- #1
Syrena
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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
[itex]\int\int[/itex]x f dS=[itex]\int\int[/itex] D f(X(s,t))[itex]||[/itex]Ts[itex]\times[/itex]Tt[itex]||[/itex]
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[itex]\leq[/itex]s[itex]\leq[/itex]2π and 0[itex]\leq[/itex]t[itex]\leq[/itex]4
S2 (bottom disk): x=sCos(t), y=sSin(t), z=0. With bounderies 0[itex]\leq[/itex]s[itex]\leq[/itex]3 and 0[itex]\leq[/itex]t[itex]\leq[/itex]2π
S3 (top disk): x=sCos(t), y=sSin(t), z=4. With bounderies 0[itex]\leq[/itex]s[itex]\leq[/itex]3 and 0[itex]\leq[/itex]t[itex]\leq[/itex]2π
Then I found the ||Ts[itex]\times[/itex]Tt[itex]||[/itex] for each S, that is
S1||Ts[itex]\times[/itex]Tt[itex]||[/itex] =(3Cos(s), -3Sins,0)
S2||Ts[itex]\times[/itex]Tt[itex]||[/itex] =(0,0,sCos^2 (t)+sSin^2 (t))
S3||Ts[itex]\times[/itex]Tt[itex]||[/itex] =(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[itex]\times[/itex]Tt[itex]||[/itex].
S2 and S3 will then be 0 ( since ||Ts[itex]\times[/itex]Tt[itex]||[/itex] have both 0 in the x position)
The last one S1
[itex]\int\int[/itex] X^2 dS=[itex]\int\int[/itex] (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