How to find the area element dA in this situation?

[Laudator]

Homework Statement

A plane in 3D Cartesian coordinate system, defined by these 3 points (a,0,0), (0,2a,0), (0,0,a), find the area bounded in the 1st octant.

Relevant Equations

No particular equation in mind.

I know how to find the area of a plane which is parallel to the xy-, yz-, or, xz-plane, those are the easiest case. I also tried to find the area of a plane which is only perpendicular to 1 particular axis plane, like the one passing through points (0,0,a), (a,0,a), (0,2a,0), in which case, dA=\sqrt 5 dx dy. However, when it comes to planes that are not this special, I don't know where to start.

 

[DrClaude]

This falls in the category of homework problems, so I've moved the thread. You'll have to provide an attempt at a solution.

 

[PeroK]

A plane in 3D Cartesian coordinate system, defined by these 3 points (a,0,0), (0,2a,0), (0,0,a), I want to find the area bounded in the 1st octant, how to find the area element dA in this situation and how to carry out the integral?

You need to post the full problem statement. You've posted too little information here.

And, you should be using the homework template.

 

[Laudator]

@DrClaude, @PeroK, thanks for the info, I've already edited the main post, hope that's enough.

 

[PeroK]

Problem Statement: A plane in 3D Cartesian coordinate system, defined by these 3 points (a,0,0), (0,2a,0), (0,0,a), find the area bounded in the 1st octant.
Relevant Equations: No particular equation in mind.

I know how to find the area of a plane which is parallel to the xy-, yz-, or, xz-plane, those are the easiest case. I also tried to find the area of a plane which is only perpendicular to 1 particular axis plane, like the one passing through points (0,0,a), (a,0,a), (0,2a,0), in which case, dA=\sqrt 5 dx dy. However, when it comes to planes that are not this special, I don't know where to start.

I would start by calculating the area of the triangle by the simple method. That gives you the answer.

Then, do it the hard way to practice your integration skills. Maybe start with the equation of the plane?

 

[Laudator]

@PeroK, thanks for the tip, I think I got it, here's my result, I hope I got it all worked out correctly:

3 points are labelled A(a,0,0) B(0,2a,0) C(0,0,a). Two vectors are then formed using point A,C and B,C, which gives:

$$AC=a\vec x - a\vec z$$
$$BC=2a\vec y - a\vec z$$.

The area of the triangle can be calculated using the cross product: $AC \times BC = \vec C = 2a^2\vec x + a^2\vec y + 2a^2\vec z$. The area is one-half of the modulus of this vector, which is $\frac{3}{2}a^2$.

Next I found the unit vector of $\vec C$, which is $\frac{2}{3}\vec x + \frac{1}{3}\vec y + \frac{2}{3}\vec z$. The equation of the plane is then found to be $2x+y+2z=2a$, using point C(0,0,a).

I held z constant to find the relationship between $dx$ and $dy$, which is $dy=-2dx$, and the infinitesimal line segment on the xy-plane is given by $\sqrt {dx^2+2dy^2}$, which then is $\sqrt {dx^2+8dx^2}=3dx$. The limit of integration can be determined by looking at the triangle's projection on the xz-plane, which can be expressed as $z=a-x$, so $x$ goes from 0 to $x=a-z$ and thus the first integral is:

$$\int_0^{a-z}3 \, dx=3(a-z)$$.

And the second integral is with respect to z and goes from 0 to a:

$$\int_0^a 3(a-z) \, dz =\frac{3}{2}a^2$$.

Both method give the same result, check. This is the first time for me to use LaTeX, probably will be messy...

 

[PeroK]

Both method give the same result, check. This is the first time for me to use LaTeX, probably will be messy...

There's a general method for calculating the surface area of a surface. See here:

http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx

 

[Laudator]

@PeroK, thanks a lot, that's of great help.

 

[LCKurtz]

You might also like the "OK, how do I use it" section in this insight:
https://www.physicsforums.com/insights/parameterization-and-surface-integrals/