945. Find the flux of the vector field $\bar{p}=x \bar{i}-y^{2} \bar{j}+\left(x^{2}+z^{2}-1\right) \bar{k}$ through the surface $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{3}}{c^{2}}=1$ (ellipsoid) from inside this surface.
Solution. According to formula (1)

$$
K=\oiint_{+\sigma} x d y d z-y^{2} d x d z+\left(x^{2}+z^{2}-1\right) d x d y
$$

We decompose this surface integral (of the second type) into three component integrals and, using the given equation of the ellipsoid (b), reduce their computation to the computation of double integrals.

1) $K_{1}=\oiint_{+\sigma} x d y d z=\iint_{\sigma_{1}} x d y d z+\iint_{\sigma_{z}} x d y d z$

where $\sigma_{1}$ and $\sigma_{2}$ are parts of the given ellipsoid located on opposite sides of the $y O z$ plane (see Fig. 98), which have
different explicit equations:

$$
x_{\sigma_{1}}=-a \sqrt{1-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}} \quad \text { and } \quad x_{\sigma_{1}}=a \sqrt{1-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}}.
$$

Transforming these surface integrals into double integrals (using the formula specified in § 11 of the previous chapter), we get:

$$
\iint_{\sigma_{1}} x d y d z=-\iint_{\left(\sigma_{\sigma_{1}}\right)_{y z}}-a \sqrt{1-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}} d y d z
$$

since the surface $\sigma_{1}$ is oriented towards the negative direction of the $O x$ axis;

$$
\iint_{\sigma_{2}} x d y d z=\iint_{\left(\sigma_{2}\right)_{y z}} a \sqrt{1-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}} d y d z
$$

since the surface $\sigma_{2}$ is oriented towards the positive direction of the $O x$ axis.

The projections $\left(\sigma_{1}\right)_{y z}$ and $\left(\sigma_{2}\right)_{y z}$ of the surfaces $\sigma_{1}$ and $\sigma_{2}$ onto the $y O z$ plane represent the same ellipse $\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} \leqslant 1$. Therefore,

$$
\begin{aligned}
K_{1} & =2 a \int_{\left(\sigma_{1}\right)_{y z}} \sqrt{1-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}} d y d z= \\
& =2 a \int_{-b}^{b} d y \int_{-z_{1}}^{z_{1}} \sqrt{1-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}} d z
\end{aligned}
$$

where $z_{1}$ is the positive value of $z$ from the equation of the ellipse $\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$.

Evaluating the double integral, we find $K_{1}=\frac{4}{3} \pi a b c$. (The inner integral can be easily found using formula (B), Chapter IV, § 5, by setting $\left.a=\sqrt{1-\frac{y^{2}}{b^{2}}}, t=\frac{2}{c} \cdot\right)$

$$
\text { 2) } K_{2}=\oiint_{+\sigma} y^{2} d x d z=\iint_{\sigma_{3}} y^{2} d x d z+\iint_{\sigma_{4}} y^{2} d x d z
$$

where $\sigma_{3}$ and $\sigma_{4}$ are parts of the surface $\sigma$ located on opposite sides of the $x O z$ plane, with equations

$$
\begin{gathered}
y_{\sigma_{3}}=-b \sqrt{1-\frac{x^{2}}{a^{2}}-\frac{z^{2}}{c^{2}}} \text { and } y_{\sigma_{1}}=b \sqrt{1-\frac{x^{2}}{a^{2}}-\frac{z^{2}}{c^{2}}} \\
-335-
\end{gathered}
$$

Transforming the surface integrals into double integrals, we get

$$
\begin{aligned}
K_{2} & =-\iint_{\left(\sigma_{3}\right)_{x z}} b^{2}\left(1-\frac{x^{2}}{a^{2}}-\frac{\xi^{2}}{c^{2}}\right) d x d z+ \\
& +\iint_{\left(\sigma_{A_{1}}\right)_{x z}} b^{2}\left(1-\frac{x^{2}}{a^{2}}-\frac{z^{2}}{c^{2}}\right) d x d z=0
\end{aligned}
$$

since the projections $\left(\sigma_{3}\right)_{x z}$ and $\left(\sigma_{4}\right)_{x z}$ of the surfaces $\sigma_{3}$ and $\sigma_{4}$ onto the $x O z$ plane are the same.

3) For the same reason, due to the evenness of the integrand of the surface integral $K_{3}$ and the symmetry of the surface $\sigma$ relative to the $x O y$ plane,

$$
K_{3}=\oiint_{+\sigma}\left(x^{2}+z^{2}-1\right) d x d y=0
$$

Therefore, $K=K_{1}-K_{2}+K_{3}=\frac{4}{3} \pi a b c$.

By the Ostrogradsky-Gauss formula, this problem is solved more simply: we find the divergence of the field

$$
\operatorname{div} \vec{p}=(x)_{x}^{\prime}+\left(-y^{2}\right)_{\prime \prime}^{\prime}+\left(x^{2}+z^{2}-1\right)_{z}^{\prime}=1-2 y+2 z
$$

and substitute it into formula (3):

$$
K=\iiint_{i} \operatorname{div} p d v=\iiint_{i} \int_{1}(1-2 y+2 z) d x d y d z
$$

where the region $G$ is the ellipsoid $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} \leqslant 1$.

We decompose the resulting triple integral:

$$
K=\iiint_{G} d x d y d z-2 \iint_{C_{r z}} d x d z \int_{-y_{1}}^{y_{1}} y d y+2 \iint_{G_{x y}} d x d y \int_{-z_{1}}^{z_{1}} z d z
$$

where

$$
y_{1}=b \sqrt{1-\frac{x^{2}}{a^{2}}-\frac{2^{2}}{c^{2}}}, \quad z_{1}=c \sqrt{1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}}
$$

The first integral is equal to the volume of the region $G$, i.e., the volume of the ellipsoid $K_{1}=\frac{4}{3} \pi a b c$ (Chapter V, §4).

The second and third integrals are zero, because their specified inner simple integrals are zero, as integrals of odd functions (Chapter V, § 2).

Therefore, as in the first solution, $K=\frac{4}{3} \pi a b c$.