# Area of plane in first octant

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!

## 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!)

quietrain
i realise 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?

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