Area between plane and coordinate plane

Question

I want to calculate area between plane

$$x+{\frac{1}{\sqrt{3}}}y+{\frac{1}{6}}z = 1$$

and coordinate planes.

$\frac{\delta f}{\delta x}=1, \frac{\delta f}{\delta y}=\frac{\sqrt3}{3}, \frac{\delta f}{\delta z}=\frac{1}{6}$

How do I proceed from here?

The pictures dimensions are not real.

Answer 1

Apply the formula $$V=(1/3)Bh$$ where $B$ is the area of the base which is the area of the triangle to be found.

To find the volume we consider the base on the $x-y$ plane and height to be $6$ so the volume is $$V=2(1/2)(1)\sqrt 3 =\sqrt 3$$

In order to find $h$ we find the distance from the origin to the plane, which is $6/7$

Thus we have $$\sqrt 3 =(1/3)(B)(6/7)$$

Solve for $B$ and you get $B=7\sqrt 3 /2 $

Answer 2

The simpler method is to use

$$V=\frac13 A\cdot H=\frac13\cdot \left(\frac12 \cdot 1\cdot \sqrt 3\right)\cdot 6=\sqrt 3$$

otherwise by integration, that is

$$V=\int_0^1 dx\int_0^{\sqrt 3(1-x)} dy \int_0^{6\left(1-x-\frac y{\sqrt 3}\right)} dz=$$

$$=6\int_0^1 dx\int_0^{\sqrt 3(1-x)} \left(1-x-\frac y{\sqrt 3}\right)dy=$$

$$=6\int_0^1 \frac{\sqrt 3}2(1-x)^2dx=3\sqrt 3\left[-\frac13(1-x)^3\right]_0^1=\sqrt 3$$

Answer 3

The area of the triangle is half of the area of the paralellogram defined by two of its sides. This in turn is equal to the norm of the cross product of the vectors that define those sides. So, if the three vertices of the triangle are $A$, $B$ and $C$, the triangle’s area is $\frac12\lVert(B-A)\times(C-A)\rVert$. You can choose any of the three points as a common vertex instead of $A$, of course.

Here, we can read the plane’s coordinate axis intercepts directly from the equation. They are $(1,0,0)$, $(0,\sqrt3,0)$ and $(0,0,6)$. Choosing the first point as the common vertex looks convenient, so compute $\frac12\lVert(-1,\sqrt3,0)\times(-1,0,6)\rVert$.

However, you’ve tagged your question as having to do with integration and calculus, so presumably you’re meant to do something like integrate to find the volume of the tetrahedron and divide by its height (the distance between the origin and plane), as in this other answer. Then again, in this instance one can perform this computation without any calculus at all, too.