Problem 7 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{-2}^{2} \int_{3}^{6} \int_{0}^{2} d x d y d z$$

Step-by-Step Solution

Answer

**Question:** Evaluate the volume of the rectangular region in three-dimensional space described by the triple integral $$\int_{-2}^{2} \int_{3}^{6} \int_{0}^{2} dx \, dy \, dz.$$ **Answer:** The volume of the rectangular region is 24 cubic units.

Step 1: Evaluate the innermost integral with respect to x

Since there is no function inside the integral, the inner integral will be evaluated as: $$\int_{0}^{2} dx = [x]_{0}^{2} = 2 - 0 = 2$$

Step 2: Substitute the result into the second integral

Now we substitute the result we found in step 1 into the second integral: $$\int_{3}^{6} 2 dy$$

Step 3: Evaluate the second integral with respect to y

Now evaluate the second integral: $$\int_{3}^{6} 2 dy = [2y]_{3}^{6} = 12 - 6 = 6$$

Step 4: Substitute the result into the third integral

Now substitute the result we found in step 3 into the third integral: $$\int_{-2}^{2} 6 dz$$

Step 5: Evaluate the third integral with respect to z

Finally, evaluate the third integral: $$\int_{-2}^{2} 6 dz = [6z]_{-2}^{2} = 12 - (-12) = 24$$ Thus, the volume of the region defined by this triple integral is 24 cubic units.