Part 1Evaluate the following integral.Integral from nothing to nothing ModifyingBelow Integral from nothing to nothing With Upper D Integral from nothing to nothing left parenthesis xy plus xz plus yz right parenthesis dV∫∫D∫(xy+xz+yz) dV; Dequals={(x,y,z): minus−33less than or equals≤xless than or equals≤33, minus−44less than or equals≤yless than or equals≤44, minus−11less than or equals≤zless than or equals≤11} Part 1Integral from nothing to nothing ModifyingBelow Integral from nothing to nothing With Upper D Integral from nothing to nothing left parenthesis xy plus xz plus yz right parenthesis dV∫∫D∫(xy+xz+yz) dVequals=[input]enter your response here (Simplify your answer.)

Question

Part 1Evaluate the following integral.Integral from nothing to nothing ModifyingBelow Integral from nothing to nothing With Upper D Integral from nothing to nothing left parenthesis xy plus xz plus yz right parenthesis dV∫∫D∫(xy+xz+yz) dV​; Dequals=​{(x,y,z): minus−33less than or equals≤xless than or equals≤33​, minus−44less than or equals≤yless than or equals≤44​, minus−11less than or equals≤zless than or equals≤11​} Part 1Integral from nothing to nothing ModifyingBelow Integral from nothing to nothing With Upper D Integral from nothing to nothing left parenthesis xy plus xz plus yz right parenthesis dV∫∫D∫(xy+xz+yz) dVequals=[input]enter your response here ​(Simplify your​ answer.)

BlackTom AI · Accepted Answer

First, I will restate the problem to ensure clarity: we are evaluating the triple integral of the function xy + xz + yz over the rectangular region D defined by x in [-3, 3], y in [-4, 4], and z in [-1, 1], i.e., ∭_D (xy + xz + yz) dV.

Consider the integrand term by term. The integrand is a sum of three terms: xy, xz, and yz.

- For the term xy: if we integrate with respect to x over a symmetric interval [-3, 3], the integral of x times a fixed y is zero because x is odd about 0 while y is treated as a constant during the x-integration. Formally, ∫_{-3}^{3} xy dx = y ∫_{-3}^{3} x dx = y [x^2/2]_{-3}^{3} = y (9/2 - 9/2) = 0. Since this vanishes for every fixed y, the triple integral contribution from xy is zero.

- For the term xz: similarly, integrating with respect to x over [-3, 3] yields ∫_{-3}^{3} xz dx = z ∫_{-3}^{3} x dx = z [x^2/2]_{-3}^{3} = 0. This means the xz component contributes zero to the triple integral as well.

- For the term yz: here y and z are constants with respect to x, but we still integrate with respect to x first; the integral over x of yz is simply yz ∫_{-3}^{3} dx = yz (6). However, note that when we then integrate over y and z, the yz term becomes odd in y and odd in z across symmetric bounds, leading to the total contribution being zero after integrating over y and z as well. More formally, yz as a function in y and z integrated over symmetric [-4, 4] and [-1, 1] produces zero because the integrand is odd in either y or z when paired with the symmetric limits.

Putting it together, each component of the integrand is odd in at least one variable and evaluated over symmetric bounds. Therefore, the overall triple integral of xy + xz + yz over D vanishes: the positive and negative contributions cancel out in each term.

If one were to compute step-by-step via iterated integrals, you would find that:
- ∫_{-3}^{3} ∫_{-4}^{4} ∫_{-1}^{1} xy dV = 0,
- ∫_{-3}^{3} ∫_{-4}^{4} ∫_{-1}^{1} xz dV = 0,
- ∫_{-3}^{3} ∫_{-4}^{4} ∫_{-1}^{1} yz dV = 0,
and summing these gives 0.

A potential pitfall to avoid is trying to treat the yz term as if x integration would produce a nonzero constant; remembering that the full symmetry of the region applies across all three variables is crucial for concluding zero. Another common mistake is to ignore the odd/even nature of the integrand and assume a nonzero result without checking the symmetry; symmetry checks are often the quickest way to evaluate such integrals.

In summary, exploiting the symmetry of the region and the oddness of each term in at least one variable leads to the conclusion that the triple integral over D of (xy + xz + yz) dV is zero.

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