Surface Integral From Div, Grad, Curl and all that

samedi 31 août 2013

First, this is my first time actually posting anything so hi PF!

Second, I have been working out of Div, Grad, Curl and all that. This problem has me stumped for some reason. My answer never comes out to be the same as the books. If you could help me figure out where I am going wrong I would greatly appreciate it.

1. The problem statement, all variables and given/known data

Evaluate the surface integral using the following equation.

2. Relevant equations

[itex]\int[/itex][itex]\int[/itex][itex]\vec{F}[/itex](x,y,z)[itex]\hat{n}[/itex]=[itex]\int[/itex][itex]\int[/itex]-F[itex]_{x}[/itex]∂S[itex]/[/itex]∂x-F[itex]_{y}[/itex]∂S[itex]/[/itex]∂y+F[itex]_{z}[/itex]dS

Where,

S(x,y,f(x,y))=z=1-x[itex]^{2}[/itex]-y[itex]^{2}[/itex]

above the xy plane

and

[itex]\vec{F}[/itex](x,y,z)=[itex]\hat{j}[/itex]y+[itex]\hat{k}[/itex]

3. The attempt at a solution

After setting up the integral I attempt to switch to polar:

[itex]\int[/itex][itex]\int[/itex]-F[itex]_{x}[/itex]∂S[itex]/[/itex]∂x-F[itex]_{y}[/itex]∂S[itex]/[/itex]∂y+F[itex]_{z}[/itex]dS=[itex]\int[/itex][itex]\int[/itex]2y[itex]^{2}[/itex]+1 dxdy

=[itex]\int[/itex][itex]\int[/itex]2((rSin[θ])[itex]^{2}[/itex]+1)rdrdθ

from r=0 to r=1 and θ=0 to θ=2[itex]\pi[/itex]

=[itex]\int[/itex]Sin[itex]^{2}[/itex][θ][itex]/2[/itex]dθ+[itex]\pi[/itex]

[itex]\int[/itex]Sin[itex]^{2}[/itex][θ]dθ=[itex]\int[/itex](1-Cos[2θ])[itex]/[/itex]2dθ

=θ[itex]/[/itex]2-Sin[2θ][itex]/[/itex]4

∴

[itex]\int[/itex]Sin[itex]^{2}[/itex][θ][itex]/2[/itex]dθ+[itex]\pi[/itex]=[itex]\pi[/itex][itex]/[/itex]4+[itex]\pi[/itex]

The book says the answer is [itex]\pi[/itex][itex]/[/itex]2. Looking at what I have I I could say 1=r[itex]^{2}[/itex]

Then [itex]\int[/itex][itex]\int[/itex]2((rSin[θ])[itex]^{2}[/itex]+1)rdrdθ=[itex]\int[/itex][itex]\int[/itex]2((rSin[θ])[itex]^{2}[/itex]+r[itex]^{2}[/itex])rdrdθ and thus equal to [itex]\pi[/itex][itex]/[/itex]2.

Where am I going wrong? Thanks in advance.