How to Solve a Surface Integral with Given Limits and Variables?

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In summary, the problem is to evaluate the double integral of (x^2 + 4y + z) over the portion of the plane 3y + 2z = 6 with 0 < x < 3. The author attempts to solve the integral by choosing different variable combinations, but concludes that the integral does not exist due to the infinite extent of the surface. The possibility of integrating over the first octant is also mentioned, but the problem provided does not specify this condition.
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Homework Statement



Evaluate [tex]\int[/tex][tex]\int[/tex] (x^2 +4y + z)dS where S is the portion of the plane 3y +2z = 6 with 0<x<3

Homework Equations


choosing x, y as variables
dS = [tex]\sqrt{}(partial dz/dx)^2 + (partial dz/dy)^2 +1[/tex]

The Attempt at a Solution


If i choose x, y as variables or x, z as variables, then when i choose limit of intergral, x will go from 0 to 3 whereas y or z will go from -infinite to +infinite, and solving the intergral results of infinite result (seem to be not right to me)
if choosing y,z as variables, then i can't "translate" f = x^2 + 4y + z into parameterized f since x does not depend on y and z in the plane 3y +2z = 6 (in the common problems, the plane involves x, so i just forming x = g(y,z) and then sub x into f which is in the intergral)
Any help would highly appriciated.
 
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  • #2
Are you sure you have read the problem completely? If the only condition is that "0< x< 3", then, yes, the surface is of infinite extent and the integral does not exist. Is it possible that you are asked to integrate over the part of the plane in the first octant with 0< x< 3?
 
  • #3
the problem is given exactly above, nothin say about first obtant (i also thought about 1st obtant, too when i read the word "portion of", but it seems not that) so i guess the intergral is infinite is right, maybe the instructor gave the wrong problem. I wasn't sure so i asked, thank you for pointing out.
 

FAQ: How to Solve a Surface Integral with Given Limits and Variables?

1. What is a surface integral?

A surface integral is a type of multiple integral used in mathematics and physics to calculate the flux of a vector field through a surface. It is used to measure the amount of a quantity flowing through a surface rather than just at a single point.

2. How is a surface integral different from a regular integral?

A surface integral is different from a regular integral because it is performed over a two-dimensional surface rather than a one-dimensional interval. It also involves a vector field and a normal vector to the surface.

3. What are some common applications of surface integrals?

Surface integrals have many applications in physics, such as calculating the electric or magnetic flux through a surface, determining the mass or center of mass of a curved object, and finding the flow of a fluid through a surface.

4. What is the general formula for a surface integral?

The general formula for a surface integral is ∬SF(x,y,z)·dS, where F is the vector field, dS is the differential of the surface area, and the integral is taken over the surface S.

5. How can I evaluate a surface integral?

To evaluate a surface integral, you first need to determine the limits of integration, which depend on the shape of the surface. Then, you can use various techniques such as parametrization, Green's theorem, or Stokes' theorem to simplify the integral and solve for the desired quantity.

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