Question

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${\mathbf{(}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^2\mathbf{\le }\mathbf{1}$

Short answer

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

Step-by-step solution

Given Condition

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

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

Concept of surface area

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.

Draw the diagram

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

Calculate the angle

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.

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

Calculate the area

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$A=\sqrt{2}\mathrm{\pi }$ .