Problem 13 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\begin{aligned} &\iiint_{D}(x y+x z+y z) d V ; \quad D=\\{(x, y, z):-1 \leq x \leq 1\\\ &-2 \leq y \leq 2,-3 \leq z \leq 3\\} \end{aligned}$$

Step-by-Step Solution

Answer

The given triple integral evaluates to 0 for the function \((xy + xz + yz)\) over the region D defined by the inequalities \( -1 \leq x \leq 1, -2 \leq y \leq 2, -3 \leq z \leq 3\). The integral was separated into three 1-dimensional integrals, and after integrating each of them, we found that the value is 0.

Step 1: Separate the Integral

Let's separate the given integral into three individual 1-dimensional integrals. We can rewrite the integral as follows: $$\iiint_{D}(x y+x z+y z) d V = \int_{-1}^1\int_{-2}^2\int_{-3}^3 (xy + xz + yz) \,dz\,dy\,dx$$

Step 2: Integrate with respect to z

Let's now integrate the innermost integral with respect to z: $$ \int_{-3}^3 (xy + xz + yz) \,dz = xy \int_{-3}^3\,dz + x\int_{-3}^3 z\,dz + y\int_{-3}^3 z\,dz $$ Since x and y are constant with respect to z, we can take them outside the integral. $$ = xy(z\big|_{-3}^3) + x(\frac{1}{2}z^2\big|_{-3}^3) + y(\frac{1}{2}z^2\big|_{-3}^3) $$ $$ = 6xy + 0 + 0 = 6xy $$

Step 3: Integrate with respect to y

Now, let's integrate the result with respect to y: $$ \int_{-2}^2 (6xy) \,dy = 6x \int_{-2}^2 y\,dy $$ Since x is constant with respect to y, we can take it outside the integral. $$ = 6x(\frac{1}{2}y^2\big|_{-2}^2) = 6x(0) = 0 $$ The integral is equal to zero.

Step 4: Integrate with respect to x

Now, let's integrate the result with respect to x: $$ \int_{-1}^1 (0) \,dx = 0 $$ So, the final value of the triple integral is 0.