# Homework Help: Surface Integral of a Cylinder!

1. Nov 25, 2012

### Syrena

1. The problem statement, all variables and given/known data
Let S denote the closed cylinder with bottom given by z=0, top given by z=4, and lateral surface given by the equation x^2 + y^2 = 9. Orient S with outward normals. Determine the indicated scalar and vector surface integral to ∫∫ x^2 i dS (I have tried to solve this problem, but i dont think i have done it correct, since i get the answer 0. Please if anybody can help, this task is very importaint to get right)

2. Relevant equations
$\int\int$x f dS=$\int\int$ D f(X(s,t))$||$Ts$\times$Tt$||$

3. The attempt at a solution
Since this is a cylinder, (i think) we can slice it in to 3 parts,
S1 (lateral cylindrical surface), S2 (bottom disk) and S3 (top disk).
I parametrized the three smooth pieces as follows:
S1 (lateral cylindrical surface): x=3Cos(t), y=3Sin(t), z=t With bouderies 0$\leq$s$\leq$2π and 0$\leq$t$\leq$4

S2 (bottom disk): x=sCos(t), y=sSin(t), z=0. With bounderies 0$\leq$s$\leq$3 and 0$\leq$t$\leq$2π

S3 (top disk): x=sCos(t), y=sSin(t), z=4. With bounderies 0$\leq$s$\leq$3 and 0$\leq$t$\leq$2π

Then I found the ||Ts$\times$Tt$||$ for each S, that is
S1||Ts$\times$Tt$||$ =(3Cos(s), -3Sins,0)

S2||Ts$\times$Tt$||$ =(0,0,sCos^2 (t)+sSin^2 (t))

S3||Ts$\times$Tt$||$ =(0,0,sCos^2 (t)+sSin^2 (t))

Then I thought i could set in all that is in the parametrizised in the x^2 positision, dotted with the ||Ts$\times$Tt$||$.
S2 and S3 will then be 0 ( since ||Ts$\times$Tt$||$ have both 0 in the x position)

The last one S1
$\int\int$ X^2 dS=$\int\int$ (3^2 Cos^2 (s), 0, 0) ×(3Cos(s),3Sin(s),0) ds dt (boundaries 0 and 4(dt), 0 and 2π(ds))

=∫∫12 Cos^3 (s) dsdt (boundaries 0 and 4(dt), 0 and 2π(ds))

=12∫∫ Cos^3 (s) dsdt (boundaries 0 and 4(dt), 0 and 2π(ds))

=12∫ -3Sin(s)Cos^2 (s) dt (input 0 and 2π( for s)) (boundaries 0 and 4(dt)

This gives =12∫ dt wich gives 0 <-- and i dont think this is the right answer. If anybody can help

2. Nov 25, 2012

### HallsofIvy

Was this a typo? You should have x= 3Cos(s), y= 3sin(s), not "t".

It is a bit confusing to use the same letters as before for the parameters, but change their meaning.

Okay, so the "t" above was a typo. This is correct for the lateral surface.

S2||Ts$\times$Tt$||$ =(0,0,sCos^2 (t)+sSin^2 (t))[/quote]
You do know that $s cos^2(t)+ s sin^2(t)= s$, right? :tongue:

Actually, 0 is correct! This integral measures the flow of the vector quantity $x^2$i into the cyinder. Over the top and bottom, which are parallel to the xy-plane, there is no flow through the surface. For the lateral surrface, the flow into the cylinder on the x< 0 side is cancelled by flow out of the cylinder on the x> 0 side.

3. Nov 25, 2012

### Syrena

I Thank Thee!
I did find though that I had some mistakes in my calculations:

=∫∫12 Cos^3 (s) dsdt (boundaries 0 and 4(dt), 0 and 2π(ds))
is ∫12 (1/12(9sin(s)+sin(3s)) dt =∫(9sin(s)+sin(3s))dt instead of =12∫ -3Sin(s)Cos^2 (s) dt which I wrote at the beginning.
But =∫(9sin(s)+sin(3s))dt ( with boundaries 0 to 2π) also gives the answer 0.