1.18. Calculate the areas of the surfaces:

a) of the cylinder $x^{2}+y^{2}=R^{2}$, bounded by the planes $z=0(z \geqslant 0), z-y=0$

b) of the cylinder $y^{2}+z^{2}=R^{2}, z \geqslant 0$, bounded inside the cylinder $x^{2}+y^{2}=R^{2}$;

c) of the paraboloid $x^{2}+y^{2}=6 z$, bounded inside the cylinder $x^{2}+y^{2}=27$

d) of the sphere $x^{2}+y^{2}+z^{2}=3 a^{2}$, bounded inside the paraboloid $x^{2}+y^{2}=2 a z, a>0$.
The problem is solved. If $z=f(x, y) ;(x, y) \in \Omega_{\text {Hoch }} \subset R_{2}$ is a piecewise-smooth surface, then its area is calculated using the formula (1.83)

$$
\mu(S)=\iint_{\Omega} \sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y
$$

In the general case, when the surface is given by the implicit equation $F(x, y, z)=0$, it is necessary to first determine which of the three coordinate planes it projects onto uniquely and then transition to an explicit equation.

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-066.jpg?height=294&width=435&top_left_y=396&top_left_x=701)

Figure 1.40 of this surface. This process is described in more detail later when solving the given problems.

a) The surface is shown in Figure 1.40.

In this case, the cylindrical surface given implicitly by $x^{2}+y^{2}=R^{2}$ projects uniquely onto the $X O Z$ plane, and its projection $\Omega_{z o X}=\left\{(\dot{x}, z): x^{2}+z^{2} \leqslant R^{2}, z \geqslant 0\right\}$. The explicit equation of this surface is $y=\sqrt{R^{2}-x^{2}}$. We will calculate its area:

$$
\begin{gathered}
\mu(S)=\iint_{\int_{z o x}} \sqrt{1+\left(y_{x}^{\prime}\right)^{2}+\left(y_{z}^{\prime}\right)^{2}} d x d z= \\
=\iint_{\Omega_{z o x}} \frac{R d x d z}{\sqrt{R^{2}-x^{2}}}=R \int_{-R}^{R} d x \int_{0}^{\sqrt{R^{2}-x^{2}}} \frac{d z}{\sqrt{R^{2}-x^{2}}}=2 R^{2}
\end{gathered}
$$

b) The cylinder $y^{2}+z^{2}=R^{2}(z \geqslant 0)$, intersected by the cylinder $x^{2}+y^{2}=R^{2}$, projects uniquely onto the circle $\Omega_{\text {Hoch }}$ : $x^{2}+y^{2} \leqslant R^{2}$ on the $X O Y$ plane. Therefore, the area of this surface is

$$
\begin{aligned}
\mu(S)= & \iint_{\Omega_{X O Y}} \sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y=\iint_{\Omega_{X O Y}} \frac{R d x d y}{\sqrt{R^{2}-y^{2}}}= \\
& =R \int_{-R}^{R} d y \int_{-\sqrt{R^{2}-y^{2}}}^{\sqrt{R^{2}-y^{2}}} \frac{d x}{\sqrt{R^{2}-y^{2}}}=4 R^{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-067.jpg?height=326&width=374&top_left_y=130&top_left_x=170)

Figure 1.41

c) The surface is shown in Figure 1.41.

First, we determine the projection of the paraboloid of revolution $6 z=x^{2}+y^{2}$, intersected by the cylinder $x^{2}+y^{2}=27$, onto the $X O Y$ plane. Clearly, this is the circle $\Omega_{\text {Hoch }}$ : $x^{2}+y^{2} \leqslant 27$, with

$$
0 \leqslant z \leqslant \frac{27}{6}=\frac{9}{2}
$$

We calculate the area

$$
\mu(S)=\iint_{\Omega_{X O Y}} \sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y=\iint_{\Omega_{X O Y}} \sqrt{1+\frac{x^{2}}{9}+\frac{y^{2}}{9}} d x d y
$$

We introduce polar coordinates $x=\rho \cos \varphi ; y=\rho \sin \varphi$;

$$
\begin{gathered}
D=\{(\varphi, \rho): \varphi \in[0,2 \pi], \quad 0 \leqslant \rho \leqslant 3 \sqrt{3}\} . \\
\mu(S)=\iint_{D} \sqrt{1+\frac{\rho^{2}}{9}} \rho d \rho d \varphi= \\
=\int_{0}^{2 \pi} d \varphi \int_{0}^{3 \sqrt{3}} \sqrt{1+\frac{\rho^{2}}{9}} \rho d \rho=\left.2 \pi \cdot 3 \cdot\left(1+\frac{\rho^{2}}{9}\right)^{\frac{3}{2}}\right|_{0} ^{3 \sqrt{3}}=42 \pi .
\end{gathered}
$$

d) To find the projection of the given surface onto the $X O Y$ plane, substitute $x^{2}+y^{2}=2 a z$ into the equation of the sphere $x^{2}+y^{2}+z^{2}=3 a^{2}$. We obtain a quadratic equation in $z$:

$$
z^{2}+2 a z-3 a^{2}=0
$$

From this, we determine $z=a(a \geqslant 0, z \geqslant 0$ by the condition). Therefore, the sphere projects onto the circle $\Omega_{\text {Hoch }}: x^{2}+y^{2} \leqslant 2 a^{2}$. We determine the area

$$
\mu(S)=\iint_{\Omega_{X O Y}} \sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y=\iint_{\Omega_{X O Y}} \frac{\sqrt{3} a d x d y}{\sqrt{3 a^{2}-x^{2}-y^{2}}}
$$

We introduce polar coordinates $x=\rho \cos \varphi ; y=\rho \sin \varphi$; $D=\{(\varphi, \rho): \varphi \in[0,2 \pi], \quad 0 \leqslant \rho \leqslant \sqrt{2} a\}$. We obtain

$$
\begin{gathered}
\mu(S)=\sqrt{3} a \int_{0}^{2 \pi} d \varphi \int_{0}^{\sqrt{2} a} \frac{\rho d \rho}{\sqrt{3 a^{2}-\rho^{2}}}=\left.2 \sqrt{3} a \pi\left(-\sqrt{3 a^{2}-\rho^{2}}\right)\right|_{0} ^{\sqrt{2} a}= \\
=2 \sqrt{3} a \pi\left(\sqrt{3 a^{2}}-a\right)=2 \sqrt{3}(\sqrt{3}-1) \pi a^{2}
\end{gathered}
$$