Question

How do you find the volume of the pyramid bounded by the plane $2x + 3y + z = 6$ and the coordinate plane?

Answer 1

Hint : A pyramid is a polyhedron figure which has only one base. The base of the pyramid is a poly sided figure. If a region in the plane revolves about a given line, the resulting solid is a solid of revolution, and the line is called the axis of revolution. It’s also the height of a pyramid multiplied by area. Hence, to find the volume of the pyramid bounded by the plane apply integral to the given vectors and find the volume.

Complete step-by-step answer :
Volume of the pyramid bounded by the plane $2x + 3y + z = 6$
Hence, the normal vector is \left( {\begin{array}{*{20}{c}} 2 \\\ 3 \\\ 1 \end{array}} \right) , which points out in the direction of octant 1, so the volume in question is under the plane and in octant 1.
We can re-write the plane as:
On the plane $z\left( {x,y} \right) = 6 - 2x - 3y$
And on the coordinate plane, for $z = 0$ , we have
$\Rightarrow z:0 \to 6 - 2x - 3y$
Along $z = 0$ , y goes from 0 to $3y = 6 - 2x$ hence
$\Rightarrow y:0 \to 2 - \dfrac{2}{3}x$
Along $y = 0$ , $z = 0$ hence
$\Rightarrow x:0 \to 3$
The volume we need is $\int\limits_A^{} {z\left( {x,y} \right)dA}$
We are finding the volume, so $f\left( {x,y,z} \right) = 1$ , Hence the integral we get as:
$\int\limits_0^3 {\int\limits_0^{2 - \dfrac{2}{3}x} {\int\limits_0^{6 - 2x - 3y} {dzdydx} } }$
With respect to $dz$ :
= $\int\limits_{x = 0}^3 {\int\limits_{y = 0}^{2 - \dfrac{2}{3}x} {\int\limits_{z = 0}^{6 - 2x - 3y} {\left[ z \right] _0^{6 - 2x - 3y}dydx} } }$
With respect to $dy$ :

= $\int\limits_{x = 0}^3 {\int\limits_{y = 0}^{2 - \dfrac{2}{3}x} {6 - 2x - 3y \cdot dy \cdot dx} }$
With respect to $dx$ :
= $\int\limits_{x = 0}^3 {\left[ {6y - 2xy - \dfrac{3}{2}{y^2}} \right] _{y = 0}^{2 - \dfrac{2}{3}x}dx}$
= $\int\limits_{x = 0}^3 {\left( {6\left( {2 - \dfrac{2}{3}x} \right) - 2x\left( {2 - \dfrac{2}{3}x} \right) - \dfrac{3}{2}{{\left( {2 - \dfrac{2}{3}x} \right)}^2}} \right)dx}$
Simplifying the terms, we get
= $\int\limits_{x = 0}^3 {\left( {12 - 4x - 4x + \dfrac{4}{3}{x^2} - \dfrac{3}{2}\left( {4 - \dfrac{8}{3}x + \dfrac{4}{9}{x^2}} \right)} \right)dx}$
= $\int\limits_{x = 0}^3 {\left( {12 - 4x - 4x + \dfrac{4}{3}{x^2} - 6 - \dfrac{2}{3}{x^2} + 4x} \right)dx}$
= $\int\limits_{x = 0}^3 {\left( {6 + \dfrac{2}{3}{x^2} - 4x} \right)dx}$
= $\left[ {6x + \dfrac{2}{9}{x^3} - 2{x^2}} \right] _0^3$
Applying the limits, we get:
= $6\left( 3 \right) + \dfrac{2}{9}{\left( 3 \right)^3} - 2{\left( 3 \right)^2}$
= 6 cubic units.

Note : Here, the volume of the pyramid is bounded by the plane with the given vectors hence, we need to apply integrals to find the volume of the pyramid. The volume of a pyramid depends upon the type of pyramid’s base, whether it is a triangle, square or rectangle. Hence, the formula to find not only volume but also the surface area of a pyramid will be based on the structure of its base and height of the pyramid. To find the volume of a pyramid, we need to know the total capacity of the given pyramid. The formula for the pyramid’s volume is given by one-third of the product of the area of the base to its height.
Volume of pyramid is given as: $V = \dfrac{1}{3}A \times H$