Question

Find the area of the part of the cone${\mathbf{z}}^{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}$which is inside the sphere${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathbf{16}$

Short answer

$8\mathrm{\pi }$Per nappe

Step-by-step solution

Given Condition

The equation of the cone is$z^2=3(x^2+y^2)$

The equation of the sphere is$x^2+y^2+z^2=16$

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

Calculate the angle

The equation of the cone is$\varphi (x,y,z)=z^2-3x^2-3y^2$

The secant of the angle is found as follows.

$$\begin{gathered} sec\gamma =\frac{\sqrt{{\left|\nabla \varphi \right|}^2}}{\left|\partial \varphi /\partial z\right|} \\ =\frac{\sqrt{{\left|2z\hat{z}-6x\hat{x}-6y\hat{y}\right|}^2}}{2z} \\ =\frac{\sqrt{z^2+9x^2+9y^2}}{z} \end{gathered}$$

Calculate the area.

The area of one nappe of the cone cut out by the sphere is calculated as below.

$$\begin{gathered} A={\int }_0^{2x}d0{\int }_0^{2}rdr\frac{\sqrt{z^2+9r^2}}{z} \\ putz=rcot\frac{\mathrm{\pi }}{6}=\sqrt{3}r \\ A=2\mathrm{\pi }{\int }_0^2\mathrm{rdr}\frac{\sqrt{3{\mathrm{r}}^2+9{\mathrm{r}}^2}}{\sqrt{3}\mathrm{r}} \\ =2\mathrm{\pi }{\int }_0^2\mathrm{rdr}\frac{\sqrt{12}}{\sqrt{3}} \\ ={\overline{)4\mathrm{\pi }\frac{1}{2}{\mathrm{r}}^2}}_0^2 \\ =8\mathrm{\pi } \\ \mathrm{A}=8\mathrm{\pi }\mathrm{per}\mathrm{nappe}. \end{gathered}$$