91Ó°ÊÓ

The equation of the cone is${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{z}}^2$

The equation of the disk is${(x+1)}^2+y^2≤1$

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 area of integration is the shaded region given in the diagram below.

The equation of the cone is$\mathrm{Ï•}=z^2-x^2-y^2$

The secant of the angle is found as follows.

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

The angle of the cone is$\mathrm{Ï€}/4$

So, the angle will be as follows.

$$\begin{gathered} sec\mathrm{γ}=sec\frac{\mathrm{π}}{4} \\ =\sqrt{2} \end{gathered}$$

The required area is calculated as below.

$$\begin{gathered} A={∫}_x{∫}_{y}sec\mathrm{γ}dxdy \\ =\sqrt{2}{∫}_{x/2}^{x/2}d\mathrm{θ}{∫}_0^{2\cos \mathrm{θ}}rdr \\ =\sqrt{2}{∫}_{x/2}^{x/2}d\mathrm{θ}\frac{1}{2}{\left(2\cos \mathrm{θ}\right)}^2 \\ =2\sqrt{2}{∫}_{x/2}^{x/2}{\cos }^2\mathrm{θ}d\mathrm{θ} \\ =\sqrt{2}{∫}_{x/2}^{x/2}(1+\cos (2\mathrm{θ}\left)\right)d\mathrm{θ} \\ =\sqrt{2}\mathrm{π} \end{gathered}$$

Hence, the area of the part of the cone $x^2+y^2=z^2$that is over the disk ${\left(x-1\right)}^2+y^2≤1$is$A=\sqrt{2}\mathrm{π}$ .