Question

Find the volume of the given solid. Bounded by the coordinate planes and the planes \(3x + 2y + z = 6\)

Short answer

The volume of the given solid can be:

\(V = 6\).

Step-by-step solution

Find the limits

Consider the solid \(3x + 2y + z = 6\)

Convert the boundary\(3x + 2y + z = 6\)into the proper form for integrand: \(z = 6 - 3x - 2y\)

The region of integration will have two boundaries determine by the coordinate planes\(x = 0\)and\(y = 0\)

The third boundary is determined by the intersection of\(z = 0\)and top boundary.

Setting the integrand to zero and solving for y yields\(y = 3 - \frac{3}{2}x\)

Neither the integrand nor the boundaries give much reason to prefer integration on x or on y .first, So integrate first on y.

The region of integration is shown below along with a vertical line to help set up the limits of integration.

The vertical line extends from\(y = 0\)to\(y = 3 - \frac{3}{2}x\)and it can slide from\(x = 0\)to\(x = 2\).

The volume can be given by: \(V = \int\limits_0^2 {\int\limits_0^{3 - \frac{3}{2}x} {\left( {6 - 3x - 2y} \right)dydx} } \)

Find the integral

\(\begin{aligned}{l}V = \int\limits_0^2 {\int\limits_0^{3 - \frac{3}{2}x} {\left( {6 - 3x - 2y} \right)dydx} } \\V = \int\limits_0^2 {\left[ {6y - 3xy - 2{y^2}} \right]_0^{3 - \frac{3}{2}x}} dx\\V = \int\limits_0^2 {\left[ {6\left( {3 - \frac{3}{2}x} \right) - 3x\left( {3 - \frac{3}{2}x} \right) - 2{{\left( {3 - \frac{3}{2}x} \right)}^2} - 0} \right]} dx\\V = \int\limits_0^2 {\left[ {18 - 9x - 9x + \frac{9}{2}{x^2} - 9 + 9x - \frac{9}{4}{x^2}} \right]} dx\\V = \int\limits_0^2 {\left[ {\frac{9}{4}{x^2} - 9x + 9} \right]} dx\\V = \left[ {\frac{9}{4}\frac{{{x^3}}}{3} - \frac{{9{x^2}}}{2} + 9x} \right]_0^2\\V = \left[ {\frac{3}{4}{{(2)}^3} - \frac{9}{2}{{(2)}^2} + 9(2) - 0} \right]\\V = 6 - 18 + 18 - 0\\V = 6\end{aligned}\)

Hence, The volume of the given solid can be evaluated as: \(V = 6\).