(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Evaluate the surface integral ∫A[itex]\bullet[/itex][itex]\hat{n}[/itex]dS where A = z[itex]\hat{x}[/itex]+x[itex]\hat{y}[/itex]+3y^2z[itex]\hat{z}[/itex] and S is the cylinder x^2+y^2=16 for the range of x[itex]\geq[/itex]0,y[itex]\geq[/itex]0, 0[itex]\leq[/itex]z[itex]\leq[/itex]5

2. Relevant equations

I used this page as an example way to do this. I'm not good with surface integrals.

3. The attempt at a solution

I know I did this wrong because my answer seems wrong but... following that page linked to above

[itex]\Phi[/itex](θ,t) = 4cos(θ)[itex]\hat{x}[/itex]+4sin(θ)[itex]\hat{y}[/itex]+t[itex]\hat{z}[/itex]

where 0[itex]\leq[/itex]θ[itex]\leq[/itex]2∏,0[itex]\leq[/itex]t[itex]\leq[/itex]5

[itex]\partial[/itex][itex]\Phi[/itex]/[itex]\partial[/itex]θ = -4sin(θ)[itex]\hat{x}[/itex]+4cos(θ)[itex]\hat{y}[/itex]

[itex]\partial[/itex][itex]\Phi[/itex]/[itex]\partial[/itex]t = [itex]\hat{z}[/itex]

normal vector is the cross product between those two derivatives which I found it equal to

4cos(θ)[itex]\hat{x}[/itex]+4sin(θ)[itex]\hat{y}[/itex]

setting up the integral

(the integral on the left is from 0 to 5 and the second integral is from 0 to 2∏)

∫∫A([itex]\Phi[/itex](θ,t))[itex]\bullet[/itex](that cross product I got)dθdt

= ∫∫(t[itex]\hat{x}[/itex]+4cos(θ)[itex]\hat{y}[/itex]+3*16*(sin(θ))^2*t)[itex]\bullet[/itex](4cos(θ)[itex]\hat{x}[/itex]+4sin(θ)[itex]\hat{y}[/itex])dθdt

= ∫∫(4t*cos(θ)+16sin(θ)cos(θ))dθdt

=∫ (4t*sin(θ)-4cos(2θ))dt evaluated for 0 to 2π

and this is where I get zero.... where did I go wrong?

thanks!

Hover

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# Surface integral of a cylinder

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