Find the area of the part of the cone${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}{\mathbf{z}}^{\mathbf{2}}$that is over the disk${\mathbf{(}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^2\mathbf{\le }\mathbf{1}$

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

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

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$\varphi =z^2-x^2-y^2$

The secant of the angle is found as follows.

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

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

So, the angle will be as follows.

$$\begin{gathered} sec\gamma =sec\frac{\mathrm{\pi }}{4} \\ =\sqrt{2} \end{gathered}$$

The required area is calculated as below.

$$\begin{gathered} A={\int }_x{\int }_{y}sec\gamma dxdy \\ =\sqrt{2}{\int }_{x/2}^{x/2}d\theta {\int }_0^{2\cos \theta }rdr \\ =\sqrt{2}{\int }_{x/2}^{x/2}d\theta \frac{1}{2}{\left(2\cos \theta \right)}^2 \\ =2\sqrt{2}{\int }_{x/2}^{x/2}{\cos }^2\theta d\theta \\ =\sqrt{2}{\int }_{x/2}^{x/2}(1+\cos (2\theta \left)\right)d\theta \\ =\sqrt{2}\mathrm{\pi } \end{gathered}$$

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