Question

Find the area of the paraboloid${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{z}$inside the cylinderrole="math" localid="1659151613290" ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{9}$

Short answer

The area of the paraboloid $x^2+y^2=z$inside the cylinder ${\mathrm{x}}^2+{\mathrm{y}}^2=9$is$A=\frac{\mathrm{\pi }}{6}({37}^{3/2}-1)$

Step-by-step solution

Given Condition

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

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

The shaded area is the area of integration.

Calculate the angle

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

The secant of the angle is found as follows.

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

Calculate the area.

The area of the surface cut out by the cylinder is calculated as below.

$$\begin{gathered} A={\int }_x{\int }_{y}sec\gamma dxdy \\ ={\int }_0^{2x}d\theta {\int }_0^3\sqrt{4x^2+4y^2+1}rdr \\ ={\int }_0^{2x}d\theta {\int }_0^3\sqrt{4r^2+1}d(4r^2+1)\frac{1}{8} \\ ={\overline{)\frac{\mathrm{\pi }}{4}\frac{2}{3}{(4r^2+1)}^{3/2}}}_0^3 \\ =\frac{\mathrm{\pi }}{6}({37}^{3/2}-1) \end{gathered}$$

Hence, the area of the paraboloid $x^2+y^2=z$inside the cylinder $x^2+y^2=9$is$A=\frac{\mathrm{\pi }}{6}({37}^{3/2}-1)$ .