over the region above it. You have$$z(x,y)=\max\{x^3,y^{3/2}\}=\begin{cases}x^3,&y<x^2\\y^{3/2},&y\ge x^2\end{cases}$$The required integral is$$\begin{align*}I&=\int_0^1\left[\int_0^{x^2} x^3~dy+\int_{y^2}^1 y^{3/2}~dy\right]~dx\\&=\underbrace{\int_0^1\int_0^{x^2} x^3~dy~dx}_{\text{below }y=x^2 }+\underbrace{\int_0^1\int_{y^2}^1 y^{3/2}~dy~dx}_{\text{above }y=x^2}\end{align*}$$
