Problem 8 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{-1}^{1} \int_{-1}^{2} \int_{0}^{1} 6 x y z d y d x d z$$

Step-by-Step Solution

Answer

Answer: The value of the triple integral is 0.

Step 1: Integrate with respect to x

First, we will find the integral of \(6xyz\) with respect to x over the given limits, keeping y and z constant: $$\int_{-1}^{1} 6xyz dx = 6yz\int_{-1}^{1}x dx$$ Since the integral of x is \(\frac{1}{2}x^2\), we have: $$6yz\left[\frac{1}{2}x^2\right]_{-1}^{1} = 6yz\left(\frac{1}{2}(1)^2 - \frac{1}{2}(-1)^2\right) = 0$$

Step 2: Integrate with respect to y

Since the result of the first integration is 0, it eliminates the need to proceed to the next integration. We found that the value of the integral over the x range is 0 which means that the product of integrals over the other ranges will also be 0. Hence, $$\int_{-1}^{1} \int_{-1}^{2} \int_{0}^{1} 6 x y z d y d x d z = \int_{-1}^{1} \int_{-1}^{2} 0 dy dz = 0$$

Step 3: Conclusion

The triple integral is evaluated to be zero: $$\int_{-1}^{1} \int_{-1}^{2} \int_{0}^{1} 6 x y z d y d x d z = 0$$