   Chapter 16.8, Problem 18E

Chapter
Section
Textbook Problem

Evaluate ∫c (y + sin x) dx + (z2 + cos y) dy + x3 dz where C is the curve r(t) = (sin t, cos t, sin 2t⟩, 0 ≤ t ≤ 2π. [Hint: Observe that C lies on the surface z = 2xy.]

To determine

To evaluate: The value of line integral C(y+sinx)dx+(z2+cosy)dy+x3dz .

Explanation

Given data:

Line integral is C(y+sinx)dx+(z2+cosy)dy+x3dz and curve is r(t)=sint,cost,sin2t,0t2π .

Formula Used:

Consider a vector field F(x,y)=P(x,y)i+Q(x,y)j+R(x,y)k and P, Q and R have continuous partial derivatives.

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

curlF=|ijkxyzPQR|

curlF=(RyQz)i(RxPz)j+(QxPy)k (1)

Write the expression for the Stokes’ theorem.

CFdr=ScurlFdS (2)

Here,

S is the surface.

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

ScurlFdS=D(PgxQgy+R)dA (3)

Here,

A is the area.

Consider the surface S as a region curve r(t)=sint,cost,sin2t,0t2π . And hence, the surface S, D={0r10θ2π .

The parametric equations of curve are,

x=sinty=cost

z=sin2t (4)

Modify equation (4).

z=2sintcost{sin2x=2sinxcosx}

Substitute x for sint and y for cost .

z=2xy

Hence the equation is in the form of z=g(x,y) and the orientation of surface S is in downward direction.

Compare the expressions CFdr and C(y+sinx)dx+(z2+cosy)dy+x3dz .

F=(y+sinx)i+(z2+cosy)j+x3k

Find the value of curlF by using equation (1).

curlF=[((x3)y(z2+cosy)z)i((x3)x(y+sinx)z)j+((z2+cosy)x(y+sinx)y)k]=(0(2z+0))i(3x20)j+(0(1+0))k{t(k)=0,t(t)=1,t(tn)=ntn1}=2zi3x2jk

Compare the equations curlF=Pi+Qj+Rk and curlF=2zi3x2jk .

P=2zQ=3x2R=1

Find the value of (PgxQgy+R) .

PgxQgy+R=(2z)(2xy)x(3x2)(2xy)y+(1)=(2(2xy))(2y(1))+(3x2)(2x(1))1{z=2xy}=8xy2+6x31

Substitute rcosθ for x and rsinθ for y,

PgxQgy+R=8(rcosθ)(rsinθ)2+6(rcosθ)31=8r3cosθsin2θ+6r3cos3θ1

Re-modify the equation (3)

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