Question

Find the area of the plane $\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{5}\mathbf{z}\mathbf{=}\mathbf{13}$ cut out by the cylinder ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{9}$.

Short answer

The area of the plane $x-2y+5z=13$cut out by the cylinder $x^2+y^2=9$ is$A=9\mathrm{\pi }\frac{\sqrt{30}}{5}.$

Step-by-step solution

Given Condition

The equation of the plane is $x-2y+5z=13$ and the equation of the cylinder is $x^2+y^2=9$.

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 secant of the angle $\gamma$is given by,

$$\begin{gathered} sec\gamma =\frac{\sqrt{{\left|\nabla \varphi \right|}^2}}{\left|\partial \varphi /\partial z\right|} \\ =\frac{\sqrt{{\left|\hat{x}-2\hat{y}+5\hat{z}\right|}^2}}{5} \\ =\frac{\sqrt{1+4+5}}{5} \\ =\frac{\sqrt{30}}{5} \end{gathered}$$

Calculate the area

The area of the plane cut by the cylinder can be found by,

$$\begin{gathered} A={\int }_x{\int }_{y}sec\gamma dxdy \\ =\frac{\sqrt{30}}{5}\left(9\mathrm{\pi }\right) \\ =9\mathrm{\pi }\frac{\sqrt{30}}{5} \end{gathered}$$

Hence, the required area is $A=9\mathrm{\pi }\frac{\sqrt{30}}{5}$.