Solve Surface Integral for xy-Plane Projection

In summary, the conversation discusses converting a surface integral into a double integral and determining the appropriate limits of integration for a given region. The speaker suggests projecting the region onto the xy-plane and notes that for the given paraboloid between z = 0 and z = 1, the projection is a circle of radius 1 at z = 1. The speaker also mentions that the limits of integration for the surface integral may change if the region is between z = 1 and z = 2.
  • #1
awvvu
188
1

Homework Statement


I'm not sure how to convert this surface integral into a double integral for evaluation.

[tex]\iint_S \frac{1}{1 + 4(x^2 + y^2)} dS[/tex]

S is the portion of the paraboloid [itex]z = x^2 + y^2[/itex] between z = 0 and z = 1.

The Attempt at a Solution


How do you project this onto the xy-plane? From z = 0 to z = 1, we get circles of increasing radius. So do I just take z = 1, which is a circle of radius 1, and make the region:

[tex]D = \{(r, \theta) | 0 <= r <= 1, 0 <= \theta <= 2 \pi\}[/tex]

Also, I know:

[tex]dS = \sqrt{1 + (2x)^2 + (2y)^2} dA = \sqrt{4r^2 + 1} dA[/tex]

[tex]\iint_S \frac{1}{1 + 4(x^2 + y^2)} dS = \iint_D \frac{r}{1 + 4r^2} \sqrt{4r^2 + 1} dA[/tex]

I just don't know what the limits should be. The region I chose gives me the right answer but I'm not sure exactly why the projection was when z = 1. What if it was the paraboloid between z = 1 and z = 2 then? Do I have to split up the surface integral to be the circle at z = 1 and the paraboloid surface? I'm really confused about this projection stuff.
 
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  • #2
The plane z=a cuts the paraboloid in the circle a=x^2+y^2. So the domain in the x,y plane is x^2+y^2<=a or r^2<=a. If the limits were z=1 to z=2, then the r limits would be r=1 to r=sqrt(2).
 

Related to Solve Surface Integral for xy-Plane Projection

1. What is the definition of a surface integral?

A surface integral is a mathematical concept used to calculate the area of a 3-dimensional surface. It involves integrating a function over a two-dimensional surface to find the total value of the function over that surface.

2. How do you set up a surface integral for a xy-plane projection?

To set up a surface integral for a xy-plane projection, you first need to determine the bounds of the integral. This can be done by identifying the limits of the x and y variables. Then, you need to express the function in terms of x and y, and multiply it by the differential area element dA = dx dy. Finally, integrate the function over the bounds that were determined earlier.

3. What is the purpose of solving a surface integral for a xy-plane projection?

The purpose of solving a surface integral for a xy-plane projection is to find the total value of a function over a two-dimensional surface. This can be useful in many scientific fields such as physics, engineering, and economics.

4. What are some common applications of solving surface integrals for xy-plane projections?

Surface integrals for xy-plane projections are commonly used in physics to calculate the electric or magnetic flux through a surface, in engineering to calculate the flow of a fluid through a surface, and in economics to calculate the surface area of a production possibility frontier.

5. Can a surface integral for xy-plane projection be solved using any integration technique?

Yes, a surface integral for xy-plane projection can be solved using various integration techniques such as double integrals, polar coordinates, or triple integrals. The choice of technique depends on the complexity of the function and the shape of the surface.

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