91Ó°ÊÓ

The given triple integral has the following limits of integration: $$\int_{-2}^{2} \int_{3}^{6} \int_{0}^{2} d x d y d z$$ These limits represent the range in each dimension of Cartesian space: - x ranges from -2 to 2 - y ranges from 3 to 6 - z ranges from 0 to 2 The integration is performed with respect to x, y and z variables.

Since there is no function to evaluate within the integral, the integrand is considered as 1: $$\int_{-2}^{2} \int_{3}^{6} \int_{0}^{2} 1\, d x d y d z$$ To solve the triple integral, we start by evaluating the innermost integral first with respect to x: $$\int_{-2}^{2} \int_{3}^{6} [x]_{0}^{2} d y d z$$ Since x ranges from 0 to 2, the inner integral evaluates to 2: $$\int_{-2}^{2} \int_{3}^{6} 2 d y d z$$ Now we'll evaluate the second integral with respect to y: $$\int_{-2}^{2} [2y]_{3}^{6} d z$$ Substituting the limits, we get: $$\int_{-2}^{2} (12 - 6) d z$$ This simplifies to 6: $$\int_{-2}^{2} 6 d z$$ Finally, evaluate the outer integral with respect to z: $$[6z]_{-2}^{2}$$ Substitute the limits: $$6(2) - 6(-2)$$ This evaluates to 24. Hence, the volume of the region of integration, represented by the given triple integral, is 24.