   Chapter 16, Problem 27RE

Chapter
Section
Textbook Problem

Evaluate the surface integral.27. ∬S z dS, where S is the part of the paraboloid z = x2 + y2 that lies under the plane z = 4

To determine

To Evaluate: The surface integral of SzdS .

Explanation

Given data:

z=x2+y2 , z=4

Formula used:

Write the expression for surface integral.

Sf(x,y,z)dS=Df(r(u,v))|ru×rv|dA (1)

Write the expression for rx×ry .

rx×ry=zxizyj+k (2)

Write the expression for |rx×ry| .

|rx×ry|=(zx)2+(zy)2+1 (3)

Write the required differentiation formulae.

ddx(x2)=2xddy(x2)=0ddy(y2)=2yddx(y2)=0

Consider the expression as follows.

z=x2+y2 (4)

Differentiate equation (4) with respect to x .

zx=2x

Differentiate equation (4) with respect to y .

zy=2y

Substitute 2x for zx and 2y for zy in equation (2),

rx×ry=2xi2yj+k

Substitute 2x for zx and 2y for zy in equation (3),

|rx×ry|=(2x)2+(2y)2+1=4x2+4y2+1

Modify equation (1) as follows.

SzdS=Dz|rx×ry|dA

Substitute x2+y2 for z and 4x2+4y2+1 for |rx×ry| ,

SzdS=D(x2+y2)(4x2+4y2+1)dA

SzdS=D(x2+y2)(4(x2+y2)+1)dA (5)

Find the value of limits as follows.

Consider the circle equation as follows.

x2+y2=4 (6)

Write the expression for circle equation.

x2+y2=r2 (7)

Compare equation (6) and (7).

r2=4r=2

Area of circle depends upon radius (r) and angle (θ) . The total angle required to complete one circle is 0 to 2π .

Substitute r2 for x2+y2 in equation (5),

SzdS=Dr2(4r2+1)dA

Apply limits and modify equation as follows.

SzdS=02π02r2(4r2+1)rdrdθ=02π02r3(4r2+1)drdθ=02r3(4r2+1)dr[θ]02π=02r3(4r2+1)dr[2π0]

Simplify equation as follows.

SzdS SzdS=2π02r3(4r2+1)dr

SzdS=2π02r2(4r2+1)rdr (8)

Consider the expression for u as follows

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