Surface integral of a cylinder

In summary, the conversation is about evaluating a surface integral involving a cylinder and the use of Gauss's theorem to simplify the solution. The expert suggests using the original method of calculation to practice skills and using Gauss's theorem as a check for the answer.
  • #1
hover
343
0

Homework Statement



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

Homework Equations


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


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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  • #2
hover said:

Homework Statement



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

Homework Equations


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


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

I guess the easiest way is to convert into a volume integral, [itex]\int A\cdot\hat{n}\,dS=\int \nabla\cdot A\,dV[/itex], then you realize the divergence of A is really simple...
 
  • #3
sunjin09 said:
I guess the easiest way is to convert into a volume integral, [itex]\int A\cdot\hat{n}\,dS=\int \nabla\cdot A\,dV[/itex], then you realize the divergence of A is really simple...

Well if that equation you have there is equal then it would be very simple. The divergence of A is 3y^2. This is a triple integral and the answer would be 1280... even if I screwed up this integral... this method seems way too easy. What is the logic behind that equation you gave me?
 
  • #4
hover said:
Well if that equation you have there is equal then it would be very simple. The divergence of A is 3y^2. This is a triple integral and the answer would be 1280... even if I screwed up this integral... this method seems way too easy. What is the logic behind that equation you gave me?

I guess you haven't learned Gauss's theorem, which states exactly that identity I wrote:
http://en.wikipedia.org/wiki/Divergence_theorem
 
  • #5
sunjin09 said:
I guess you haven't learned Gauss's theorem, which states exactly that identity I wrote:
http://en.wikipedia.org/wiki/Divergence_theorem

Well I can't argue with that. Just one last question. Is my new answer right? I'll show my work if you want me too.
 
  • #6
hover said:
Well I can't argue with that. Just one last question. Is my new answer right? I'll show my work if you want me too.

Since you haven't known Gauss's theorem, the problem you have in hand is obviously not intended for you to use Gauss's theorem, but to practice your skill of straightforward calculation. So stick to your original method and make sure every step is right, you can use the other method to check your answer, which I can't do for you. Also note that in Gauss's theorem the surface integral also includes the two end surfaces of the cylinder.
 

1. What is the formula for calculating the surface integral of a cylinder?

The formula for calculating the surface integral of a cylinder is S = 2πrh + 2πr^2, where r is the radius of the cylinder and h is the height of the cylinder.

2. How is the surface integral of a cylinder related to its volume?

The surface integral of a cylinder is related to its volume through the formula V = πr^2h. This means that the surface area of a cylinder is directly proportional to its volume, with the constant of proportionality being 2.

3. Can the surface integral of a cylinder be negative?

No, the surface integral of a cylinder cannot be negative. This is because surface area is a measure of the amount of space occupied by an object, and it cannot be negative.

4. What is the significance of the surface integral of a cylinder in real-world applications?

The surface integral of a cylinder has many real-world applications, such as in engineering and construction. It is used to calculate the amount of material needed to construct a cylinder-shaped object, such as a pipe or a can. It is also used in physics and fluid mechanics to calculate the pressure and force exerted by fluids on the surface of cylinders.

5. Can the surface integral of a cylinder be used to find the surface area of other shapes?

No, the surface integral of a cylinder is specific to cylinders and cannot be used to find the surface area of other shapes. Each shape has its own unique formula for calculating surface area.

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