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In the original order of integration,In the reversed order, the below-axis contribution is ${\int }_{-1}^{0}y(1+y)dy=-1/6$, and the above-axis contribution is ${\int }_0^{1}y(1-\sqrt{y})dy=1/10$. Their sum is again $-1/15$. A negative answer is possible because the integrand $y$ changes sign; the integral is not the area of the region.

The integration region lies between the line $y=x$ and the parabola $y=x^2$, with $-1\le x\le 0$. It has a portion below the $x$-axis bounded above by $y=0$, and a portion above it bounded below by $y=0$. Horizontal slices change form at $y=0$; keeping that split is essential when reversing the order of integration.