Find the area of the part of the cone$\mathbf{2}{\mathbf{z}}^{\mathbf{2}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}$in the first octant cut out by the planes y = 0 and$\mathbf{y}\mathbf{=}\frac{\mathbf{x}}{\sqrt{\mathbf{3}}}$,and the cylinder

The equation of the cone is$2z^2=x^2+y^2$

The planes are$y=0$and$y=\frac{x}{\sqrt{3}}$

The equation of the cylinder is$x^2+y^2=4$

An intersection curve consists of the common points of two transversally intersecting surfaces. The surface area of a three-dimensional object is the total area of all its faces.

The shaded area is the area of integration.

The equation of the cone is$\varphi (x,y,z)=2z^2-x^2-y^{}$

The secant of the angle is found as follows.

$$\begin{gathered} sec\gamma =\frac{\sqrt{{\left|\nabla \varphi \right|}^2}}{\left|\partial \varphi /\partial z\right|} \\ =\frac{\sqrt{{\left|4z\hat{z}-2x\hat{x}-2y\hat{y}\right|}^2}}{4z} \\ =\frac{\sqrt{16z^2+4x^2+4y^2}}{4z} \end{gathered}$$

From the equation of the cone, we have the following.

$$\begin{gathered} z^2=\frac{1}{2}(x^2+y^2) \\ =\frac{1}{2}{r}^2 \\ \Rightarrow r^2=2z^2 \end{gathered}$$

Substitute this back into angle equation.

$$\begin{gathered} sec\gamma =\frac{\sqrt{16z^2+4x^2+4y^2}}{4z} \\ =\frac{\sqrt{16z^2+4r^2}}{4z} \\ =\frac{\sqrt{6}}{2} \end{gathered}$$

The required area is calculated as below.

$$\begin{gathered} A={\int }_x{\int }_{y}sec\gamma dsdy \\ ={\int }_0^{x/6}d\theta {\int }_0^{2}rdrsec\gamma \\ =\frac{\sqrt{6}}{2}\frac{1}{12}\left(2^2\mathrm{\pi }\right) \\ =\frac{\sqrt{6}}{6}\mathrm{\pi } \end{gathered}$$

Hence, the area of the part of the cone $2z^2=x^2+y^2$in the first octant cut out by the planes $y=0$and$y=\frac{x}{\sqrt{3}}$ , and the cylinder $x^2+y^2=4$is$A=\frac{\sqrt{6}}{6}\mathrm{\pi }$ .