Evaluating a Surface Integral for S: How Can the Given Formula be Used?

In summary: So, for this problem, you would need to know the following: the equation of the surface (something you could find from a geometry textbook or online), the coordinates of a point on the surface, the direction of the vector from that point to the origin, and the length of the vector.
  • #1
wifi
115
1
Problem:

Use the fact that [tex]\int_S \vec{v} \cdot d\vec{S}=\int_S \vec{v} \cdot \frac{\nabla f}{\partial f/\partial x} dy\ dz[/tex]

to evaluate the integral for ##S=\{(x,y,z):y=x^2 ; 0 \geq x \geq 2; 0 \geq z \geq 3 \}## and ##\vec{v}=(3z^2, 6, 6xz)##.

Attempt at a Solution:

I'm having trouble setting up this integral. If I knew what ##f## was, I could easily calculate the gradient, as well as the partial wrt to x. I'd still need to figure out the limits of integration though.
 
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  • #2
f(x,y,z) ought to be the scalar function describing the surface S through the relationship:
f(x,y,z)=0
 
  • #3
arildno said:
f(x,y,z) ought to be the scalar function describing the surface S through the relationship:
f(x,y,z)=0

Hmm. I'm not sure you'd get this from the info given. Any hints?
 
  • #4
Because
1. It fits. Sort of (I admit I haven't seen closely, though).
2. Otherwise, it would be utterly meaningless, since you would have no way to evaluate the second expression due to lack of knowledge of f.
 
  • #5
More specifically I meant, how would you find ##f## analytically from this info given?
 
  • #6
wifi said:
More specifically I meant, how would you find ##f## analytically from this info given?

From here:

wifi said:
##S=\{(x,y,z):y=x^2 ; 0 \geq x \geq 2; 0 \geq z \geq 3 \}##

If [itex]y = x^2[/itex] then [itex]y - x^2 = 0[/itex].

(Your inequalities are the wrong way round: you want [itex]0 \leq x \leq 2[/itex] etc.)
 
  • #7
pasmith said:
From here:



If [itex]y = x^2[/itex] then [itex]y - x^2 = 0[/itex].

(Your inequalities are the wrong way round: you want [itex]0 \leq x \leq 2[/itex] etc.)

So ##f(x,y,z)=y-x^2##?
 
  • #8
I would parameterize the surface as ##\vec R(x,z)## and use the formula$$
\iint \vec v\cdot d\vec S =\pm \iint_{x,z}\vec v \cdot \vec R_x\times \vec R_z~dxdz$$where the choice of signs depends on the orientation of the surface which, by the way, you need to specify.
 

1. What is a surface integral?

A surface integral is an extension of the concept of a definite integral to functions of two or more variables defined on a surface in three-dimensional space. It is used to calculate the total amount of a scalar or vector quantity over a given surface.

2. What is the purpose of evaluating a surface integral?

The purpose of evaluating a surface integral is to find the total value of a function over a given surface. This is useful in many applications, such as calculating the flux of a vector field through a surface or finding the mass or center of mass of an object with a varying density over its surface.

3. How is a surface integral calculated?

A surface integral is calculated by first parameterizing the surface into a set of two variables, typically u and v, which represent the coordinates on the surface. Then, the function being integrated is multiplied by the magnitude of the cross product of the partial derivatives of the parameterization. The resulting expression is integrated over the ranges of u and v to obtain the final value.

4. What are some common applications of surface integrals?

Surface integrals have many applications in physics, engineering, and mathematics. They are commonly used to calculate the flux of a vector field through a surface, calculate the work done by a force over a surface, and determine the total mass or center of mass of an object with a varying density. They are also used in surface area calculations and in parametric equations for curves and surfaces.

5. Are there any practical tips for evaluating a surface integral?

One practical tip for evaluating a surface integral is to carefully choose the parameterization of the surface. This can greatly simplify the integral and make it easier to solve. Additionally, breaking the surface into smaller, simpler regions and then summing the integrals over each region can also make the process more manageable. It is also important to pay attention to the ranges of u and v and ensure they are consistent with the orientation of the surface.

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