   Chapter 16.8, Problem 14E

Chapter
Section
Textbook Problem

Verify that Stokes’ Theorem is true for the given vector field F and surface S.14. F(x, y, z) = -2yz i + y j + 3x k,S is the part of the paraboloid z = 5 - x2 - y2 that lies above the plane z = 1, oriented upward

To determine

To verify: Whether Stokes’ theorem is true for given vector filed F and surface S.

Explanation

Given data:

The field is F(x,y,z)=2yzi+yj+3xk and

Consider the expression for surface that is part parabolid above the plane is,

z=5x2y2 (1)

Formula Used:

Write the expression for curl of F(x,y,z)=Pi+Qj+Rk .

curlF=|ijkxyzPQR|

curlF=(RyQz)i(RxPz)j+(QxPy)k (2)

Write the expression for the Stokes’ theorem.

CFdr=ScurlFdS (3)

Here,

S is surface.

Consider surface S, z=g(x,y) is in upward orientation. Write the expression for surface integral of F over surface S.

ScurlFdS=D(PgxQgy+R)dA (4)

Here,

A is area.

The parabolid intersects a plane z=1 .

Substitute 1 for z in equation (1),

1=5x2y2x2+y2=51x2+y2=4

Consider the parametric equations since the orientation of surface is upward.

x=2costy=2sintz=1,0t2π

Write the expression for vector function r(t) .

r(t)=xi+yj+zk

Substitute 2cost for x, 2sint for y, 1 for z,

r(t)=2costi+2sintj+(1)k=2costi+2sintj+k

Differentiate the expression with respect to t.

r(t)=ddt(2costi+2sintj+k)=ddt(2cost)i+ddt(2sint)j+ddt(1)k=2(sint)i+(2cost)j+(0)k=2sinti+2costj

Find the value of F(r(t)) .

F(r(t))=2(2sint)(1)i+(2sint)j+3(2sint)k=4sinti+2sintj+6sintk

Find the value of F(r(t))r(t) .

F(r(t))r(t)=(4sinti+2sintj+6sintk)(2sinti+2costj)=(4sint)(2sint)+(2sint)(2cost)+(6sint)(0)=8sin2t+4sintcost+0=8sin2t+4sintcost

Write the expression for CFdr .

CFdr=02πF(r(t))r(t)dt

Substitute 8sin2t+4sintcost for F(r(t))r(t) ,

CFdr=02π(8sin2t+4sintcost)dt=02π(8(12(1cos2t))+2sin2t)dt{cos2t=12sin2t,sin2t=2sintcost}=02π(4(1cos2t)+2sin2t)dt=[4(tsin2t2)+2(cos2t2)]02π

CFdr=[4(2πsin2(2π)2)+2(cos2(2π)2)4(0sin2(0)2)+2(cos2(0)2)]=8π0+2(12)02(12)=8π1+1=8π

Find the value of curlF by using equation (2).

curlF=((3x)y(y)z)i((3x)x(2yz)z)j+((y)x(2yz)y)k=(00)i(3(1)(2y(1)))j+(0(2z(1)))k=(32y)j+2zk

The expression for surface S, D={(r,θ)|0r2,0θ2π} in polar coordinate system is,

z=5x2y2

Here,

x=rcosθandy=rsinθ

Hence the equation is in the form of z=g(x,y)

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