Bounds for a region of integration

Question

I want to determine the bounds for integral $\displaystyle\int\int_R f(x,y)dxdy$, or $\displaystyle\int\int_R f(x,y)dydx$ so that I change the order of the iteration. $R$ is the region lying on the second and third quadrants bounded by $y=2-x^2, y=x$ and $x=0$.

Here is how I think:

Let us first start iterating by $x$. I think that we must divide the region into two parts, namely $R_1$ is the region in the second quadrant and $R_2 $ is the third quadrant. Otherwise there would be a problem. So

Integral on $R_1$ is $\displaystyle\int_{0}^{2}\int_{-\sqrt {2-y}}^{0} fdxdy$ and the integral on $R_2$ is $\displaystyle \int_{-2}^{0}\int_{-\sqrt{2-y}}^{y}f dxdy$.

Now if we start from $y$ if I am not wrong, we have the following: $\displaystyle\int_{-2}^{0}\int_{x}^{2-x^2} fdydx$.

Is there any mistake? If so, can you explain me how to write the integral in the two forms? Thanks.

Answer 1

This looks good except for one small (sign) mistake; added in red:

Integral on $R_1$ is $\displaystyle\int_{0}^{2}\int_{ \color{red}{-} \sqrt {2-y}}^{0} fdxdy$ and the integral on $R_2$ is $\displaystyle \int_{-2}^{0}\int_{-\sqrt{2-y}}^{y}f dxdy$.

Note that in the region of integration, $x$ is always negative.