For a National Board Exam Review:
What is the volume of the solid bounded by the plane $3x+4y+6z=12$ and the coordinate axes?
Answer is $4$.
I am looking for a quick and intuitive way to solve this without calculus; it's not that I have no knowledge of triple/double integrals; it's also freely posted on the net; but I was hoping there is some sort of quick formula for the solid it formed? I don't know what is the name of the volume.
On
Geometrical Approach
Notice, we have the equation of the plane: $3x+4y+6z=12$. The normal distance say $h$ of the plane from the origin $(0, 0, 0)$ is given as $$=\frac{|3(0)+4(0)+6(0)-12|}{\sqrt{3^2+4^2+6^2}}=\frac{|-12|}{\sqrt{61}}=\frac{12}{\sqrt {61}}$$
Now, the equation of the plane can be expressed in the intercept form as follows $$\frac{3x}{12}+\frac{4y}{12}+\frac{6z}{12}=1$$ $$\frac{x}{4}+\frac{y}{3}+\frac{z}{2}=1$$ Above equation represents that the given plane intercepts the x-axis at $(4, 0, 0)$, y-axis at $(0, 3, 0)$ & z-axis at $(0, 0, 2)$ .
Edit
The triangle having vertices at these points of intersection of plane with the coordinate axes has its sides $5$, $\sqrt{13}$ & $2\sqrt{5}$. Hence, area can calculated by Hero's formula as $\sqrt{61}$
Solid bounded by the given plane: $3x+4y+6z=12$ is a pyramid having triangular base of area $A=6$ & the apex point at the origin $(0, 0, 0)$ at a normal height $h=\frac{12}{\sqrt{61}}$ from the base. Hence the volume bounded is given as $$=\frac{1}{3}\times \text{(area of triangular base)}\times \text{(normal height)}$$ $$=\frac{1}{3}\times (\sqrt{61})\times \left(\frac{12}{\sqrt{61}}\right)$$ $$=4$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{volume of solid bounded by plane}=\color{blue}{4}}}$$
You can treat it as a pyramid and the volume is found by $V = \dfrac{B\cdot h}{3}=\dfrac{\dfrac{xy}{2}\cdot z}{ 3}= \dfrac{xyz}{6}=\dfrac{4\times 3\times 2}{6}= 4$, with $x,y,z$ are the plane's $x,y,z$ intercepts with the coordinate planes.