Example 5. Find the flux of the vector field $\mathbf{a} = \boldsymbol{y}^{2} \mathbf{j} + z \mathbf{k}$ through the part of the surface $z = x^{2} + y^{2}$, cut off by the plane $z = 2$. The normal is taken outward with respect to the region bounded by the paraboloid.

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-24.jpg?height=285&width=248&top_left_y=171&top_left_x=106)
Solution. The given surface (a paraboloid of revolution) is projected one-to-one onto the plane $x O y$ as a circle $D_{x y}$ (Fig. 19). We find the unit normal $\boldsymbol{n}^{0}$ to the surface $S$:

$$
n^{\prime \prime}=\frac{\operatorname{grad}\left(z-x^{2}-y^{2}\right)}{\left|\operatorname{grad}\left(z-x^{2}-y^{2}\right)\right|}= \pm \frac{-2 x \mathbf{i}-2 y \mathbf{j}+\mathbf{k}}{\sqrt{4 x^{2}+4 y^{2}+1}}
$$

According to the problem, the normal $\mathbf{n}^{0}$ forms an obtuse angle $\gamma$ with the $O z$ axis, so we should take the negative sign in the fraction. Thus,

$$
n^{0}=\frac{2 x \mathbf{i}+2 y \mathbf{j}-\mathbf{k}}{\sqrt{4 x^{2}+4 y^{2}+1}}
$$

Fig. 19

From this,

$$
\cos \gamma=-\frac{1}{\sqrt{4 x^{2}+4 y^{2}+1}}<0
$$

and, therefore,

$$
d S=\frac{d x d y}{|\cos \gamma|}=\sqrt{4 x^{2}+4 y^{2}+1} d x d y
$$

We find the scalar product

$$
\left(a, n^{0}\right)=\frac{2 y^{3}-z}{\sqrt{4 x^{2}+4 y^{2}+1}}
$$

The sought flux, by formula (I), is

$$
\mathrm{n}=\iint_{S}\left(2, \mathrm{n}^{0}\right) d S=\left.\iint_{D_{x y}}\left(2 y^{3}-z\right)\right|_{z=x^{2}+y^{2}} d x d y=\iint_{D_{x y}}\left(2 y^{3}-y^{2}-x^{2}\right) d x d y
$$

The integration region $D_{x y}$ is a circle centered at the origin with radius $\boldsymbol{R}=\sqrt{2}$. Introducing polar coordinates $\boldsymbol{x}=\rho \cos \varphi, y=\rho \sin \varphi$, we have

$$
\begin{aligned}
& \mathrm{II}=\iint_{D_{x y}}\left(2 \rho^{3} \sin ^{3} \varphi-\rho^{2}\right) \rho d \rho d \varphi= \\
&=\int_{0}^{2 \pi} d \varphi \int_{0}^{\sqrt{2}}\left(2 \rho^{4} \sin ^{3} \varphi-\rho^{3}\right) d \rho=-\left.2 \pi \frac{\rho^{4}}{4}\right|_{0} ^{\sqrt{2}}=-2 \pi
\end{aligned}
$$