In chapter 13 of Stewart's calculus the integral $$\iiint_A 3y \, dx \,dy \,dz,$$ where $A$ is the region bounded by the parabolic cylinder $z = 1-x^2$ and the planes $z = 0,$ $y = 0$ and $y+z=2,$ is computed by projecting $A$ on the $XZ$ plane, which yields $184/35$ as a result.
Now, I've tried to compute the integral by using the projection on the $XY$ plane, but I get a different result. I guess the way I've set up the integral is mistaken, but I can't see why. It seems to me that the projection of $A$ on the $XY$ plane is $\pi_{XY}(A) = [-1, 1] \times [0, 2]$ and that, for each $(x, y) \in \pi_{XY}(A),$ the corresponding section is $$ A_{(x, y)} = \begin{cases} [0, 1-x^2] &\text{ if } 0\leq y \leq 1, \\ [0, 2-y] &\text{ if } 1< y \leq 2.\end{cases}$$ Hence $$ \iiint_A 3y \, dx \, dy \, dz = \int_{-1}^1 \int_0^1 \int_0^{1-x^2}3y \, dz \, dy \, dx+\int_{-1}^1 \int_1^2 \int_0^{2-y} 3y \, dz \, dy \, dx.$$
Edit: There must definitely be something wrong with the limits of integration because my section is not continuous at $y=1,$ but I still can't picture the situation clearly.
You are right about the orthogonal projection on the $XY$ plane. But then your integral should be$$\int_{-1}^1\int_0^1\int_0^{1-x^2}3y\,\mathrm dz\,\mathrm dy\,\mathrm dx+\int_{-1}^1\int_1^2\int_0^{\min\{1-x^2,2-y\}}3y\,\mathrm dz\,\mathrm dy\,\mathrm dx.$$It is easy to check that$$\int_{-1}^1\int_0^1\int_0^{1-x^2}3y\,\mathrm dz\,\mathrm dy\,\mathrm dx=2.$$On the other hand\begin{align}\int_{-1}^1\int_1^2\int_0^{\min\{1-x^2,2-y\}}3y\,\mathrm dz\,\mathrm dy\,\mathrm dx&=\int_{-1}^1\int_1^23y\min\{1-x^2,2-y\}\,\mathrm dy\,\mathrm dx\\&=\int_{-1}^1\int_1^{1+x^2}3y(1-x^2)\,\mathrm dy\,\mathrm dx+\\&{}\quad+\int_{-1}^1\int_{1+x^2}^23y(2-y)\,\mathrm dy\,\mathrm dx\\&=\int_{-1}^1-\frac{3 x^6}{2}-\frac{3 x^4}{2}+3 x^2\,\mathrm dx+\\&{}\quad+\int_{-1}^1x^6-3 x^2+2\,\mathrm dx\\&=\int_{-1}^1-\frac{x^6}{2}-\frac{3 x^4}{2}+2\,\mathrm dx\\&=\frac{114}{35}.\end{align}And $2+\frac{114}{35}=\frac{184}{35}$.