   Chapter 16.6, Problem 62E

Chapter
Section
Textbook Problem

The figure shows the surface created when the cylinder y2 + z2 = 1 intersects the cylinder x2 + z2 = 1. Find the area of this surface. To determine

To find: The area of the surface that creates by the intersection of two cylinders y2+z2=1 and x2+z2=1 .

Explanation

Given data:

The equation of first cylinder is given as follows.

y2+z2=1

The equation of second cylinder is given as follows.

x2+z2=1

The two cylinders y2+z2=1 and x2+z2=1 intersect and create the surface area, that is to be determined.

Formula used:

Write the expression to find the surface area of the plane with the vector equation r(x,z) .

A(S)=D|rx×rz|dA (1)

Here,

rx is the derivative of vector equation r(x,z) with respect to the parameter x and

rz is the derivative of vector equation r(x,z) with respect to the parameter z .

Write the expression to find rx .

rx=x[r(x,z)] (2)

Write the expression to find rz .

rz=z[r(x,z)] (3)

As the required area of the surface is created by the interaction of two cylinders, the total surface area is the four times of surface area of any one face.

Calculate the area for the face of the surface that intersects the y-axis and multiply with 4 to find the total area of the required surface.

Rewrite the equation of first cylinder y2+z2=1 as follows.

y=1z2

Consider the x and z as parameters and parameterize the first cylinder y2+z2=1 as follows.

x=x,y=1z2,z=z

Write the vector equation of the first cylinder from the parametric equations as follows.

r(x,z)=xi+(1z2)j+zk

Calculation of rx :

Substitute xi+(1z2)j+zk for r(x,z) in equation (2),

rx=x[xi+(1z2)j+zk]=x(x)i+x(1z2)j+x(z)k=(1)i+(0)j+(0)k=(1)i+(0)j+(0)k

Calculation of rz :

Substitute xi+(1z2)j+zk for r(x,z) in equation (3),

rz=z[xi+(1z2)j+zk]=z(x)i+z(1z2)j+z(z)k=(0)i+121z2(02z)j+(1)k=(0)iz1z2j+(1)k

Calculation of rx×rz :

Substitute (1)i+(0)j+(0)k for rx and (0)iz1z2j+(1)k for rz in the expression rx×rz ,

rx×rz=[(1)i+(0)j+(0)k]×[(0)iz1z2j+(1)k]

Rewrite and compute the expression as follows

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