Using stokes' or divergence theorem to solve integral

In summary: I didn't see that! Thanks!In summary, the problem involves using either Stokes' theorem or the divergence theorem to evaluate an integral over a closed surface of a tin can. The vector V is given by 2xyi - y^2j + (z + xy)k, and the divergence of V is 2y - 2y + 1. Using the divergence theorem, the integral can be transformed into a volume integral. The volume element, dτ, is equal to dx dy dz in cartesian coordinates and r dr dθ dz in cylindrical coordinates. After integrating over the cylinder, the 2x and 2y terms drop out, leaving only the constant term of 1.
  • #1
leroyjenkens
616
49

Homework Statement



Use either Stokes' theorem or the divergence theorem to evaluate this integral in the easiest possible way.

∫∫V [itex]\cdot[/itex]ndσ over the closed surface of the tin can bounded by x2+y2=9, z = 0, z = 5, if V = 2xyi - y2j + (z + xy)k

The bolded letters are vectors.


Homework Equations



∫∫( xV) [itex]\cdot[/itex] ndσ = ∫V [itex]\cdot[/itex]dr Stokes' theorem

∫∫∫ [itex]\cdot[/itex]Vdτ = ∫∫V [itex]\cdot[/itex]ndσ Divergence theorem


The Attempt at a Solution



So what I did was use the divergence theorem to turn this into a volume integral. I did the ∇xV to get 2y-2y+1. But here's where I'm stuck. I can input the formula for the volume of a cylinder, right? Is that dτ? I have to integrate with respect to τ. So if 2y-2y+1 = 1, then I'm left with τ after I integrate, and is τ just ∏r2h?

Thanks
 
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  • #2
You mean you calculated ##\nabla\cdot \vec{V}##, not the curl. ##d\tau## is the volume element. For cartesian coordinates, it's equal to ##d\tau = dx\,dy\,dz##. The fact that you're integrating over a cylindrical volume is reflected in the limits. If I were you, however, I'd switch to cylindrical coordinates, where ##d\tau = r\,dr\,d\theta\,dz##.

EDIT: Never mind the last sentence. I misread your expression for the divergence and thought you would need to actually evaluate a non-trivial integral. Your work looks fine. If the divergence weren't simply a constant, you'd probably be better off switching to cylindrical coordinates.
 
  • #3
leroyjenkens said:

Homework Statement



Use either Stokes' theorem or the divergence theorem to evaluate this integral in the easiest possible way.

∫∫V [itex]\cdot[/itex]ndσ over the closed surface of the tin can bounded by x2+y2=9, z = 0, z = 5, if V = 2xyi - y2j + (z + xy)k

The bolded letters are vectors.

Homework Equations



∫∫( xV) [itex]\cdot[/itex] ndσ = ∫V [itex]\cdot[/itex]dr Stokes' theorem

∫∫∫ [itex]\cdot[/itex]Vdτ = ∫∫V [itex]\cdot[/itex]ndσ Divergence theorem

The Attempt at a Solution



So what I did was use the divergence theorem to turn this into a volume integral. I did the ∇xV to get 2y-2y+1. But here's where I'm stuck. I can input the formula for the volume of a cylinder, right? Is that dτ? I have to integrate with respect to τ. So if 2y-2y+1 = 1, then I'm left with τ after I integrate, and is τ just ∏r2h?

Thanks

That all sounds reasonable. You can switch to cylindrical co-ordinates to make integrating easier I believe.
 
  • #4
vela said:
You mean you calculated ##\nabla\cdot \vec{V}##, not the curl. ##d\tau## is the volume element. For cartesian coordinates, it's equal to ##d\tau = dx\,dy\,dz##. The fact that you're integrating over a cylindrical volume is reflected in the limits. If I were you, however, I'd switch to cylindrical coordinates, where ##d\tau = r\,dr\,d\theta\,dz##.

EDIT: Never mind the last sentence. I misread your expression for the divergence and thought you would need to actually evaluate a non-trivial integral. Your work looks fine. If the divergence weren't simply a constant, you'd probably be better off switching to cylindrical coordinates.

But the divergence isn't a constant. It's 2x-2y+1. After you integrate over the cylinder then the 2x and 2y drop out. But you should say why.
 
  • #5
That's the same mistake I made originally, but the first term is really 2y, not 2x.
 
  • #6
vela said:
That's the same mistake I made originally, but the first term is really 2y, not 2x.

Ooo. Yes, it is.
 

1. What is Stokes' Theorem and how is it used to solve integrals?

Stokes' Theorem is a mathematical theorem that relates a surface integral over a surface to a line integral around the boundary of the surface. It can be used to solve integrals by converting a difficult surface integral into a simpler line integral, which can then be evaluated using standard techniques.

2. What is the difference between Stokes' Theorem and Divergence Theorem?

Stokes' Theorem is used to relate a surface integral to a line integral, while the Divergence Theorem relates a volume integral to a surface integral. Essentially, Stokes' Theorem is a special case of the Divergence Theorem, where the volume being integrated over is a surface.

3. When is it appropriate to use Stokes' Theorem to solve an integral?

Stokes' Theorem is most commonly used when the surface being integrated over is a closed surface, or when the boundary of the surface can be easily defined. It is also useful when the surface is defined by a vector field, as the line integral can then be evaluated using the fundamental theorem of calculus.

4. Can Stokes' Theorem be used to solve integrals in any coordinate system?

Yes, Stokes' Theorem can be used in any coordinate system, as long as the surface and boundary are defined in terms of the same coordinates. However, it is often easier to use Stokes' Theorem in Cartesian coordinates, as the vector operations are simpler.

5. What are some practical applications of Stokes' Theorem in real-world problems?

Stokes' Theorem has many applications in physics and engineering, particularly in fluid mechanics, electromagnetism, and heat transfer. It is used to calculate fluxes, forces, and work done by vector fields, and is also used in the study of fluid flow past boundaries and in the analysis of electric and magnetic fields.

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