   Chapter 16.7, Problem 12E

Chapter
Section
Textbook Problem

Evaluate the surface integral.12. ∫∫s y dS, S is the surface z   =   2 3 ( x 3 / 2   +   y 3 / 2 ) , 0   ≤   x   ≤   1 ,   0   ≤   y   ≤   1

To determine

To find: The value of SydS .

Explanation

Given data:

z=23(x32+y32) , 0x1 and 0y1 .

Formula used:

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

Find zx .

zx=x(23(x32+y32))=x[23(x32)]+x[23(y32)]=(23)x(x32)+[23(y32)]x(1)=(23)(32x12)+[23(y32)](0)

zx=x

Find zy .

zy=y[23(x32+y32)]=y[23(x32)]+y[23(y32)]=[23(x32)]y(1)+(23)y(y32)=(0)+(23)[32y12]

zy=y

Find SydS .

Modify equation (1) as follows.

SydS=Dy(zx)2+(zy)2+1dA

Apply limits and substitute x for zx and y for zy ,

SydS=0101y(x)2+(y)2+1dxdy=0101yx+y+1dxdy=01y[(x+y+1)3232]01dy=01y[23((10)+y+1)32]dy

SydS=0123y[(y+2)32(y+1)32]dy

SydS=2301y(y+2)32dy2301(y+1)32dy (2)

Consider u=y+2 that is y=u2

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