   Chapter 15, Problem 45RE

Chapter
Section
Textbook Problem

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

To determine

To find: The area of the given region.

Explanation

Formula used:

The surface area with equation z=f(x,y),(x,y)D, where fx and fy are continuous, is A(S)=D[fx(x,y)]2+[fy(x,y)]2+1dA.

Here, D is the given region.

Property used:

a2+u2dx=u2a2+u2+a22ln(u+a2+u2).

Given:

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

Calculation:

From the given conditions, it is observed that the equations of the triangle are y=0,y=22x and x varies from 0 to 1.

The partial derivatives fx and fy are,

fx=2xfy=1

Then, by the formula mentioned above, the area of the surface is given by,

A(S)=D(2x)2+(1)2+1dA=D4x2+1+1dA=D4x2+2dA=01022x4x2+2dydx

Integrate with respect to y and apply the limit of it.

A(S)=014x2+2[y]022xdx=014x2+2[22x0]dx=014x2+2[22x]dx=2014x2+2dx012x4x2+2dx

Integrate with respect to x. To integrate the second integrand, substitute t=4x2+2,dt=8xdx and to integrate the first integrand, use the above mentioned property as u=2x,a=2 which implies, du=2dx

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