The figure shows the surface created when the cylinder ${y}^{2}+{Z}^{2}=1$ intersects the cylinder ${x}^{2}+{Z}^{2}=1$. Find the area of this surface.

The figure is something like:

sodni3
2020-12-16
Answered

The figure shows the surface created when the cylinder ${y}^{2}+{Z}^{2}=1$ intersects the cylinder ${x}^{2}+{Z}^{2}=1$. Find the area of this surface.

The figure is something like:

You can still ask an expert for help

stuth1

Answered 2020-12-17
Author has **97** answers

Note that the given cylinder will intersect at four congruent surfaces. e. equal in area. So we need to find the area of only one surface and then multiply it by 4.

First, we find the area of the face of the surface that intersect the positive y axis. So the surface lies above the region${x}^{2}+{z}^{2}=1$ then $z=f(x,z)=\sqrt{1-{z}^{2}}$

$=\int {\int}_{{x}^{2}+{z}^{2}\le 1}\sqrt{{f}_{x}^{2}+{f}_{z}^{2}+1}dA$

$=\int {\int}_{{x}^{2}+{z}^{2}\le 1}\sqrt{\frac{{z}^{2}}{1-{z}^{2}}+1}dA$

$={\int}_{-1}^{1}{\int}_{-\sqrt{1-{z}^{2}}}^{\sqrt{1+{z}^{2}}}\frac{1}{\sqrt{1-{z}^{2}}}dxdz$

$=4{\int}_{0}^{1}{\int}_{0}^{\sqrt{1-{z}^{2}}}\frac{1}{\sqrt{1-{z}^{2}}}dxdz$

$=\underset{t\to 1}{lim}+4{\int}_{0}^{t}{\int}_{0}^{\sqrt{1-{z}^{2}}}\frac{1}{\sqrt{1-{z}^{2}}}dxdz$

$=\underset{t\to 1}{lim}+4{\int}_{0}^{t}1dt$

$=4$

Hence the total area of the intersected surface is$4\times 4=16$

First, we find the area of the face of the surface that intersect the positive y axis. So the surface lies above the region

Hence the total area of the intersected surface is

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