# Calculating the Area of a Plane in the First Octant

• quietrain
In summary, you were able to find the area of the triangle using the following equation: (d2 sqrt(a2+b2+c2))/2abc.
quietrain

## Homework Statement

consider the area of a plane in first octant where a,b,c,d are postive real numbers
show that the area is [d2 sqrt(a2+b2+c2)] / 2abc

ax+by+cz = d

## The Attempt at a Solution

ok i tried to search for formulas for areas of planes but there seem to be none !
i found one at paula's calculus notes that is about using surface integrals but it seems not for this problem?

can anyone point me to the right direction? thanks!

quietrain said:

## Homework Statement

consider the area of a plane in first octant where a,b,c,d are postive real numbers
show that the area is [d2 sqrt(a2+b2+c2)] / 2abc

ax+by+cz = d

## The Attempt at a Solution

ok i tried to search for formulas for areas of planes but there seem to be none !
i found one at paula's calculus notes that is about using surface integrals but it seems not for this problem?

can anyone point me to the right direction? thanks!

Hey there.

I'm going to assume that the area is finite.

Since planes are linear objects you can break them up into simple shapes and then add up the areas.

First you need to find the boundary of the shape. Have you got any ideas how to do that?

(Hint think of the axis that bound the plane!)

i realize that it is a triangle?

can i use the cross product to get area of parallelogram and then half it to get the triangle area?

i got 3 coordinates for the 3 places of intercept

namely
0,0,c -------- for x=0, y=0, z=c
0,d/b,0
d/a,0,0

i calculated out but somehow i get a very complicated solution. and i can't get it to fit into the elegant form in the first post ><

i even use half base x height by using projection to get height and then using base, but the expression is horrible too...

anywhere i did wrong?

oH i solved it... it was a careless mistake ><... thanks!

## 1. What is the definition of the "first octant"?

The first octant is a three-dimensional space that consists of all points with positive x, y, and z coordinates. It is the first of eight octants in a standard three-dimensional coordinate system.

## 2. How do you find the area of a plane in the first octant?

To find the area of a plane in the first octant, you can use the formula A = bh/2, where b is the base of the triangle formed by the plane and the x-y plane, and h is the height of the triangle.

## 3. Can the area of a plane in the first octant be negative?

No, the area of a plane in the first octant cannot be negative. Since the first octant consists of positive coordinates, the area of a plane in this region will always be positive.

## 4. What is the difference between the area of a plane in the first octant and the area of a plane in the entire octant?

The area of a plane in the first octant only takes into account the portion of the plane that lies in the first octant, while the area of a plane in the entire octant would include all eight octants. Therefore, the area in the entire octant would be larger than the area in the first octant.

## 5. How is the area of a plane in the first octant related to the volume of a solid in the first octant?

The area of a plane in the first octant is used to calculate the volume of a solid in the first octant. By integrating the area function over the z-axis, you can find the volume of a solid in the first octant bounded by the x-y plane and the plane in question.

• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
10
Views
2K
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
6K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
11
Views
961
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
3K