Solve Surface Integral: \vec{F}\cdot\vec{n}\, dS

In summary, the surface integral \vec{F}\cdot\vec{n}\, dS is being evaluated, where \vec{F}=<-y,x,0> and S is the part of the plane z=8x-4y-5 that lies below the triangle with vertices at (0,0,0,), (0,1,0,) and (1,0,0). The orientation of S is given by the upward normal vector. The process involves parametrizing S, finding the partial derivatives, and using them to simplify the integral. However, there was an error in the integration limits, which was corrected to be over the triangle formed by (0,0,0), (0,1
  • #1
Saladsamurai
3,020
7

Homework Statement


Evaluate the surface integral [tex]\vec{F}\cdot\vec{n}\, dS[/tex]

where [itex]\vec{F}=<-y,x,0>[/itex] and S is the part of the plane [itex]z=8x-4y-5[/itex] that lies below the triangle with vertices at (0,0,0,), (0,1,0,) and (1,0,0). The orientation of S is given by the upward normal vector. answer: 2


I am not sure if I am just making a careless mistake or a conceptual one.

The Attempt at a Solution



i.) Parametrizing S gives [itex]\vec{r}(x,y)=<x, y, 8x-4y-5>[/itex]

ii.) Finding [tex]\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y}=<1,0,8>\times<0,1,-4>=-8,4,1[/tex]

iii.) Thus, [itex] \vec{F}(\vec{r}(x,y))\cdot (\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y})=<-y,x,0>\cdot<-8,4,1>=<4x+8y>[/itex]

iv) Therefore [tex]I=\int\int_D (4x+8y)\, dA[/tex]

[tex]=\int_{x=0}^1 \int_{y=0}^x (4x+8y)\,dy\, dx[/tex]

I believe that if I made an error, it was made somewhere in here and not in my integration.

Any major blunders here?

Thanks,
Casey
 
Physics news on Phys.org
  • #2
Look at your integral limits. You are integrating over the triangle formed by (0,0,0), (0,1,0) and (1,1,0).
 
  • #3
Dick said:
Look at your integral limits. You are integrating over the triangle formed by (0,0,0), (0,1,0) and (1,1,0).

Not sure I follow. My x bounds are correct right?
 
  • #4
Is my upper y bound supposed to be (1-x)?
 
  • #5
Saladsamurai said:
Is my upper y bound supposed to be (1-x)?

It sure is!
 
  • #6
:redface: That's what I get for thinking I can do everything in my head...Teehee...
 

What is a surface integral?

A surface integral is a mathematical technique used to calculate the flux, or flow, of a vector field over a surface. It is commonly used in physics and engineering to solve problems related to fluid flow, electromagnetism, and other physical phenomena.

What does \vec{F} represent in the equation?

\vec{F} represents a vector field, which is a function that assigns a vector to each point in space. It can be thought of as a collection of arrows, each with a specific magnitude and direction. In the context of the surface integral, \vec{F} represents the quantity being integrated over the surface.

What is \vec{n} in the equation?

\vec{n} represents the unit normal vector to the surface at a given point. It is perpendicular to the surface and is used to determine the direction of the flux through the surface.

How do you calculate a surface integral?

To calculate a surface integral, you first need to determine the limits of integration, which are the boundaries of the surface being integrated over. Then, you need to evaluate the dot product of \vec{F} and \vec{n} at each point on the surface. Finally, you integrate this dot product over the surface, typically using a double integral.

What are some real-world applications of surface integrals?

Surface integrals have many real-world applications, including calculating the flow of fluids through pipes, determining the electric field around a charged object, and measuring the heat transfer on a surface. They are also used in computer graphics to create realistic 3D models of objects.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
474
  • Calculus and Beyond Homework Help
Replies
8
Views
809
  • Calculus and Beyond Homework Help
Replies
4
Views
751
  • Calculus and Beyond Homework Help
Replies
6
Views
776
  • Calculus and Beyond Homework Help
Replies
5
Views
536
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
369
  • Calculus and Beyond Homework Help
Replies
20
Views
351
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
426
Back
Top