   Chapter 16.7, Problem 10E

Chapter
Section
Textbook Problem

Evaluate the surface integral.10. ∫∫s xz dS, S is the part of the plane 2x + 2y + z = 4 that lies in the first octant

To determine

To find: The value of SxzdS .

Explanation

Given data:

2x+2y+z=4 (1)

Formula used:

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

Rearrange equation (1).

z=42x2y (3)

Find the x limits by considering y and z as 0 in equation (1).

0=42x2(0)2x=4x=42x=2

Find the y limits by considering z as 0 in equation (1).

0=42x2y2y=42x2y=2(2x)y=2x

The regions of D is 0x2 and 0y2x .

Find zx .

Take partial differentiation for equation (3) with respect to x.

zx=x(42x2y)=x(4)2x(x)2yx(1)=(0)2(1)2y(0)=2

Find zy .

Take partial differentiation for equation (3) with respect to y.

zy=y(42x2y)=y(4)2xy(1)2y(y)=(0)2x(0)2(1)=2

Find SxzdS .

Modify equation (2) as follows.

SxzdS=Dxz(zx)2+(zy)2+1dA

Apply limits and substitute 42x2y for z , 2 for zx and 2 for zy ,

SxzdS=0202xx(42x2y)(2)2+(2)2+1dydx=02<

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