Reverse the order of integration of $\int_{-1}^4\int_{y-4}^{4y-y^2} (y+1)dxdy$ and then evaluate

Question

The answer key says that the original integral is equivalent to the following sum of iterated integrals:

$$\int_{-5}^0\int_{2-\sqrt{4-x}}^{x+4} \ (y+1) \ dydx+ \int^4_0 \int_{2-\sqrt{4-x}}^{2+\sqrt{4-x}} \ (y+1) \ dydx$$

But where do these limits come from?!?

I understand why $[-5,4]$ needs to be split up into two (the region is not vertically simple otherwise) and why $y \leq x+4$ when $-5\leq x \leq 0$. But how can we determine the other limits in terms of the inequality $(1)$?

$$4-y \leq x\leq 4y-y^2 \tag {1}$$

Answer 1

It is the region bounded by the line from the picture below. So:

• If $-5\leqslant x\leqslant0$, then $2-\sqrt{4-x}\leqslant y\leqslant x+4$;
• If $0\leqslant x\leqslant4$, then $2-\sqrt{4-x}\leqslant y\leqslant2+\sqrt{4-x}$.