Region of integration for double integral:

Question

When tasked with reversing the order of integration in the following integral:

$$\int^1_0\int^{y^2}_{-y}x\ \mathrm{d}x\mathrm{d}y$$

It seems sensible to sketch out our situation:

we have that $-y\le x\le y^2$ and that $0 \le y \le 1$ which gives the following enclosed region:

However, where is the fallacy in writing:

$$\int^1_0\int^{y^2}_{-y}x\ \mathrm{d}x\mathrm{d}y = \int^1_0\left[\int^{0.1}_{-y}x\ \mathrm{d}x + \int^{y^2}_{0.1}x\ \mathrm{d}x\right]\mathrm{d}y$$

(Where 0.1 is chosen arbitrarily) which will then give an enlarged region and thus a larger answer:

This seems clearly wrong, but there's something I'm missing when we have both upper and lower limits of an "inner" integral being a function of $y$. If anyone can spot the exact assumption I'm making which is wrong, that would be appreciated.

Answer 1

The outer integral is evaluated for $y \in [0,1]$, and the inner integral's limits are on $x \in [-y, y^2]$, which you partition into $x \in [-y, 0.1] \cup (0.1, y^2]$. So if $y \in [0, \sqrt{0.1})$, the integral over the second interval is negative, because the upper limit of integration is smaller than the lower limit of integration. For example, if $y = 0.2$, then $y^2 = 0.04 < 0.1$, and the two inner integrals in your expression become $$\begin{align} \int_{x=-y}^{0.1} x \, dx + \int_{x=0.1}^{y^2} x \, dx &= \int_{x=-0.2}^{0.1} x \, dx + \int_{x=0.1}^{0.04} x \, dx \\ &= \int_{x=-0.2}^{0.1} x \, dx - \int_{x=0.04}^{0.1} x \, dx \\ &= \int_{x=-0.2}^{0.04} x \, dx. \end{align}$$ This resolves your contradiction.

Answer 2

If you switch the order, the integral is given by $$ \int_{-1}^0\left(\int_{-x}^1 x dy\right) dx +\int_0^1\left(\int_{\sqrt{x}}^1xdy\right) dx $$

Answer 3

I think your second image is actually showing the integral of:

$$\int^1_0\left[\int^{0}_{-y}x\ \mathrm{d}x + \int^{y^2+0.1}_{0}x\ \mathrm{d}x\right]\mathrm{d} \neq \int^1_0\int^{y^2}_{-y}x\ \mathrm{d}x\mathrm{d}yy$$

I think rather than shifting $$\mathbb{x = y^2}$$ you need to add a line $$\mathbb{x}=0.1$$ which splits R vertically