Discussion
Let \(f : D \to \mathbf{R}\) be defined as \(f(x, y, z) = xyz\); where \(D = \{(x, y, z) \in \mathbf{R}^{3}/x^{2} + y^{2} + z^{2} \lt 1\}\). Let P denote the plane \(\{(x, y, z)/2x + 3y + 5z = 0\}\). Then \(\displaystyle\int_{D \cap P} f\, dx\, dy\, dz\) is :
Answer: B
The integral is 0.
\(D \cap P\) is a flat disc: the intersection of the open unit ball with a plane through the origin. As a subset of \(\mathbf{R}^{3}\) it has three-dimensional volume zero, and the integral of a bounded function with respect to \(dx\,dy\,dz\) over a set of volume zero is 0.
(Even as a surface integral over the disc the answer would be 0, since \(xyz\) is odd: \((x, y, z)\) and \(-(x, y, z)\) both lie in the disc and \(f\) changes sign between them.) The integral is well defined, so 'undefined' is wrong, and \(\pi\) and \(2\pi\) confuse it with an area.
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