   Chapter 16, Problem 25RE

Chapter
Section
Textbook Problem

Find the area of the part of the surface z = x2 + 2y that lies above the triangle with vertices (0, 0), (1, 0), and (1, 2).

To determine

To Find: The area of the part of the surface z=x2+2y .

Explanation

Given data:

z=x2+2y ,

Triangle vertices are (0,0) , (1,0) and (1,2) .

Formula used:

Write the expression for surface area.

A(S)=D1+(zx)2+(zy)2dA (1)

Write the required differentiation formulae.

ddx(x2)=2xddy(2y)=2

Consider the expression as follows.

z=x2+2y (2)

Differentiate equation (2) with respect to x .

zx=2x

Differentiate equation (2) with respect to y .

zy=2

Substitute 2x for zx and 2 for zy in equation (1),

A(S)=D1+(2x)2+(2)2dA=D1+4x2+4dA

A(S)=D5+4x2dA (3)

Find the value of limits as follows.

Draw the region D enclosed by C as shown in Figure 1.

Refer Figure 1, x -axis changes with 0 to 1 and y -axis changes with 0 to 2x .

Apply limits and modify equation (3) as follows.

A(S)=0102x(5+4x2)dxdy=01(5+4x2)dx[y]02x=01(5+4x2)dx[2x0]=012x(5+4x2)dx

Integrate with respect to x and simplify as follows

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