   Chapter 16.6, Problem 63E

Chapter
Section
Textbook Problem

Find the area of the part of the sphere x2 + y2 + z2 = a2 that lies inside the cylinder x2 + y2 = ax.

To determine

To find: The area of the part of the sphere x2+y2+z2=a2 that lies inside the cylinder x2+y2=ax .

Explanation

Given data:

The equation of the part of the sphere is given as follows.

x2+y2+z2=a2

The equation of the cylinder is given as follows.

x2+y2=ax

Formula used:

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

A(S)=D|rx×ry|dA (1)

Here,

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

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

Write the expression to find rx .

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

Write the expression to find ry .

ry=z[r(x,y)] (3)

The required total surface area of the part of the sphere is the sum of the area of above and below portions of the cylinder.

As the above and below portions of the cylinder is symmetrical, find the area of any one of the portions and multiply with 2 to get the total area of the part of the sphere.

Rewrite the equation of sphere x2+y2+z2=a2 as follows.

z=a2x2y2

Consider the x and y as parameters and parameterize the sphere x2+y2+z2=a2 as follows.

x=x,y=y,z=a2x2y2

Write the vector equation of the part of the sphere from the parametric equations as follows.

r(x,y)=xi+yj+(a2x2y2)k

Calculation of rx :

Substitute xi+yj+(a2x2y2)k for r(x,y) in equation (2),

rx=x[xi+yj+(a2x2y2)k]=x(x)i+x(y)j+x(a2x2y2)k=(1)i+(0)j+[12a2x2y2(02x0)]k=(1)i+(0)j+(xa2x2y2)k

Calculation of ry :

Substitute xi+yj+(a2x2y2)k for r(x,y) in equation (3),

ry=y[xi+yj+(a2x2y2)k]=y(x)i+y(y)j+y(a2x2y2)k=(0)i+(1)j+[12a2x2y2(002y)]k=(0)i+(1)j+(ya2x2y2)k

Calculation of rx×ry :

Substitute (1)i+(0)j+(xa2x2y2)k for rx and (0)i+(1)j+(ya2x2y2)k for ry in the expression rx×ry ,

rx×ry=[(1)i+(0)j+(xa2x2y2)k]×[(0)i+(1)j+(ya2x2y2)k]

Rewrite and compute the expression as follows.

rx×ry=|ijk10xa2x2y201ya2x2y2|=|0xa2x2y21ya2x2y2|i|1xa2x2y20ya2x2y2|j+|1001|k=[0(xa2x2y2)]i(ya2x2y20)j+(10)k=(xa2x2y2)i+(ya2x2y2)j+(1)k

Substitute (xa2x2y2)i+(ya2x2y2)j+(1)k for rx×ry in equation (1),

A(S)=D|(xa2x2y2)i+(ya2x2y2)j+(1)k|dA=D[(xa2x2y2)2+(ya2x2y2)2+(1)2]dA=D(x2a2x2y2+y2a2x2y2+1)dA=D(x2+y2+a2x2y2a2x2y2)dA

Simplify the expression as follows

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