📝 Properties of Triple Integrals (13 MCQs)

📖 Explanation: Since $E$ is symmetric about the $xy$-plane, for every point $(x,y,z)$ in $E$, the point $(x,y,-z)$ is also in $E$. Because $f$ is odd in $z$ ($f(x,y,-z) = -f(x,y,z)$), the contributions from the upper and lower halves cancel exactly. The volume being 12 is a distractor; the integral of an odd function over a symmetric domain is always zero, regardless of the volume.

📖 Explanation: Fubini's theorem states that for a continuous (or absolutely integrable) function over a rectangular box, the iterated integral is independent of the order of integration. However, the statement as phrased is too broad; it fails if the function is not absolutely integrable or if the region is non-rectangular and the limits are not properly adjusted. But among the options, the core idea that the order can be changed under suitable conditions makes (A) the best representation of the property, though it requires caveats.

📖 Explanation: For the tetrahedron $x\ge0, y\ge0, z\ge0, x+y+z\le2$, the limits can be set in multiple valid orders. Option A gives $z$ from 0 to $2-x-y$, $y$ from 0 to $2-x$, $x$ from 0 to 2. Option B gives $x$ from 0 to $2-y-z$, $y$ from 0 to $2-z$, $z$ from 0 to 2. Both correctly describe the same region. Option C has incorrect order of differentials relative to limits. This tests the student's ability to visualize and set up limits in different orders.

📖 Explanation: For a rectangular box with constant limits, Fubini's theorem allows any order of integration without changing the integrand or limits. The integrand is symmetric in $x,y,z$, but even if it were not, the order of variables in the integrand does not change; you simply integrate with respect to the new order of differentials. Thus option A is correct; the limits remain 0 to 1 for each variable.

📖 Explanation: The integrand $x^2+y^2$ is always non-negative and is even in both $x$ and $y$. Symmetry about the axes does not make the integral zero; it only allows us to integrate over a quarter or half and multiply by 4 or 2. The correct value is indeed $192\pi$ (computed as $\int_0^{2\pi}\int_0^2\int_0^3 r^2 \cdot r\,dz\,dr\,d\theta = 192\pi$). The second student confused even symmetry with odd symmetry. This tests error analysis in applying symmetry properties.

📖 Explanation: The integral $\iiint_E x\,dV$ is by definition the first moment of the solid about the yz-plane (since distance from yz-plane is $x$). The centroid's x-coordinate is $\bar{x} = \frac{1}{V}\iiint_E x\,dV$, so the integral equals $V\bar{x}$. Thus both A and B are correct interpretations. This question links the geometric meaning of the integral with its physical interpretation, requiring multi-step reasoning.

📖 Explanation: The region is a paraboloid $z=1-r^2$ with $r$ from 0 to 1. To change order to $dr\,dz$, we invert $z=1-r^2$ to get $r=\sqrt{1-z}$. For a fixed $z$ (from 0 to 1), $r$ ranges from 0 to $\sqrt{1-z}$. Thus the correct alternative is $\int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{\sqrt{1-z}} f(r,\theta,z)\,r\,dr\,dz\,d\theta$. Option B is correct. Option C misses the Jacobian $r$, and A has incorrect limits for $r$. This tests the ability to manipulate limits in cylindrical coordinates.

📖 Explanation: The outer limits give $x$ from 0 to 2 and $y$ from 0 to $\sqrt{4-x^2}$, which is the first-quadrant portion of the circle $x^2+y^2=4$. The inner limit gives $z$ from 0 to $x^2+y^2$, which is the paraboloid $z=x^2+y^2$. Thus the solid is the region under the paraboloid and above the quarter-disk in the first quadrant. This requires interpreting the geometry from the limits, not just computing.

📖 Explanation: The property $\iiint_E f\,dV = \iiint_E |f|\,dV$ is false in general; it holds only if $f\ge0$ everywhere. The other options are standard linearity and monotonicity properties. This is a direct recall question but framed as error identification to test conceptual understanding of the absolute value property.

📖 Explanation: The integral $\iiint_E z\,dV$ over the unit ball is 0 because $z$ is an odd function and the ball is symmetric about the xy-plane. The student incorrectly replaced the integrand $z$ by 1 (thinking volume) and also assumed the average value is 1. Additionally, even if they computed in spherical coordinates, they would need the Jacobian $\rho^2\sin\phi$, but the main error is symmetry. Thus all options point to different aspects of the same misconception: treating the integral of a function as its volume.

📖 Explanation: The surface $z=4-x^2-y^2$ is a downward paraboloid. Its intersection with the plane $z=0$ gives $x^2+y^2=4$, which is a circle of radius 2. Thus the projection is a disk of radius 2, so $r$ ranges from 0 to 2. All three options are correct graph-based checks: (A) is the algebraic check, (B) is the shape interpretation, (C) is the projection. This question requires interpreting the graph to validate the limits.

📖 Explanation: By Fubini's theorem, for continuous functions over a rectangular box, all permutations of the order of integration give the same result. This is a fundamental property of triple integrals. Options C and D are false restrictions; option B is false because any order is valid. This is a direct conceptual recall question.

📖 Explanation: The region is a spherical shell between radii 1 and 2. Spherical coordinates use $\rho$ from 1 to 2, $\phi$ from 0 to $\pi$, $\theta$ from 0 to $2\pi$. The Jacobian is $\rho^2\sin\phi$. Option A has the correct limits and Jacobian. Option B uses $\rho$ up to 4 (wrong radius), C uses $\rho\sin\phi$ (wrong Jacobian for volume), D omits $\sin\phi$. This tests multi-step reasoning: recognizing the geometry, choosing coordinates, setting limits, and applying the Jacobian correctly.