Projecting a surface integral in 3-D onto a plane (2-D)

In summary, the conversation discusses a method of descent for solving the 2D heat equation from the 3D heat equation. One step involves changing the surface integral over a ball in 3D to the surface integral over its projection onto the xy plane. The surface of the projection is represented as (x1-y1)^2 + (x2-y2)^2 + (y3)^2 = t^2. The book then uses this to show that dy1dy2dy3 can be written as [(sigma y3/sigma y1)^2 + (sigma y3/sigma y2)^2 + 1]^1/2 dy1dy2. The speaker is unsure how this step is made and asks for
  • #1
Fractal20
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Homework Statement


So the context is this arises in the method of descent, for finding a solution for the 2D heat equation from the 3d heat equation. Anyway, in one step, we must change the surface integral over a ball in 3d, to the surface integral over it's projection into a plane. In this case it is to the xy plane. So I understand that for the surface of a ball B(x1,x2,x3,t) can be represented as (y1, y2, y3) such that

(x1-y1)2 + (x2-y2)2 + (x3 - y3)2 = t2

Moreover, with the projection there is no x3 dependence so we can set it to zero and get that the surface of the projection is:

(x1-y1)2 + (x2-y2)2 + (y3)2 = t2

So my book used this and then says that dy1dy2dy3 can be written as:

[([itex]\sigma[/itex]y3/[itex]\sigma[/itex]y1)2 + ([itex]\sigma[/itex]y3/[itex]\sigma[/itex]y2)2 + 1]1/2 dy1dy2

I don't understand this step. I want to take the derivative of the expression for y3 but this certainly won't result in this expression for dy3. Can anybody explain how this step is made, or even just suggest what to look up in a calculus book? I was consulting my old calculus book but only found things describing how to integrate over a 3d surface by considering the 2d domain, so kind of the reverse of what I wanted here. Thanks!

PS is there a command so I can just type equations using the formatting from latex?


Homework Equations





The Attempt at a Solution


 
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  • #2
I've tried to take the derivative of y3, but the result doesn't match up with what I'm supposed to get.
 

Related to Projecting a surface integral in 3-D onto a plane (2-D)

1. What is the purpose of projecting a surface integral in 3-D onto a plane (2-D)?

The purpose of projecting a surface integral in 3-D onto a plane (2-D) is to simplify complex mathematical calculations and visualize the results in a more comprehensible manner. It allows us to represent a three-dimensional surface as a two-dimensional shape, making it easier to analyze and understand.

2. How is a surface integral projected in 3-D onto a plane (2-D)?

To project a surface integral in 3-D onto a plane (2-D), we use a process called parametrization. This involves choosing a set of parameters, such as u and v, to represent the surface in terms of a two-dimensional coordinate system. This allows us to transform the three-dimensional surface into a two-dimensional plane.

3. What are the applications of projecting a surface integral in 3-D onto a plane (2-D)?

The applications of projecting a surface integral in 3-D onto a plane (2-D) are numerous in the field of mathematics and in various branches of science such as physics, engineering, and computer graphics. It is used to solve complex surface integrals, calculate areas and volumes, and visualize curved surfaces in a simplified manner.

4. What are some challenges in projecting a surface integral in 3-D onto a plane (2-D)?

One of the main challenges in projecting a surface integral in 3-D onto a plane (2-D) is choosing an appropriate parametrization for the given surface. If the parameters are not chosen correctly, it can lead to incorrect results. Another challenge is dealing with surfaces that have complicated shapes or singular points, which may require more advanced techniques for projection.

5. Are there any limitations to projecting a surface integral in 3-D onto a plane (2-D)?

Yes, there are some limitations to projecting a surface integral in 3-D onto a plane (2-D). This method is not applicable to every type of surface and may not yield accurate results for highly irregular or discontinuous surfaces. Additionally, the projection may lose some information about the original surface, which may be important in certain applications.

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