Calculating Surface Integrals with Stokes' Theorem

In summary, the conversation discussed the use of Stokes' Theorem to evaluate integrals, with the first problem involving a cone and the second involving a hemisphere. The conversation also touched on parameterizing curves and determining orientation. Ultimately, it was found that the answers for both problems were relatively easy, with the first being 2187*pi/4 and the second being 0.
  • #1
Theelectricchild
260
0
I expected Stokes theorem to make my life easier but these problems are even harder than the normal ones I've been doing.


Use Stokes' Theorem to evaluate [tex]\int\int_ScurlFdS[/tex]

where F(x, y, z) = < x^2*y^3*z, sin(xyz) ,xyz >

S: Part of cone [tex]y^2 = x^2 + z^2[/tex] that lies between the planes y = 0 and y = 3 oriented in the direction of the positive y - axis.

not sure about this one... I parameterized the curve C as r(t) = < 3cost, 3, -3sint > to get a curve with positive orientation induced by S having normal vector in pos-y direction...a circle in plane y = 3,
answer: 2187*pi/4

Use Stokes' Theorem to evaluate [tex]\int\int_ScurlFdS[/tex]

where [tex]F(x, y, z)=< e^{xy}*cos(z) ,x^2*z, xy >[/tex]

S: Hemisphere [tex]x = sqrt{1-y^2-z^2}[/tex] oriented in the direction of the positive x - axis.

curve C : r(t) = <0, cost, sint> ... a circle again in plane x=0, hmmm, wonder if I'm messing these up...?
answer: 0 (this seemed way too easy...everything was zero!)
 
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  • #2
I assume you're talking about this verson of Stokes' theorem:
[tex]\iint_{S}\nabla\times\vec{F}\cdot{d}\vec{S}=\oint_{C}\vec{F}\cdot{d}\vec{r}[/tex]

In that case, your last calculation is correct; even if it is indecently easy.. :smile:
 
  • #3
Even for the first one? The answer just seemed completely off considering its size.
 
  • #4
At a glance, the first one ought to be zero as well..
 
  • #5
Sorry, first glance wrong..
 
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  • #6
If 3^{7}=2187, then the first should be correct as well.
 
  • #7
thank you :)
 

1. What is a surface integral?

A surface integral is a mathematical concept that extends the idea of a regular integral to surfaces in three-dimensional space. It involves calculating the flux, or flow, of a vector field across a given surface.

2. Why are surface integrals difficult?

Surface integrals can be difficult because they require a good understanding of vector calculus and parametric equations, as well as the ability to visualize and manipulate three-dimensional surfaces. Additionally, the complexity of the surface being integrated over can greatly affect the difficulty of the calculation.

3. What are some common applications of surface integrals?

Surface integrals have many applications in physics and engineering, including calculating electric flux, calculating the surface area of a three-dimensional object, and determining the mass and center of mass of an irregularly shaped object.

4. How do you set up a surface integral?

To set up a surface integral, you first need to determine the bounds of integration by finding the limits of the parameters that describe the surface. Then, you need to calculate the magnitude and direction of the vector field at each point on the surface and integrate it over the surface using a double integral.

5. What are some tips for solving difficult surface integrals?

One tip for solving difficult surface integrals is to break up the surface into smaller, more manageable pieces and then use the linearity property of integrals to add up the individual integrals. Another tip is to carefully choose the orientation of the surface and the direction of the normal vector to simplify the calculation. Additionally, practice and familiarity with different types of surfaces and vector fields can greatly improve your ability to solve difficult surface integrals.

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