2.16. Compute the surface integral of the second kind

$$
\iint_{S} x d y d z+y d z d x+z d x d y
$$

where $S--$ the inner side of the ellipsoid $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 ;$
$z \in[-H, H];$
Solution. The ellipsoid $S$ can be written in parametric form as:

$$
\begin{aligned}
& x=a \sin \theta \cos \varphi \\
& y=b \sin \theta \sin \varphi, \quad \theta \in[0, \pi], \varphi \in[0,2 \pi] \\
& z=c \cos \theta
\end{aligned}
$$

We calculate the tangent vectors

$$
\tau_{\theta}=\left(x_{\theta}, y_{\theta}, z_{\theta}\right), \quad \tau_{\varphi}=\left(x_{\varphi}, y_{\varphi}, z_{\varphi}\right),
$$

the element of area $d S=\left|\tau_{\theta} \times \tau_{\varphi}\right| d \theta d \varphi$, and the normal vector

$$
\ddot{n}= \pm \frac{\tau_{\theta} \times \tau_{\varphi}}{\left|\tau_{\theta} \times \tau_{\varphi}\right|}
$$

(the sign is chosen depending on the side of the ellipsoid). We obtain:

$$
\begin{aligned}
& \widetilde{\tau_{\theta}}=(a \cos \theta \cos \varphi, b \cos \theta \sin \varphi, c \sin \theta) \\
& \cdots \tau_{\varphi}=(-a \sin \theta \sin \varphi, b \sin \theta \cos \varphi, 0) \\
& \tau_{\theta} \times \tau_{\varphi}=\left(b c \sin ^{2} \theta \cos \varphi, a c \sin ^{2} \theta \sin \varphi, a b \sin \theta \cos \theta\right)
\end{aligned}
$$

To calculate the surface integral of the second kind, we use formula (2.54), which essentially reduces it to a surface integral of the first kind. In our case, $P=x$, $Q=y$, $R=z$,

$$
\ddot{n} d S=-\frac{\overrightarrow{\tau_{\theta}} \times \overrightarrow{\tau_{\varphi}}}{\left|\overrightarrow{\tau_{\theta}} \times \overline{\tau_{\varphi}}\right|} \cdot\left|\dddot{\tau_{\theta}} \times \tau_{\varphi}\right| d \theta d \varphi=
$$

$=-\left(b c \sin ^{2} \theta \cos \varphi d \theta d \varphi, a c \sin ^{2} \theta \sin \varphi d \theta d \varphi, a b \sin \theta \cos \theta d \theta d \varphi\right)$.

(The minus sign corresponds to the inner side of the ellipsoid.) Thus, we obtain:

$$
\iint_{S} x d y d z+y d z d x+z d x d y=
$$

$$
\begin{aligned}
& =-a b c \iint_{\substack{0 \leqslant \theta \leqslant 2 \pi \\
0 \leqslant \varphi \leqslant 2 \pi}}\left(\sin ^{3} \theta \cos ^{2} \varphi+\sin ^{3} \theta \sin ^{2} \varphi+\sin \theta \cos ^{2} \theta\right) d \theta d \varphi= \\
& =a b c \int_{0}^{2 \pi} \cos ^{2} \varphi d \varphi \int_{0}^{\pi} \sin ^{2} \theta d(\cos \theta)+a b c \int_{0}^{2 \pi} \sin ^{2} \varphi d \varphi \int_{0}^{\pi} \sin ^{2} \theta d(\cos \theta)+
\end{aligned}
$$

$$
\begin{aligned}
& +2 \pi a b c \int_{0}^{\pi} \cos ^{2} \theta d(\cos \theta)=a b c \int_{0}^{2 \pi} \frac{1+\cos 2 \varphi}{2} d \varphi \int_{1}^{-1}\left(1-t^{2}\right) d t+ \\
& +a b c \int_{0}^{2 \pi} \frac{1-\cos 2 \varphi}{2} d \varphi \int_{1}^{-1}\left(1-t^{2}\right) d t=2 \pi a b c \int_{1}^{-1} t^{2} d t=-4 \pi a b c
\end{aligned}
$$