Surface Integral of a Cylinder

[Syrena]

Homework Statement

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 don't 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)

Homework Equations

\int\int_x f dS=\int\int _D f(X(s,t))||T_s\timesT_t||

The Attempt at a Solution

Since this is a cylinder, (i think) we can slice it into 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\leqs\leq2π and 0\leqt\leq4

S₂ (bottom disk): x=sCos(t), y=sSin(t), z=0. With bounderies 0\leqs\leq3 and 0\leqt\leq2π

S₃ (top disk): x=sCos(t), y=sSin(t), z=4. With bounderies 0\leqs\leq3 and 0\leqt\leq2π

Then I found the ||T_s\timesT_t|| for each S, that is
S₁||T_s\timesT_t|| =(3Cos(s), -3Sins,0)

S₂||T_s\timesT_t|| =(0,0,sCos^2 (t)+sSin^2 (t))

S₃||T_s\timesT_t|| =(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 ||T_s\timesT_t||.
S₂ and S₃ will then be 0 ( since ||T_s\timesT_t|| 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 which gives 0 <-- and i don't think this is the right answer. If anybody can help

 

[HallsofIvy]

Homework Statement

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 don't 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)

Homework Equations

\int\int_x f dS=\int\int _D f(X(s,t))||T_s\timesT_t||

The Attempt at a Solution

Since this is a cylinder, (i think) we can slice it into 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\leqs\leq2π and 0\leqt\leq4

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

S₂ (bottom disk): x=sCos(t), y=sSin(t), z=0. With bounderies 0\leqs\leq3 and 0\leqt\leq2π

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

S₃ (top disk): x=sCos(t), y=sSin(t), z=4. With bounderies 0\leqs\leq3 and 0\leqt\leq2π

Then I found the ||T_s\timesT_t|| for each S, that is
S₁||T_s\timesT_t|| =(3Cos(s), -3Sins,0)

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

S₂||T_s\timesT_t|| =(0,0,sCos^2 (t)+sSin^2 (t))[/quote]
You do know that s cos^2(t)+ s sin^2(t)= s, right?

S₃||T_s\timesT_t|| =(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 ||T_s\timesT_t||.
S₂ and S₃ will then be 0 ( since ||T_s\timesT_t|| have both 0 in the x position)
I'm not sure what you mean by the "x^2" position.

For the lateral sides, x^2= (3 cos(s))^2= 9cos^2(t) and idS itex is 3cos(s)ds dt so the integral is 27\int\int cos^3(s)dsdt

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 which gives 0 <-- and i don't think this is the right answer. If anybody can help

Actually, 0 is correct! This integral measures the flow of the vector quantity x^2i 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 canceled by flow out of the cylinder on the x> 0 side.

 

[Syrena]

Actually, 0 is correct! This integral measures the flow of the vector quantity x^2i 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 canceled by flow out of the cylinder on the x> 0 side.

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.