Surface Integral From Div, Grad, Curl and all that

[SoCoToAh]

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.

Homework Statement

Evaluate the surface integral using the following equation.

Homework Equations

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

Where,
S(x,y,f(x,y))=z=1-x$^{2}$-y$^{2}$

above the xy plane

and

$\vec{F}$(x,y,z)=$\hat{j}$y+$\hat{k}$

The Attempt at a Solution

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

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

=$\int$$\int$2((rSin[θ])$^{2}$+1)rdrdθ
from r=0 to r=1 and θ=0 to θ=2$\pi$
=$\int$Sin$^{2}$[θ]$/2$dθ+$\pi$

$\int$Sin$^{2}$[θ]dθ=$\int$(1-Cos[2θ])$/$2dθ
=θ$/$2-Sin[2θ]$/$4
∴
$\int$Sin$^{2}$[θ]$/2$dθ+$\pi$=$\pi$$/$4+$\pi$

The book says the answer is $\pi$$/$2. Looking at what I have I I could say 1=r$^{2}$
Then $\int$$\int$2((rSin[θ])$^{2}$+1)rdrdθ=$\int$$\int$2((rSin[θ])$^{2}$+r$^{2}$)rdrdθ and thus equal to $\pi$$/$2.

Where am I going wrong? Thanks in advance.

 

[vela]

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.

Homework Statement

Evaluate the surface integral using the following equation.

Homework Equations

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

Where,
S(x,y,f(x,y))=z=1-x$^{2}$-y$^{2}$

above the xy plane

and

$\vec{F}$(x,y,z)=$\hat{j}$y+$\hat{k}$

The Attempt at a Solution

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

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

=$\int$$\int$2((rSin[θ])$^{2}$+1)rdrdθ
from r=0 to r=1 and θ=0 to θ=2$\pi$
=$\int$Sin$^{2}$[θ]$/2$dθ+$\pi$

$\int$Sin$^{2}$[θ]dθ=$\int$(1-Cos[2θ])$/$2dθ
=θ$/$2-Sin[2θ]$/$4
∴
$\int$Sin$^{2}$[θ]$/2$dθ+$\pi$=$\pi$$/$4+$\pi$

You should get $3\pi/2$. The first term works out to $\pi/2$, not $\pi/4$.

The book says the answer is $\pi$$/$2. Looking at what I have I I could say 1=r$^{2}$
Then $\int$$\int$2((rSin[θ])$^{2}$+1)rdrdθ=$\int$$\int$2((rSin[θ])$^{2}$+r$^{2}$)rdrdθ and thus equal to $\pi$$/$2.

Where am I going wrong? Thanks in advance.

To get an answer of $\pi/2$, you need to include the contribution from the bottom surface. What exactly did the problem ask you to calculate?

 

[SoCoToAh]

Actually, after looking, it appears I may have looked at the wrong answer. 3(pi)/2 is the answer. Sorry for my carelessness.