# How to find the area element dA in this situation?

• Laudator
In summary, you can calculate the area of a plane by integrating the equation of the plane or by using the surface area formula.
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.

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

Laudator said:
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.

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

Laudator said:
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?

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

## 1. What is the definition of an area element dA?

The area element dA is a small unit of area that is used to measure the size of a two-dimensional surface. It is often represented as a small rectangle with sides dx and dy, and its value is infinitesimally small.

## 2. How is the area element dA calculated?

The area element dA can be calculated by multiplying the length of the base (dx) by the length of the height (dy). In some cases, the area element may also be calculated using other mathematical formulas, depending on the shape of the surface.

## 3. What is the significance of the area element dA in scientific calculations?

The area element dA is an important concept in many scientific calculations, particularly in the fields of physics and mathematics. It allows for precise measurements of two-dimensional surfaces and is crucial in understanding various physical phenomena and mathematical concepts.

## 4. How do you find the area element dA in a specific situation?

The method for finding the area element dA in a specific situation will depend on the shape of the surface and the given information. In general, it involves breaking down the surface into smaller, more manageable shapes and using mathematical formulas to calculate the area of each element.

## 5. Can the area element dA be negative?

No, the area element dA cannot be negative. It represents a physical quantity (area) and therefore cannot have a negative value. If a calculation results in a negative dA, it may indicate an error in the calculation or an incorrect assumption about the shape of the surface.

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