Triple integral volume by equations I have trouble setting up a triple integral to find volume bound by equations, such as:
$$z = x^2 + 3;\quad y = 3 - x^2;\quad x + y = 2;\quad z = 0.$$
I'm not sure how to figure how to find the boundaries for the $z$-axis. How would I go about setting up the triple integral?

You have $y=3-x^2$ and $x+y=2$. Set them equal to each other and find the intersection points. Namely, $3-x^2=2-x$ gives $x=\frac{1\pm\sqrt{5}}2$. Our lower $x$-limit is $\frac{1-\sqrt{5}}2$ and our upper $x$-limit is $\frac{1+\sqrt{5}}2$.
Furthermore, $3-x^2 \ge 2-x$ on the interval $[\frac{1-\sqrt{5}}2,\frac{1+\sqrt{5}}2]$. Our lower $y$-limit is $2-x$ and our upper $y$-limit is $3-x^2$.
We also see that our lower $z$-limit is $0$ and our upper $z$-limit is $x^2+3$.
Our volume $V$ is therefore represented by the triple integral $$V=\int_\frac{1-\sqrt{5}}2^\frac{1+\sqrt{5}}2 \int_{2-x}^{3-x^2} \int_0^{x^2+3} dz \, dy \, dx.$$
