91Ó°ÊÓ

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$\mathrm{Ï•}(x,y,z)=2z^2-x^2-y^{}$

The secant of the angle is found as follows.

$$\begin{gathered} sec\mathrm{γ}=\frac{\sqrt{{\left|∇\mathrm{ϕ}\right|}^2}}{\left|∂\mathrm{ϕ}/∂z\right|} \\ =\frac{\sqrt{{\left|4z\hat{z}-2x\hat{x}-2y\hat{y}\right|}^2}}{4z} \\ =\frac{\sqrt{16z^2+4x^2+4y^2}}{4z} \end{gathered}$$

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

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

Substitute this back into angle equation.

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

The required area is calculated as below.

$$\begin{gathered} A={∫}_x{∫}_{y}sec\mathrm{γ}dsdy \\ ={∫}_0^{x/6}d\mathrm{θ}{∫}_0^{2}rdrsec\mathrm{γ} \\ =\frac{\sqrt{6}}{2}\frac{1}{12}\left(2^2\mathrm{π}\right) \\ =\frac{\sqrt{6}}{6}\mathrm{π} \end{gathered}$$

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$A=\frac{\sqrt{6}}{6}\mathrm{Ï€}$ .