Question

Evaluating triple integrals: Each of the triple integrals that follow represents the volume of a solid. Sketch the solid and evaluate the integral.

${\int }_1^5{\int }_{-2}^0{\int }_{-1}^{3}dzdydx$

Short answer

The obtained integral is 32.

Step-by-step solution

Step 1. GIven information.

Consider the given information.

${\int }_1^5{\int }_{-2}^0{\int }_{-1}^{3}dzdydx$

Step 2. Graph the solid.

By using the limits, sketch the given solid.

$\begin{array}{l}-1\le z\le 3\\ -2\le y\le 0\\ 1\le x\le 5\end{array}$

And,

Step 3. Calculate the integral.

Integrate by using limit and solve.

$\begin{array}{rcl}{\int }_1^5{\int }_{-2}^0{\int }_{-1}^{3}dzdydx& =& {\int }_1^5{\int }_{-2}^0{\left[z\right]}_{-1}^{3}dydx\\ & =& {\int }_1^5{\int }_{-2}^0{\left[3-\left(-1\right)\right]}_{-1}^{3}dydx\\ & =& 4{\int }_1^5{\left[y\right]}_{-2}^{0}dx\\ & =& 4{\int }_1^5\left[0-\left(-2\right)\right]dx\\ & =& 8{\left[x\right]}_1^5\\ & =& 8\left[5-1\right]\\ & =& 32\end{array}$