   Chapter 16.7, Problem 26E

Chapter
Section
Textbook Problem

Evaluate the surface integral ∫∫s F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.26. F(x, y, z) = y i - x j + 2z k, S is the hemisphere x2 + y2 + z2 = 4, z ≥ 0, oriented downward

To determine

To find: The value of SFdS for F(x,y,z)=yixj+2zk .

Explanation

Given data:

F(x,y,z)=yixj+2zk (1)

x2+y2+z2=4 (2)

Formula used:

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

Rearrange equation (2).

z=4x2y2

Consider g=z .

Substitute 4x2y2 for z ,

g=4x2y2

The D is disk {x2+y24} . Since S has downward orientation, modify equation (3).

SFdS=D(PgxQgy+R)dA

Substitute y for P , x for Q , 2z for R and 4x2y2 for g ,

SFdS=D[(y)(4x2y2)x(x)(4x2y2)y+2z]dA=D[(y)12(4x2y2)12(2x)(x)12(4x2y2)12(2y)+2z]dA=D(xy4x2y2xy4x2y2+2z)dA

Substitute 4x2y2 for z ,

SFdS=D(xy4x2y2xy4x2y2+24x2y2)dA

SFdS=D24x2y2dA (4)

Let x=rcosθ , y=rsinθ and limits for r and θ are 0 to 2 and 0 to 2π

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

In Exercises 1-6, solve for y. (x2)2+(y+1)2=9

Calculus: An Applied Approach (MindTap Course List)

Expand each expression in Exercises 122. (3x+1)(2x2x+1)

Finite Mathematics and Applied Calculus (MindTap Course List)

True or False: The function graphed at the right is one-to-one.

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

The polar form for the graph at the right is:

Study Guide for Stewart's Multivariable Calculus, 8th 