Answer: Arcs at larger radii subtend a greater physical length for the same angle $d\theta$, so the volume element grows with $r$

Geometric interpretation: In cylindrical coordinates, a small change $d\theta$ in angle corresponds to an arc length of $r\, d\theta$ at distance $r$ from the $z$-axis. Hence the cross-sectional area element is $r\, dr\, d\theta$. Why the correct answer works: The factor $r$ accounts for the fact that the physical width of an angular strip increases linearly with the distance from the axis, making the volume element $r\, dr\, d\theta\, dz$. Why distractors fail: Option A is nonsensical—the $z$-axis is straight. Option C dismisses the geometric meaning. Option D incorrectly attributes $r$ to height changes, but $dz$ already handles the height.

Answer: $\dfrac{\pi}{4}$

Set up the integral: The region is $0 \le \theta \le 2\pi$, $0 \le z \le 1$, $0 \le r \le z$. The integral is $\int_0^{2\pi}\int_0^1\int_0^z z \cdot r\, dr\, dz\, d\theta$. Evaluate the innermost integral: $\int_0^z z \cdot r\, dr = z \cdot \frac{r^2}{2}\Big|_0^z = \frac{z^3}{2}$. Evaluate the remaining integrals: $\int_0^1 \frac{z^3}{2}\, dz = \frac{1}{2} \cdot \frac{z^4}{4}\Big|_0^1 = \frac{1}{8}$. Then $\int_0^{2\pi} \frac{1}{8}\, d\theta = \frac{2\pi}{8} = \frac{\pi}{4}$. Why distractors fail: Option A ($\pi/10$) and Option B ($\pi/5$) arise from incorrect powers of $z$ during integration. Option D ($\pi/8$) results from omitting the factor of $2\pi$ or using $\theta \in [0,\pi]$ instead of $[0,2\pi]$.

Answer: $\displaystyle\int_0^{2\pi}\int_0^2\int_0^5 r\, dz\, dr\, d\theta$

Identify the correct volume element: In cylindrical coordinates, $dV = r\, dz\, dr\, d\theta$. Why the correct answer works: The bounds for the cylinder $r=2$, $0 \le z \le 5$ are: $0 \le \theta \le 2\pi$, $0 \le r \le 2$, $0 \le z \le 5$. With $dV = r\,dz\,dr\,d\theta$, this gives $20\pi$. Why distractors fail: Option A omits the Jacobian factor $r$. Option C uses $r^2$, which is incorrect for cylindrical coordinates. Option D uses $r^2\sin\phi$, which is the spherical Jacobian.

Answer: $\rho^2\sin\phi$

Set up the Jacobian matrix: The $3\times3$ matrix has entries $\frac{\partial x}{\partial \rho} = \sin\phi\cos\theta$, $\frac{\partial x}{\partial \phi} = \rho\cos\phi\cos\theta$, etc. for each of $x$, $y$, $z$. Compute the determinant: After expanding the determinant (using cofactor expansion or properties of the matrix), the result simplifies to $\rho^2\sin\phi$. This is a standard result. Why distractors fail: Option A ($\rho\sin\phi$) is missing a factor of $\rho$. Option B ($\rho^2$) omits $\sin\phi$. Option D ($\rho^2\cos\phi$) incorrectly uses $\cos\phi$ instead of $\sin\phi$.

Answer: It is unnecessarily complicated; the entire integral can be done in spherical coordinates with $1 \le \rho \le 3$, $0 \le \phi \le \pi/2$, $0 \le \theta \le 2\pi$

Express the region in spherical coordinates: The spherical shell $1 \le \rho \le 3$ with $z \ge 0$ translates to $\phi \in [0, \pi/2]$. All bounds become constant. Why the correct answer works: Spherical coordinates handle both the radial bounds and the $z \ge 0$ condition directly. Splitting into two systems adds unnecessary complexity. Why distractors fail: Option A: while not strictly wrong, it's never more efficient here. Option C: you can technically split a region and use different systems, so this is too absolute. Option D: there is no such restriction on the integrand.

Answer: Yes, the setup is correct

Verify the bounds: The paraboloid is $z = r^2$ and the plane is $z = 4$. They intersect when $r^2 = 4$, i.e., $r = 2$. So $r$ ranges from $0$ to $2$, $z$ ranges from $r^2$ to $4$, and $\theta$ from $0$ to $2\pi$. Verify the integrand: The volume element is $r\, dz\, dr\, d\theta$ with integrand $1$ (for volume), so the integrand is just $r$. This is correct. Why distractors fail: Option A: $r=2$ is correct since $r^2=4$ at the intersection. Option B: $r$ is the correct cylindrical Jacobian, not $r^2$. Option C: $z$ goes from the paraboloid ($r^2$) up to the plane ($4$), not the other way.

Answer: $\dfrac{4}{3}\pi a^3$

Set up the integral: $V = \int_0^{2\pi}\int_0^{\pi}\int_0^a \rho^2\sin\phi\, d\rho\, d\phi\, d\theta$. Evaluate step by step: $\int_0^a \rho^2\, d\rho = a^3/3$, $\int_0^{\pi}\sin\phi\, d\phi = 2$, $\int_0^{2\pi} d\theta = 2\pi$. Thus $V = (a^3/3)(2)(2\pi) = \frac{4}{3}\pi a^3$. Why distractors fail: Option A has $a^2$ instead of $a^3$—this is a surface area-like expression. Option C is missing the $1/3$ factor. Option D is half the correct answer, as if $\phi$ only ranged from $0$ to $\pi/2$.

Answer: The Jacobian factor $r$ is missing from the integrand

Check the volume element: In cylindrical coordinates, the volume element is $dV = r\, dz\, dr\, d\theta$, not simply $dz\, dr\, d\theta$. Why the correct answer works: The student omitted the factor $r$. The correct integral is $\int_0^{2\pi}\int_0^R\int_0^H f\cdot r\, dz\, dr\, d\theta$. Why distractors fail: Option A: the limits $0 \le r \le R$, $0 \le z \le H$, $0 \le \theta \le 2\pi$ are correct for a cylinder. Option C: the given order is valid. Option D: there is an error—the missing $r$.

Answer: $r = 1$

Find the intersection: Setting $z = r$ equal to $z = \sqrt{2 - r^2}$: $r = \sqrt{2 - r^2}$. Squaring: $r^2 = 2 - r^2$, so $2r^2 = 2$, hence $r = 1$. Why the correct answer works: The two surfaces intersect at $r = 1$, so the projection onto the $r\theta$-plane is the disk $0 \le r \le 1$. Why distractors fail: Option A ($\sqrt{2}$) is the maximum $\rho$ in spherical coordinates for the upper surface. Option C ($2$) and Option D ($\sqrt{2}/2$) arise from algebraic errors in solving the intersection.

Answer: $\displaystyle\int_0^{2\pi}\int_0^{\pi}\int_1^4 \sin\phi\, d\rho\, d\phi\, d\theta$

Combine the integrand with the Jacobian: The integrand is $1/\rho^2$ and the spherical volume element is $\rho^2\sin\phi\, d\rho\, d\phi\, d\theta$. Multiplying: $\frac{1}{\rho^2} \cdot \rho^2\sin\phi = \sin\phi$. Why the correct answer works: After simplification, the integrand becomes simply $\sin\phi$, with bounds $1 \le \rho \le 4$, $0 \le \phi \le \pi$, $0 \le \theta \le 2\pi$. Why distractors fail: Option A omits the Jacobian entirely. Option C does not incorporate $1/\rho^2$ into the integrand (it's as if $f=1$). Option D keeps $1/\rho^2$ and $\sin\phi$ but fails to include $\rho^2$ from the Jacobian.

Answer: $\dfrac{2\pi}{3}$

Evaluate the innermost integral: $\int_0^r r\, dz = r \cdot r = r^2$. Evaluate the remaining integrals: $\int_0^1 r^2\, dr = 1/3$. Then $\int_0^{2\pi} (1/3)\, d\theta = 2\pi/3$. Why distractors fail: Option A ($\pi/3$) uses $\theta$ from $0$ to $\pi$ instead of $2\pi$. Option C ($\pi$) results from integrating $r^2$ incorrectly as $1/2$ instead of $1/3$. Option D ($\pi/2$) arises from a different combination of errors.

Answer: $r\, dr\, d\theta\, dz$

Recall the Jacobian for cylindrical coordinates: When converting from Cartesian $(x,y,z)$ to cylindrical $(r,\theta,z)$, the Jacobian determinant $\left|\frac{\partial(x,y,z)}{\partial(r,\theta,z)}\right| = r$. Why the correct answer works: The volume element is $dV = r\, dr\, d\theta\, dz$, which includes the factor $r$ from the Jacobian. Why distractors fail: Option A omits the Jacobian factor $r$. Option C uses $r^2 \sin\phi$, which is the Jacobian for spherical coordinates. Option D uses $r^2$, which is not the correct Jacobian for cylindrical coordinates.

Answer: $\rho^2\sin^2\phi \cdot \rho^2\sin\phi\, d\rho\, d\phi\, d\theta$

Apply the spherical volume element: The volume element in spherical coordinates is $\rho^2\sin\phi\, d\rho\, d\phi\, d\theta$. The integrand $x^2+y^2$ becomes $\rho^2\sin^2\phi$. Why the correct answer works: The full integrand is $\rho^2\sin^2\phi \cdot \rho^2\sin\phi = \rho^4\sin^3\phi$, which matches Option B's expression before simplification. Why distractors fail: Option A uses $\rho$ instead of $\rho^2\sin\phi$ for the Jacobian. Option C omits the $\sin\phi$ from the Jacobian. Option D has $\sin^2\phi$ instead of $\sin^3\phi$—it dropped the $\sin\phi$ from the Jacobian.

Answer: The claim is incorrect; the integrand stays the same, but the limits of integration must be adjusted to describe the same region

Fubini's theorem: By Fubini's theorem, for continuous functions over bounded regions, the order of integration can be changed without altering the integrand $f(x,y,z)$. However, the limits of integration must be rewritten to correctly describe the same region $E$ in the new order. Why the correct answer works: The function being integrated does not change—only the limits need adjustment because the geometric description of the region depends on which variable is integrated first. Why distractors fail: Option A and C are wrong because the integrand and Jacobian (which is $1$ in Cartesian) do not change. Option D is wrong because the limits generally do change.

Answer: $8\pi$

Set up the integral in cylindrical coordinates: The paraboloid $z = 4 - x^2 - y^2 = 4 - r^2$ meets $z = 0$ when $r = 2$. The volume is $V = \int_0^{2\pi}\int_0^2\int_0^{4-r^2} r\, dz\, dr\, d\theta$. Evaluate the integral: $V = \int_0^{2\pi}\int_0^2 r(4-r^2)\, dr\, d\theta = \int_0^{2\pi}\int_0^2 (4r - r^3)\, dr\, d\theta$. Inner integral: $[2r^2 - r^4/4]_0^2 = 8 - 4 = 4$. Then $V = \int_0^{2\pi} 4\, d\theta = 8\pi$. Why distractors fail: Option A ($4\pi$) omits the $\theta$-integration correctly but may result from a factor error. Option C ($16\pi$) doubles the result. Option D ($2\pi$) comes from a significant arithmetic error.

Answer: Because volume is always non-negative, and a negative Jacobian would indicate a reversal of orientation that does not affect volume

Orientation vs. volume: The sign of the Jacobian determinant indicates whether the transformation preserves or reverses orientation. However, volume is a non-negative quantity. Why the correct answer works: Taking the absolute value ensures the volume element $dV$ remains positive, even if the transformation reverses the handedness of the coordinate system. Why distractors fail: Option A is false—some valid transformations yield a negative Jacobian. Option C dismisses the theoretical reason. Option D is incorrect; the Jacobian is well-defined without absolute value.

Answer: First with respect to $z$, then $y$, then $x$

Interpreting the limits of integration: The innermost integral has limits that depend on the most variables, and the outermost integral has constant limits. Why the correct answer works: Since the $z$-bounds depend on both $x$ and $y$, $z$ is integrated first. The $y$-bounds depend on $x$, so $y$ is integrated second. Finally, $x$ has constant bounds and is integrated last. The order is $dz\, dy\, dx$. Why distractors fail: Option A reverses the order entirely. Option C integrates $y$ first, but $y$'s bounds depend on $x$, which has not yet been resolved as the outer variable. Option D is false because the bounds are not all constant.

Answer: The angle measured from the positive $z$-axis

Standard convention for spherical coordinates: In the standard mathematics convention, $\rho$ is the radial distance, $\phi$ is the polar angle from the positive $z$-axis, and $\theta$ is the azimuthal angle in the $xy$-plane. Why the correct answer works: $\phi$ is measured from the positive $z$-axis downward, ranging from $0$ (along the $+z$ axis) to $\pi$ (along the $-z$ axis). Why distractors fail: Option A describes $\theta$, not $\phi$. Option C describes $\rho$. Option D is incorrect because $\phi$ is measured from the positive, not negative, $z$-axis.

Answer: $\rho^2 \sin\phi\, d\rho\, d\phi\, d\theta$

Recall the Jacobian for spherical coordinates: When converting from Cartesian to spherical coordinates, the Jacobian determinant is $\left|\frac{\partial(x,y,z)}{\partial(\rho,\phi,\theta)}\right| = \rho^2 \sin\phi$. Why the correct answer works: The volume element is $dV = \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta$, incorporating the full Jacobian. Why distractors fail: Option A uses $\rho$ (the cylindrical Jacobian). Option B omits the $\sin\phi$ factor. Option D uses $\cos\phi$ instead of $\sin\phi$, which is incorrect.

Answer: $6$

Compute the Jacobian matrix: The Jacobian matrix is diagonal: $\frac{\partial x}{\partial u}=2$, $\frac{\partial y}{\partial v}=3$, $\frac{\partial z}{\partial w}=1$, with all off-diagonal entries zero. Compute the determinant: For a diagonal matrix, the determinant is the product of diagonal entries: $2 \cdot 3 \cdot 1 = 6$. Why distractors fail: Option A ($5$) adds $2+3$ instead of multiplying. Option C ($1$) ignores the scaling. Option D ($36$) may arise from squaring $6$.

Answer: $\dfrac{1}{2}$

Find the forward Jacobian first: The transformation $(x,y,z) \to (u,v,w)$ has $\frac{\partial(u,v,w)}{\partial(x,y,z)}$ with matrix $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}$. Compute the determinant: $\det = 1(1\cdot1 - 1\cdot0) - 1(0\cdot1 - 1\cdot1) + 0 = 1 - (-1) + 0 = 2$. Use the inverse relationship: $\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right| = \frac{1}{\left|\frac{\partial(u,v,w)}{\partial(x,y,z)}\right|} = \frac{1}{2}$. Why distractors fail: Option A ($2$) is the forward Jacobian $|\partial(u,v,w)/\partial(x,y,z)|$, not the inverse. Option B ($1$) would require the transformation to be volume-preserving. Option D ($-2$) forgets the absolute value and uses the forward Jacobian.

Answer: $\dfrac{7}{8}$

Evaluate the innermost integral: $\int_0^{x+y}(x+y+z)\,dz = \left[(x+y)z + \frac{z^2}{2}\right]_0^{x+y} = (x+y)^2 + \frac{(x+y)^2}{2} = \frac{3}{2}(x+y)^2$. Evaluate the middle integral: $\int_0^x \frac{3}{2}(x+y)^2\, dy$. Let $u = x+y$, $du = dy$. When $y=0$, $u=x$; when $y=x$, $u=2x$. So $\frac{3}{2}\int_x^{2x} u^2\, du = \frac{3}{2}\left[\frac{u^3}{3}\right]_x^{2x} = \frac{1}{2}(8x^3 - x^3) = \frac{7x^3}{2}$. Evaluate the outer integral: $\int_0^1 \frac{7x^3}{2}\, dx = \frac{7}{2} \cdot \frac{1}{4} = \frac{7}{8}$. Why distractors fail: Options A, B, and D result from arithmetic errors in evaluating the nested integrals, such as incorrect expansion of $(x+y)^2$ or miscounting the coefficient $3/2$.

Answer: $\dfrac{2\pi}{5}$

Separate the iterated integral: The integral factors as $\left(\int_0^{2\pi} d\theta\right)\left(\int_0^{\pi/2}\sin\phi\, d\phi\right)\left(\int_0^1 \rho^4\, d\rho\right)$. Evaluate each factor: $\int_0^{2\pi} d\theta = 2\pi$. $\int_0^{\pi/2}\sin\phi\, d\phi = [-\cos\phi]_0^{\pi/2} = 0 -(-1) = 1$. $\int_0^1 \rho^4\, d\rho = 1/5$. Combine: $2\pi \cdot 1 \cdot 1/5 = 2\pi/5$. Why distractors fail: Option B ($4\pi/5$) uses $\int_0^\pi \sin\phi\, d\phi = 2$ instead of the upper limit $\pi/2$. Option C ($\pi/5$) uses $\int_0^\pi d\theta = \pi$. Option D ($2\pi/3$) uses $\int_0^1 \rho^2\, d\rho = 1/3$ instead.

Answer: The integral is possible in cylindrical coordinates but the bounds for $z$ and $r$ become coupled and more complicated than in spherical

Compare coordinate systems: In spherical coordinates, the bounds are simply $0 \le \rho \le 2$, $0 \le \phi \le \pi/3$, $0 \le \theta \le 2\pi$. In cylindrical, the cone $\phi = \pi/3$ becomes $z = r/\tan(\pi/3) = r/\sqrt{3}$ and the sphere becomes $z = \sqrt{4-r^2}$, requiring careful splitting. Why the correct answer works: While cylindrical coordinates can represent this region, the bounds become coupled and the setup requires determining the intersection of the cone and sphere, making spherical coordinates far simpler. Why distractors fail: Option A: cylindrical can represent cones (e.g., $z = r$). Option C: the Jacobian $r$ is always correct for cylindrical. Option D: cylindrical coordinates have no such restriction.

Answer: $6$

Evaluate the iterated integral: Since the integrand is $1$ and the bounds are all constant, the integral equals the volume of the rectangular box: $1 \times 2 \times 3 = 6$. Why the correct answer works: $\int_0^1\int_0^2\int_0^3 dz\, dy\, dx = \int_0^1\int_0^2 3\, dy\, dx = \int_0^1 6\, dx = 6$. Why distractors fail: Option A ($5$) might come from adding the dimensions. Option C ($8$) and Option D ($12$) result from incorrect multiplication.

Answer: Spherical, because $x^2+y^2+z^2 = \rho^2$ makes the exponent $\rho^3$

Analyze the integrand and region: In spherical coordinates, $x^2+y^2+z^2 = \rho^2$, so $(x^2+y^2+z^2)^{3/2} = \rho^3$. The integrand becomes $e^{\rho^3}$, which is a function of $\rho$ alone. Why the correct answer works: Both the region (a ball $\rho \le 1$) and the integrand ($e^{\rho^3}$) separate cleanly in spherical coordinates, making the integral factorable as a product of three single-variable integrals. Why distractors fail: Option A: the Cartesian form $e^{(x^2+y^2+z^2)^{3/2}}$ is not separable. Option B: cylindrical helps with $x^2+y^2$ but still leaves $z$ coupled in the exponent. Option D: symmetry helps, but the systems are not equally efficient.

Answer: The scaling factor that accounts for the distortion of volume elements under the transformation

Definition of the Jacobian: The Jacobian determinant measures how infinitesimal volume elements change when we transform from one coordinate system to another. Why the correct answer works: It acts as a scaling factor: $dV_{xyz} = \left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right| du\, dv\, dw$, ensuring the integral's value is preserved under the change of variables. Why distractors fail: Option A confuses the Jacobian with a rotation matrix. Option C confuses it with the gradient, which is a vector, not a determinant. Option D confuses it with a geometric property of coordinate surfaces.

Answer: $\displaystyle\int_0^{2\pi}\int_0^1\int_0^{\sqrt{4-r^2}} r\, dz\, dr\, d\theta$

Analyze the region: The solid is inside the cylinder $r \le 1$ and inside the sphere $r^2 + z^2 \le 4$ with $z \ge 0$. For $r \le 1$, $z$ ranges from $0$ to $\sqrt{4-r^2}$. Why the correct answer works: The cylinder restricts $r$ to $[0,1]$, and the sphere caps $z$ at $\sqrt{4-r^2}$. With the Jacobian $r$, the integral is $\int_0^{2\pi}\int_0^1\int_0^{\sqrt{4-r^2}} r\, dz\, dr\, d\theta$. Why distractors fail: Option B uses $r$ up to $2$, which goes outside the cylinder. Option C uses constant $z = 2$, ignoring the spherical cap. Option D uses $4 - r^2$ instead of $\sqrt{4 - r^2}$ for the sphere.

Answer: $0 \le \phi \le \pi/2$

Interpret 'above the xy-plane': Above the $xy$-plane means $z \ge 0$. In spherical coordinates, $z = \rho\cos\phi \ge 0$ requires $\cos\phi \ge 0$, which holds when $0 \le \phi \le \pi/2$. Why the correct answer works: $\phi = 0$ corresponds to the positive $z$-axis and $\phi = \pi/2$ corresponds to the $xy$-plane. So $0 \le \phi \le \pi/2$ captures exactly the upper hemisphere. Why distractors fail: Option A ($0$ to $2\pi$) confuses $\phi$ with $\theta$. Option B ($0$ to $\pi$) gives the full sphere, not just above $xy$. Option D ($\pi/2$ to $\pi$) gives the region below the $xy$-plane.

Answer: Spherical coordinates

Analyze the geometry: The sphere $\rho = 3$ and the cone $\phi = \pi/4$ (since $z = \sqrt{x^2+y^2}$ gives $\cos\phi = \sin\phi$, so $\phi = \pi/4$) both have simple descriptions in spherical coordinates. Why the correct answer works: In spherical coordinates, the bounds become $0 \le \rho \le 3$, $0 \le \phi \le \pi/4$, $0 \le \theta \le 2\pi$, which are all constant—making the integral straightforward to evaluate. Why distractors fail: Option A (Cartesian) leads to complicated square-root bounds. Option B (cylindrical) can work but requires $z$ bounds involving $\sqrt{9-r^2}$ and $r$. Option D is incorrect because spherical is clearly simpler.