Volume under a plane

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

Comments

  • 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:

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

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

    mark
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