   Chapter 16.9, Problem 2E

Chapter
Section
Textbook Problem

Verify that the Divergence Theorem is true for the vector field F on the region E.2. F(x, y, z) = y2z3 i + 2yz j + 4z2 k, E is the solid enclosed by the paraboloid z = x2 + y2 and the plane z = 9

To determine

To Verify: The divergence theorem for the vector field F(x,y,z)=y2z3i+2yzj+4z2k on the region E.

Explanation

Given data:

The vector field is F(x,y,z)=y2z3i+2yzj+4z2k .

The region E is the solid enclosed by the paraboloid z=x2+y2 , and the plane z=9 .

Formula used:

Write the formula of divergence theorem.

SFdS=EdivFdV (1)

Here,

E is the solid region.

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

divF=xP+yQ+zR (2)

Write the expression for SFdS in terms of normal vector to the plane as follows.

SFdS=DFndA (3)

Here,

n is the normal vector to the plane of the surface S.

Calculation of divF :

Substitute y2z3 for P , 2yz for Q , and 4z2 for R in equation (2),

divF=x(y2z3)+y(2yz)+z(4z2)=0+2z+8z=10z

Calculation of flux EdivFdV :

Substitute 10z for divF in the expression EdivFdV as follows.

EdivFdV=E(10z)dV (4)

Write the equation of paraboloid as follows.

z=x2+y2 (5)

From the equation of paraboloid, the parameters are considered as follows.

x=rcosθ,y=rsinθ,0r3,0θ2π

Substitute rcosθ for x and rsinθ for y in equation (5),

z=(rcosθ)2+(rsinθ)2=r2(cos2θ+sin2θ)=r2(1) {cos2θ+sin2θ=1}=r2

As the surface E is bounded by the paraboloid z=x2+y2 , and the plane z=9 , the limits of z, r, and θ are written as follows.

r2z90r30θ2π

Apply the limits and rewrite the expression in equation (4) as follows.

EdivFdV=02π03r29(10z)rdzdrdθ=02π03[10r(z22)]z=r2z=9drdθ=02π03[5r(92r4)]drdθ=02π03(405r5r5)drdθ

Simplify the expression as follows.

EdivFdV=02π(1)dθ03(405r5r5)dr=[θ]02π[405(r22)5(r66)]03=[2π0]{[405(322)5(366)][405(022)5(066)]}=(2π)(3645212152)

EdivFdV=2430π (6)

Calculation of SFdS :

As the solid is enclosed by the paraboloid z=x2+y2 and the plane z=9 , compute the expression for flux SFdS as follows.

Consider the surface S1 is S1:z=x2+y2,x2+y29 with downward orientation, and the vector field F(x,y,z)=y2z3i+2yzj+4z2k .

Consider the surface S2 is S2:z=9,x2+y29 with upward orientation, and the vector field F(x,y,z)=y2z3i+2yzj+4z2k .

Calculate the flux in each surface separately and add the individual results to find the value of SFdS .

Write the expression to find the value of SFdS as follows.

SFdS=S1FdS+S2FdS (7)

Calculation of S1FdS :

As the surface S1 oriented downward, the normal vector to surface is written as follows.

n=2xi+2yj+(1)k

Substitute (y2z3i+2yzj+4z2k) for F and [2xi+2yj+(1)k] for n in equation (3),

S1FdS=D(y2z3i+2yzj+4z2k)[2xi+2yj+(1)k]dA=D[(y2z3)(2x)+(2yz)(2y)+(4z2)(1)]dA=D(2xy2z3+4y2z4z2)dA=(2)D(z)(xy2z2+2y22z)dA

As z=x2+y2 in the surface S1 , substitute (x2+y2) for z

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