Exercise · ZW-TRIPLE-INTEGRAL-008 · ★★☆ Intermediate

Find the volume of the solid in the first octant under the plane \(x+2y+3z=6\).

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  • A. \(6\)

Explanation

The axis intercepts are \(6,3,2\); the volume is \(\int_0^6\!\int_0^{(6-x)/2}\!\int_0^{(6-x-2y)/3}dz\,dy\,dx=\tfrac16\cdot6\cdot3\cdot2=6\).

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Review the concept: Triple integral

The triple integral of a function \(f\) over a solid \(E\) of space is the limit of the sums of the values of \(f\) times the volume \(\Delta V\) of ever finer cells covering \(E\).

\[\iiint_E f\,dV=\lim_{\max\Delta V_i\to0}\sum_i f(x_i^*,y_i^*,z_i^*)\,\Delta V_i\]

More practice problems: triple integral

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