   Chapter 16.9, Problem 4E

Chapter
Section
Textbook Problem

Verify that the Divergence Theorem is true for the vector field F on the region E.4. F(x, y, z) = ⟨x2, -y, z⟩, E is the solid cylinder y2 + z2 ≤ 9, 0 ≤ x ≤ 2

To determine

To Verify: The divergence theorem for the vector field F(x,y,z)=x2,y,z on the region E.

Explanation

Given data:

The vector field is F(x,y,z)=x2,y,z .

The region E is the solid cylinder y2+z29 and 0x2 .

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 as follows.

SFdS=SFndS (3)

Here,

n is the unit outward normal of the surface S.

Calculation of divF :

Substitute x2 for P , (y) for Q , and z for R in equation (2),

divF=x(x2)+y(y)+z(z)=2x1+1=2x

Calculation of flux EdivFdV :

Substitute 2x for divF in the expression EdivFdV as follows.

EdivFdV=E(2x)dV

Apply the limits of the region E and rewrite the expression as follows.

EdivFdV=y2+z29[022xdx]dA=y2+z29(4)dA=4A(D)

Here,

A(D) is the area of the circle with radius 3.

As the area of the circle with radius r is πr2 , the area with radius 3 is π(3)2 , which is 9π .

Substitute 9π for A(D) in the expression EdivFdV=4A(D) .

EdivFdV=4(9π)

EdivFdV=36π (4)

Consider the surface S1 is the front of the cylinder in the plane x=2 , surface S2 is the back of the cylinder in the yz-plane, and S3 is the lateral surface of the cylinder.

Calculate flux on each surface 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+S3FdS (5)

Calculation of S1FdS :

The surface S1 is the disk in the plane x=2 , y2+z29 .

The unit normal vector to the surface S1 is written as follows.

n=1,0,0

Substitute 2 for x in the vector field F=x2,y,z to find the vector field in the surface S1 .

F=22,y,z=4,y,z

Substitute S1 for S , 1,0,0 for n and 4,y,z for F in equation (3),

S1FdS=S14,y,z1,0,0dS=S1(4)dS=4(Surface area of S1)=4[π(3)2]

S1FdS=36π

Calculation of S2FdS :

The surface S2 is the disk in the plane x=0 , y2+z29 .

The unit normal vector to the surface S2 is written as follows.

n=1,0,0

Substitute 0 for x in the vector field F=x2,y,z to find the vector field in the surface S2 .

F=02,y,z=0,y,z

Substitute S2 for S , 1,0,0 for n and 0,y,z for F in equation (3),

S2FdS=S20,y,z1,0,0dS=S2(0)dS=0

The parametric equations of the surface S3 are written as follows

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Solve the equations in Exercises 126. 2x3x2x3x+1=0

Finite Mathematics and Applied Calculus (MindTap Course List)

Find the derivatives of the functions in Problems 1-34. 24.

Mathematical Applications for the Management, Life, and Social Sciences

Evaluating a surface Integral In Exercise 5-8, evaluate S(x2y+z)dS S:z=4x0x4,0y3,

Calculus: Early Transcendental Functions (MindTap Course List)

Find all prime ideals of .

Elements Of Modern Algebra

True or False: f(x)=x2(x+1)x is continuous for all real numbers.

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 