📝 Triple integral definition (12 MCQs)

📖 Explanation: The triple integral is formally defined as the limit of a Riemann sum over a partition of the solid region into small sub-boxes, where $(x_i^*, y_i^*, z_i^*)$ is a sample point in the $i$-th sub-box and $\Delta V_i$ is its volume. The limit is taken as the maximum diameter of the sub-boxes goes to zero (or the number of sub-boxes approaches infinity in a refined partition). Option A is wrong because it uses $\Delta x$ instead of $\Delta V$, C is a formula for iterated integrals under special conditions, and D omits the volume element and sample points.

📖 Explanation: The student's reasoning is flawed because they assumed the total volume of $E$ is 1 so that each sub-box volume is $1/n$. However, the region $E$ may have any volume $V$. If divided into $n$ equal sub-boxes, each volume would be $V/n$, and the sum $\sum 1 \cdot (V/n) = V$. The correct integral of 1 over $E$ is the volume of $E$, not necessarily 1. The student's conclusion only holds if $V=1$, which is an unstated assumption.

📖 Explanation: The iterated integral limits precisely encode the geometry of the region $E$. The outer integral (over $x$) fixes the overall range of $x$. For a fixed $x$, $y$ ranges from 0 to $1-x$, and for fixed $x$ and $y$, $z$ ranges from 0 to $1-x-y$. This corresponds to slicing the solid along the $x$-axis first, then $y$, and finally integrating over $z$. This is a direct application of the definition where the region is described by inequalities. Option B is incorrect because Fubini's theorem requires continuity or absolute integrability, and the limits are not arbitrary; they must match the region. Option C is false because the region is a tetrahedron, not a box.

📖 Explanation: Compute the incorrect integral: $\int_0^{2\pi} \int_0^2 \int_0^5 r^2 \, dz\,dr\,d\theta = 2\pi \int_0^2 5r^2 \, dr = 10\pi \cdot \frac{8}{3} = \frac{80\pi}{3}$. For correct: $\int_0^{2\pi} \int_0^2 \int_0^5 r^3 \, dz\,dr\,d\theta = 2\pi \cdot 5 \cdot \int_0^2 r^3 dr = 10\pi \cdot 4 = 40\pi$. So incorrect = $80\pi/3$, correct = $40\pi$. Option A matches this. So the correct answer is A.

📖 Explanation: Both options A and B correctly set up the triple integral for the volume of the solid. Option A uses cylindrical coordinates with the Jacobian $r$, and the limits: $\theta$ from 0 to $2\pi$, $r$ from 0 to 2 (since $r^2 = x^2+y^2$ and $z=4-r^2$ intersects $z=0$ at $r=2$), and $z$ from 0 to $4-r^2$. Option B uses rectangular coordinates: $x$ from -2 to 2, $y$ from $-\sqrt{4-x^2}$ to $\sqrt{4-x^2}$, and $z$ from 0 to $4-x^2-y^2$. Both are valid and represent the same solid. Option D has incorrect order and limits; integrating $\theta$ last is possible but the limits given are wrong because $r$ depends on $z$ and the bounds are mixed.

📖 Explanation: The Mean Value Theorem for multiple integrals does indeed state that if $f$ is continuous on a connected, compact region $E$ (like a rectangular box), then there exists a point $(\bar{x}, \bar{y}, \bar{z}) \in E$ such that $\iiint_E f \, dV = f(\bar{x}, \bar{y}, \bar{z}) \cdot \text{Volume}(E)$. This is a generalization of the single-variable Mean Value Theorem for integrals. However, the student's claim is true, but it's important to note that this point is not necessarily the centroid; it's some point in the region. So the student is correct. Option A is correct. Option B is false because the theorem extends. Option C is false because the function need not be constant. Option D is false because it holds for any continuous function, not just linear.

📖 Explanation: Both setups are correct. The spherical coordinates: $x = \rho \sin\phi \cos\theta$, so $x^2 = \rho^2 \sin^2\phi \cos^2\theta$, and the volume element is $\rho^2 \sin\phi \, d\rho\,d\phi\,d\theta$. The limits: $\rho$ from 0 to 1, $\phi$ from 0 to $\pi$, $\theta$ from 0 to $2\pi$. The rectangular setup uses the standard bounds for the unit ball: $x$ from -1 to 1, $y$ from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$, and $z$ from $-\sqrt{1-x^2-y^2}$ to $\sqrt{1-x^2-y^2}$. Both are valid and will yield the same value (by symmetry, $\iiint x^2 dV = \iiint y^2 dV = \iiint z^2 dV = \frac{1}{3} \iiint (x^2+y^2+z^2) dV = \frac{4\pi}{15}$). So option A is correct. Option B is true but incomplete; C is false because it says the rectangular setup is incorrect, but it is correct. So the best answer is A.

📖 Explanation: The setup is actually correct. The region is the solid under the paraboloid $z = 1 - x^2 - y^2$ and above the $xy$-plane. The projection is the unit disk $x^2+y^2 \le 1$. So the limits are correct: $x$ from -1 to 1, $y$ from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$, and $z$ from 0 to $1-x^2-y^2$. The integral $\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_0^{1-x^2-y^2} z \, dz\,dy\,dx$ can be evaluated using polar coordinates: $\int_0^{2\pi} \int_0^1 \int_0^{1-r^2} z \, r\, dz\,dr\,d\theta = 2\pi \int_0^1 r \cdot \frac{(1-r^2)^2}{2} dr = \pi \int_0^1 (r - 2r^3 + r^5) dr = \pi (1/2 - 1/2 + 1/6) = \pi/6$. So the correct answer is $\pi/6$. The student's answer of $\pi/4$ indicates a computational error, perhaps they forgot the $r$ or miscomputed. So the error is in evaluation, not in setup. Option D correctly identifies that the setup is right and the student's computation is wrong.

📖 Explanation: The definition of the triple integral as a Riemann sum is given by option B: partition the region into small sub-boxes (or subregions) with volume $\Delta V_{ijk}$, choose a sample point $(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)$ in each, and take the limit as the maximum diameter (or volume) of the subregions tends to 0. Option A is incorrect because it uses a rectangular grid with $\Delta x_i, \Delta y_j, \Delta z_k$ which is not general for non-rectangular regions; the subregions may not be boxes. Option C is an iterated integral that evaluates the integral, not the definition. Option D correctly identifies that B is the definition and C is the computational tool, so both are correct in different senses, but the question asks for the definition, so B is the direct answer. However, since D says both are correct representations, and C is not a definition but a computation, D is misleading. The best answer is B.

📖 Explanation: For a Riemann integral over a bounded region in $\mathbb{R}^3$, the necessary conditions are: (1) $f$ is bounded on $E$, (2) $E$ is Jordan measurable (i.e., its boundary has measure zero), and (3) the set of discontinuities of $f$ has measure zero (this is the Lebesgue criterion for Riemann integrability). Continuity is sufficient but not necessary. A function can have discontinuities on a set of measure zero and still be Riemann integrable. So option C (continuity) is not necessary. Option A is necessary, B is necessary, D is necessary. Hence the answer is C.

📖 Explanation: The region $E$ is a tetrahedron, and the function $f(x,y,z)=x+y+z$ is increasing in each variable. If we partition the bounding box $[0,1] \times [0,1] \times [0,1]$ into $n^3$ sub-boxes and only consider those entirely or partially inside $E$, but the definition requires summing over sub-boxes that partition $E$. If we take sub-boxes that cover $E$ and use upper-right corners (which lie on the plane $x+y+z=1$ or outside), we may overestimate because the function is larger at larger coordinates. However, the Riemann sum is defined for a partition of the region itself, not the bounding box. If we partition $E$ directly into subregions, the sample point choice doesn't affect the limit for continuous functions. But the question says 'partitioning $E$ into $n^3$ equal sub-boxes in the first octant' – this is ambiguous because $E$ is not a box. If we partition the bounding box and consider only the portion in $E$, it's not a partition of $E$ into boxes. So the correct interpretation: if we choose sample points as upper-right corners of sub-boxes that lie inside $E$, for increasing function, it's an overestimate. So the answer is B.

📖 Explanation: The definition of the triple integral is independent of the coordinate system. It uses a partition of the region $E$ into small subregions (not necessarily boxes), each with volume $\Delta V_i$, and then sums $f(x_i^*,y_i^*,z_i^*) \Delta V_i$. The limit as the maximum diameter of the subregions goes to zero gives the integral, provided the function is continuous (or Riemann integrable). Option A is incorrect because the definition doesn't require spherical coordinates; coordinates are for evaluation. Option C is false because subregions need not be boxes; they can be curvilinear. Option D is false because sample points can be any point in the subregion; the limit is independent of the choice for continuous functions.