   Chapter 16.7, Problem 9E

Chapter
Section
Textbook Problem

Evaluate the surface integral.9. ∬s x2yz dS, S is the part of the plane z = 1 + 2x + 3y that lies above the rectangle [0, 3] × [0, 2]

To determine

To find: The value of Sx2yzdS .

Explanation

Given data:

z=1+2x+3y

Formula used:

Sf(x,y,z)dS=Df(x,y,g(x,y))(zx)2+(zy)2+1dA (1)

Find zx .

zx=x(1+2x+3y)=x(1)+2x(x)+3yx(1)=(0)+2(1)+3y(0)=2

Find zy .

zy=y(1+2x+3y)=y(1)+2xy(1)+3y(y)=(0)+2x(0)+3(1)=3

Find Sx2yzdS .

Modify equation (1) as follows.

Sx2yzdS=Dx2yz(zx)2+(zy)2+1dA

Apply limits and substitute 1+2x+3y for z , 2 for zx and 3 for zy ,

Sx2yzdS=0302x2y(1+2x+3y)(2)2+(3)2+1dydx=0302(x2y+2x3y+3x2y2)4+9+1</

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