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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Use the Divergence Theorem to evaluate ∫∫s F · dS, where F ( x , y , z ) = z 2 x i + ( 1 3 y 3 + tan z ) j + ( x 2 z + y 2 ) k and S is the top half of the sphere x2 + y2 + z2 = 1. [Hint: Note that S is not a closed surface. First compute integrals over S1 and S2, where S1 is the disk x2 + y2 ≤ 1, oriented downward, and S2 = SS1.]

To determine

To evaluate: The expression SFdS for the vector field F(x,y,z)=z2xi+(13y3+tanz)j+(x2z+y2)k across the surface S .

Explanation

Given data:

The vector field is F(x,y,z)=z2xi+(13y3+tanz)j+(x2z+y2)k .

The surface S is the top half of the sphere x2+y2+z2=1 .

The surface S is not a closed surface.

The surface S1 is the disk x2+y21 .

The surface S2 is SS1 .

Formula used:

Write the expression to evaluate SFdS .

SFdS=DFndA (1)

Here,

n is the normal vector to the surface and

D is the region of the surface.

Write the another expression to evaluate SFdS .

SFdS=EdivFdV (2)

Here,

E is the solid region.

Write the expression to evaluate SFdS for the given data.

SFdS=S2FdSS1FdS (3)

Write the expression to find divergence of vector field F(x,y,z)=Pi+Qj+Rk .

divF=xP+yQ+zR (4)

Write the expression for spherical coordinate system.

Ef(x,y,z)dV=ρ1ρ2ϕ1ϕ2θ1θ2ρ2sinϕf(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)dρdϕdθ (5)

Here,

ρ is the radius of the sphere.

For the given data, the expression SFdS is evaluated across the surface S1 with formula in equation (1) and it is evaluated across the surface S2 with formula in equation (2).

As the surface S1 is the disk x2+y21 , consider the parametric equations of the surface as follows.

x=rcosθ,y=rsinθ,0r1,0θ2π

Calculation of S1FdS :

From the given data, the normal vector for the surface S1 is (k) .

n=k

Rewrite the normal vector as follows.

n=(0)i+(0)j+(1)k

Substitute z2xi+(13y3+tanz)j+(x2z+y2)k for F and (0)i+(0)j+(1)k for n in equation (1),

S1FdS=D[z2xi+(13y3+tanz)j+(x2z+y2)k][(0)i+(0)j+(1)k]dA=D[(z2x)(0)+(13y3+tanz)(0)+(x2z+y2)(1)]dA=D(x2zy2)dA

As the z-component is zero on the surface S1 disk, substitute 0 for z.

S1FdS=D[x2(0)y2]dA=D(y2)dA

Substitute rsinθ for y ,

S1FdS=02π01(rsinθ)2rdrdθ=02π01r3sin2θdrdθ=02πsin2θdθ01r3dr=02π(1cos2θ2)dθ[r44]01

Simplify the expression as follows

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Chapter 16 Solutions

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