# Volume under a plane

Find the volume under $x+2y+3z=6$ in the first octant.

• edited April 29

The volume in the first octant under the plane x+2y+3z=6 is bound by the values 0<=x<=6, 0<=y<=3-x/2, and 0<=z<=2-x/3-2/3y

This is a tetrahedral volume and can be evaluated using the following triple integral:

$$V=\int_0^6 \int_0^{3-\frac{x}{2}} \int_0^{2-\frac{x}{3}-\frac{2y}{3}} dzdydx.$$

Integrating with respect to z:

$$V=\int_0^6 \int_0^{3-\frac{x}{2}} 2-\frac{x}{3}-\frac{2y}{3} dydx.$$

With respect to y:

$$V=\int_0^6 \bigg[2y-\frac{xy}{3}-\frac{y^2}{3}\bigg]\Big|_0^{3-\frac{x}{2}} dx.$$

After some simplification:

$$V=\int_0^6 3-x+\frac{x^2}{12} dx.$$

And finally with respect to x:

$$V=\bigg[3x-\frac{x^2}{2}+\frac{x^3}{36}\bigg]\Big|_0^6$$

$$V=18-18+\frac{216}{36}=6$$