   Chapter 15.5, Problem 1E

Chapter
Section
Textbook Problem

Find the area of the surface.1. The part of the plane 5x + 3y - z + 6 = 0 that lies above the rectangle [1, 4] × [2, 6]

To determine

To find: The area of given surface.

Explanation

Given:

The part of the plane, 5x+3yz+6=0 .

The region D is the rectangle [1,4]×[2,6] .

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.

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (1)

Calculation:

Express the given equation as follows

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