   Chapter 15.5, Problem 4E

Chapter
Section
Textbook Problem

Find the area of the surface.4. The part of the surface 2y + 4z - x2 = 5 that lies above the triangle with vertices (0, 0), (2. 0), and (2, 4)

To determine

To find: The area of given surface.

Explanation

Given:

The part of the surafce, 2y+4zx2=5 .

The region D lies above the triangle with vertices (0,0),(2,0),(2,4) .

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.

Calculation:

The given equation becomes,

2y+4zx2=54z=52y+x2z=542y4+x24z=54y2+x24

The partial derivatives fx and fy are,

fx=2x4=x2fy=12

From the given vertices of the triangle, it is observed that x varies from 0 to 2 and y varies from 0 to 2x . Then, the area of surface is given by,

A(S)=D(x2)2+(12)2+1dA=0202xx24+14+1dydx

Integrate with respect to y

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