Question

Evaluate the iterated integral. $$ \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} x^{2} y^{2} z^{2} d x d y d z $$

Short answer

The value of the triple integral is \(\frac{8}{27}\).

Step-by-step solution

Integration with respect to x

First you integrate with respect to x by treating y and z as constants: \(\int_{-1}^{1} x^{2} \,dx\). Apply the power rule for integrals, which states that the integral of \(x^n\) equals \(\frac{x^{n+1}}{n+1}\) to get \(\frac{x^{3}}{3}\). Evaluate this from -1 to 1 to get \(\frac{1^{3}}{3} - \frac{(-1)^{3}}{3} = \frac{2}{3}\). The result is then \(\frac{2}{3} y^{2} z^{2}\)

Integration with respect to y

Next, integrate with respect to y treating z as a constant: \(\int_{-1}^{1} \frac{2}{3} y^{2} \,dy\). Again applying the power rule, this gives \(\frac{2}{3} \cdot \frac{y^{3}}{3}\). Evaluating this from -1 to 1 yields \(\frac{2}{9} \cdot 2 = \frac{4}{9} z^{2}\)

Integration with respect to z

Last, integrate with respect to z: \(\int_{-1}^{1} \frac{4}{9} z^{2} \,dz\). The power rule gives us \(\frac{4}{9} \cdot \frac{z^{3}}{3}\). Evaluating this from -1 to 1 yields \(\frac{4}{27} \cdot 2 = \frac{8}{27}\)