find the volume of the solid in the first octant bounded by the plane ( x + y + z = 6 ) and the coordinate…

Explanation:

Step1: Determine the limits of integration

The plane (x + y+z = 6) can be rewritten as (z=6 - x - y). Since we are in the first - octant ((x\geq0,y\geq0,z\geq0)), when (z = 0), we have (y=6 - x) (by setting (z = 0) in (z=6 - x - y)). And when (y = z=0), we have (x = 6). The volume (V=\int_{x = 0}^{6}\int_{y = 0}^{6 - x}\int_{z = 0}^{6 - x - y}dzdydx)

Step2: Integrate with respect to (z)

(\int_{x = 0}^{6}\int_{y = 0}^{6 - x}\left[z\right]{0}^{6 - x - y}dydx=\int{x = 0}^{6}\int_{y = 0}^{6 - x}(6 - x - y)dydx)

Step3: Integrate with respect to (y)

(\int_{x = 0}^{6}\left[6y-xy-\frac{y^{2}}{2}\right]{y = 0}^{6 - x}dx=\int{x = 0}^{6}\left(6(6 - x)-x(6 - x)-\frac{(6 - x)^{2}}{2}\right)dx) [ \begin{align*} &=\int_{x = 0}^{6}\left(36-6x-6x+x^{2}-\frac{36 - 12x+x^{2}}{2}\right)dx\ &=\int_{x = 0}^{6}\left(36-12x+x^{2}-18 + 6x-\frac{x^{2}}{2}\right)dx\ &=\int_{x = 0}^{6}\left(18-6x+\frac{x^{2}}{2}\right)dx \end{align*} ]

Step4: Integrate with respect to (x)

(\left[18x-3x^{2}+\frac{x^{3}}{6}\right]_{0}^{6}) [ \begin{align*} &=18\times6-3\times6^{2}+\frac{6^{3}}{6}\ &=108 - 108+36\ &=36 \end{align*} ]

Answer:

(36)