   Chapter 16.6, Problem 46E

Chapter
Section
Textbook Problem

Find the area of the surface.46. The part of the surface x = z2 + y that lies between the planes y = 0, y = 2, z = 0, and z = 2

To determine

To find: The area of part of the surface x=z2+y that lies between the planes y=0 , y=2 , z=0 , and z=2 .

Explanation

Given data:

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

x=z2+y

The required part of the surface lies between the planes y=0 , y=2 , z=0 , and z=2 .

Formula used:

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

A(S)=D|ry×rz|dA (1)

Here,

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

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

Write the expression to find ry .

ry=y[r(y,z)] (2)

Write the expression to find rz .

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

Consider the y and z as parameters and parameterize the surface as follows.

x=z2+y,y=y,z=z

Write the vector equation of the surface from the parametric equations as follows.

r(y,z)=(z2+y)i+yj+zk

Calculation of ry :

Substitute (z2+y)i+yj+zk for r(y,z) in equation (2),

ry=y[(z2+y)i+yj+zk]=y(z2+y)i+y(y)j+y(z)k=(0+1)i+(1)j+(0)k=(1)i+(1)j+(0)k

Calculation of rz :

Substitute (z2+y)i+yj+zk for r(y,z) in equation (3),

rz=z[(z2+y)i+yj+zk]=z(z2+y)i+z(y)j+z(z)k=(2z+0)i+(0)j+(1)k=2zi+(0)j+(1)k

Calculation of ry×rz :

Substitute (1)i+(1)j+(0)k for ry and 2zi+(0)j+(1)k for rz in the expression ry×rz ,

ry×rz=[(1)i+(1)j+(0)k]×[2zi+(0)j+(1)k]

Rewrite and compute the expression as follows

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