Find the volume of the solid that lies under the plane 3x + 2y + z = 12 and above the rectangle. $$ R = \{(x,y) | 0\le x \le {\color{red}1}, -{\color{red}2} \le y \le {\color{red}4} \} $$

Answer:

51 cubic units

Step-by-step explanation:

[tex]3x+2y+z=12\\\Rightarrow z=12-3x-2y[/tex]

Now we integrate over volume integral over the range [tex]R = \{(x,y) | 0\le x \le {1}, -{2} \le y \le {4} \}[/tex]

[tex]V=\int_0^1\int_{-2}^4 12-3x-2y dydx\\ =\int_{-2}^4 12y-3yx-y^2|_4^{-2} dx\\ =\int_{0}^1 12\times4-3\times4x-16-[12\times-2-3\times-2x-4]dx\\ =\int_{0}^1 60-18xdx\\ =60x-9x^2|_{0}^1\\ =60-9\\ =51\ \text{cubic units}[/tex]

The volume of the solid is 51 cubic units.

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3 24 38
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