Order of Integration for $\int_{0}^{2}\int_{x^2}^{x}y^2xdydx$

Question

I am trying to evaluate the integral $$I=\int_{0}^{2}\int_{x^2}^{x}y^2xdydx$$ using change of order of integration.Basically the region is $$R=\left\{(x,y):0\leq x\leq 2,x^2 \leq y\leq x\right\}$$ I have written the integral as: $$I=\int_{0}^{1}\int_{x^2}^{x}y^2xdydx-\int_{1}^{2}\int_{x}^{x^2}y^2xdydx$$ The first integral above is quite easy to change the order which is given by: $$\int_{0}^{1}\int_{y}^{\sqrt{y}}y^2xdxdy$$ But how can we do for the second integral?

Answer 1

The change of order of integration makes sense here if it is an exercise where you are supposed to do so otherwise the original order of integration is simpler. The second integral is

$ \displaystyle \int_{1}^{2}\int_{x}^{x^2}y^2x ~ dy ~ dx$

When we change the order of integration -

Given $1 \leq x \leq 2$, $x \leq y \leq x^2$ gives us $1 \leq y \leq 4$.

For $1 \leq y \leq 4$, $\sqrt y \leq 2$ and $\sqrt y \leq y$ so the lower bound of $x$ is simply $\sqrt y$ for the entire region. Now for $1 \leq y \leq 2$, upper bound of $x = y$ works but for $y \geq 2$, $x = y$ would not work as we must have $x \leq 2$. So we can write the integral as,

$ \displaystyle \int_1^4 \int_{\sqrt y}^{min(y, 2)} y^2 x ~ dx ~ dy$

and it splits into two integrals,

$ = \displaystyle \int_1^2 \int_{\sqrt y}^y y^2 x ~ dx ~ dy + \int_2^4 \int_{\sqrt y}^2 y^2 x ~ dx ~ dy$

If you sketch the region, it becomes simpler to visualize. The shaded region is the second integral.