849. Compute the triple integral $I=\int_{-1}^{1} d x \int_{x^{2}}^{1} d y \int_{0}^{z}(4+z) d z$ and sketch its region of integration.
Solution. We sequentially compute three ordinary (single) definite integrals, starting with the inner one:

$$
\begin{gathered}
I_{1}=\int_{0}^{2}(4+z) d z=\left.\frac{(4+z)^{2}}{2}\right|_{0} ^{2}=\frac{36-16}{2}=10 \\
I_{2}=\int_{x^{2}}^{1} I_{1} d y=10 \int_{x^{2}}^{1} d y=\left.10 y\right|_{x^{2}} ^{1}=10\left(1-x^{2}\right) ; \\
I_{3}=I=\int_{-1}^{1} I_{2} d x=10 \int_{-1}^{1}\left(1-x^{2}\right) d x=\left.10\left(x-\frac{x^{3}}{3}\right)\right|_{-1} ^{1}=\frac{40}{3} .
\end{gathered}
$$

Here, as when computing a double integral, a more concise notation can be used:

$$
\begin{gathered}
I=\left.\int_{-1}^{1} d x \int_{x^{2}}^{1} \frac{(4+2)^{2}}{2}\right|_{0} ^{2} d y=10 \int_{-1}^{1} d x \int_{x^{2}}^{1} d y=10 \int_{-1}^{1}\left(1-x^{2}\right) d x= \\
=\left.10\left(x-\frac{x^{3}}{3}\right)\right|_{-1} ^{1}=\frac{40}{3}
\end{gathered}
$$

To construct the region of integration for this triple integral, we first write the equations of the surfaces that bound this region. By equating the variable of integration of each integral to its limits, we obtain the following equations: $x=-1, x=1, y=x^{2}, y=1, z=0, z=2$.

By plotting in the coordinate system $0 x y z$ the surfaces corresponding to these equations (Fig. 173), we see that the region bounded by them is a right cylinder, the generators of which are parallel to the $0 z$ axis.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-288.jpg?height=454&width=375&top_left_y=1018&top_left_x=823)