Surface Integrals: Flux of F across S

In summary, the conversation discusses the evaluation of a surface integral for a given vector field and oriented surface. The vector field is defined as F(x, y, z) = xy i + yz j + zx k, and the surface is a part of the paraboloid z = 3 - x^{2} - y^{2} that lies above a square in the xy plane with bounds 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. The formula for the surface integral is given as \int \int F\cdot dS = \int \int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R
  • #1
ryoonc
5
0

Homework Statement


Evaluate the surface integral ∫∫S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.

F(x, y, z) = xy i + yz j + zx k

S is the part of the paraboloid z = 3 - x[tex]^{2}[/tex] - y[tex]^{2}[/tex] that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation.

Homework Equations


[tex]\int \int F\cdot dS = \int \int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) dA[/tex]

The Attempt at a Solution


I've gone over the examples in my calculus book two or three times now and I get confused about a couple things:

dA is r*dr*d[tex]\theta[/tex], and I replace the x's and y's with rcos[tex]\theta[/tex] and rsin[tex]\theta[/tex], respectively, and substituting z with the given equation, but I still get a wrong answer. Here's my attempt:

[tex]\int \int ( -y(-2x) -x(-2y) +3-x^{2}-y^{2})dA[/tex]

...and after substituting x and y and z for their polar coordinates and then simplifying, I get:

[tex]\int ^{\pi / 2}_{0} \int ^{1}_{0} (3+4r^{2}cos(\theta)sin(\theta)-r^{2})rdrd\theta[/tex]

I was kind of hoping this would get me the right answer, but it's not (I end up with [tex] \frac{5 \pi}{8} + \frac{1}{2}[/tex]), and I think it has to do with the r domain that I used, or perhaps my entire equation. In any case I'm not sure how to implement the square of length 1 in xy plane domain into my equation. Where have I gone wrong?

Thanks in advance
 
Physics news on Phys.org
  • #2
ryoonc said:

Homework Statement


Evaluate the surface integral ∫∫S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.

F(x, y, z) = xy i + yz j + zx k
This is your P(x,y)i + Q(x,y)j + R(x,y)k

S is the part of the paraboloid z = 3 - x[tex]^{2}[/tex] - y[tex]^{2}[/tex]

This is your z = g(x,y)

that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation.

Homework Equations


[tex]\int \int F\cdot dS = \int \int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) dA[/tex]

The Attempt at a Solution


I've gone over the examples in my calculus book two or three times now and I get confused about a couple things:

dA is r*dr*d[tex]\theta[/tex], and I replace the x's and y's with rcos[tex]\theta[/tex] and rsin[tex]\theta[/tex], respectively, and substituting z with the given equation, but I still get a wrong answer. Here's my attempt:

[tex]\int \int ( -y(-2x) -x(-2y) +3-x^{2}-y^{2})dA[/tex]

Are your P(x,y), Q(x,y) and R(x,y) terms here correct?

...and after substituting x and y and z for their polar coordinates and then simplifying, I get:

Why in the world would you try polar coordinates at this point? The integrand is simple polynomial terms in x and y and you are integrating over a square.​
 
Last edited:

Related to Surface Integrals: Flux of F across S

1. What is a surface integral?

A surface integral is a type of double integral that is used to calculate the flux (flow) of a vector field across a surface. It involves integrating a function over a two-dimensional surface in three-dimensional space.

2. How is the flux of a vector field across a surface calculated?

The flux of a vector field across a surface is calculated by taking the dot product of the vector field and the surface's normal vector, and then integrating this dot product over the surface using a surface integral.

3. What is the significance of surface integrals in physics?

Surface integrals have many applications in physics, such as calculating the flow of a fluid through a surface, the electric flux through a closed surface, or the surface area of an object. They also play a crucial role in the fundamental laws of electromagnetism, such as Gauss's Law.

4. How are surface integrals used in real-world problems?

Surface integrals are used in a variety of real-world problems, including fluid dynamics, electromagnetism, and heat transfer. For example, they can be used to calculate the amount of heat transferred through a surface, or the amount of fluid passing through a boundary.

5. Are there any applications of surface integrals outside of mathematics and physics?

Yes, surface integrals have applications in other fields such as engineering, computer graphics, and computer vision. They are used to calculate the surface area of 3D objects, perform texture mapping in computer graphics, and analyze surfaces in 3D images in computer vision.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
730
  • Calculus and Beyond Homework Help
Replies
6
Views
732
  • Calculus and Beyond Homework Help
Replies
8
Views
986
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
564
  • Calculus and Beyond Homework Help
Replies
8
Views
294
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
435
  • Calculus and Beyond Homework Help
Replies
7
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top