   Chapter 16.6, Problem 44E

Chapter
Section
Textbook Problem

Find the area of the surface.44. The part of the surface z = 4 − 2x2 + y that lies above the triangle with vertices (0, 0), (1, 0), and (1, 1)

To determine

To find: The area of the part of the surface z=42x2+y that lies above the triangle with vertices (0,0) , (1,0) and (1,1) .

Explanation

Given data:

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

z=42x2+y

The required surface lies above the triangle with vertices (0,0) , (1,0) and (1,1) .

Formula used:

Write the expression to find the surface area of the plane.

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

Write the equation of part of the surface as follows.

z=42x2+y (2)

Calculation of zx :

Take partial derivative for equation (2) with respect to x.

zx=x(42x2+y)=04x+0=4x

Calculation of zy :

Take partial derivative for equation (2) with respect to y.

zy=y(42x2+y)=00+1=1

Calculation of surface area of plane:

Substitute (4x) for zx and 1 for zy in equation (1),

A(S)=D1+(4x)2+(1)2dA=D1+16x2+1dA=D16x2+2dA

Simplify the expression as follows.

A(S)=D16x2+2dA (3)

As the required part of the surface lies above the triangle with vertices (0,0) , (1,0) and (1,1) , the triangle is a right angle triangle with base 1 and height as 1.

The region D is a triangle area with the x limit as 0 to 1 and the limit y is 0 to x. Therefore, the limit of x and y are written as follows.

0x10yx

Apply the limits and rewrite the expression in equation (3) as follows.

A(S)=010x(16x2+2)dydx=01(16x2+2)dx0x(1)dy=01(16x2+2)dx[y]0x=01(16x2+2)dx(x0)

Simplify the expression as follows

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