Answer: The total mass of the lamina

Recall the mass formula for a lamina: For a two-dimensional lamina with variable density $\rho(x,y)$, the total mass is obtained by integrating the density over the entire region: $m = \iint_R \rho(x,y)\,dA$. Why the correct answer works: Option B correctly identifies this integral as computing the total mass. Density times area element, summed over the region, yields mass. Why distractors fail: Option A would be correct only if $\rho = 1$ (uniform unit density), giving $\iint_R dA$. Option C is the moment of inertia $I_x = \iint y^2 \rho\,dA$, which includes an extra $y^2$ factor. Option D is $\bar{x} = M_y/m$, a ratio involving a first moment, not the integral shown.

Answer: $\frac{3}{2}$

Set up the double integral: $m = \int_0^1 \int_0^1 6xy\,dy\,dx$. Evaluate the inner integral: $\int_0^1 6xy\,dy = 6x \cdot \frac{y^2}{2}\Big|_0^1 = 3x$. Evaluate the outer integral: $\int_0^1 3x\,dx = \frac{3x^2}{2}\Big|_0^1 = \frac{3}{2}$. Why distractors fail: Option A ($\frac{1}{2}$) omits the coefficient $6$. Option C ($3$) forgets to divide by $2$ in the outer integral. Option D ($6$) treats $\rho$ as constant over the unit square.

Answer: $M_y = \iint x(x+y)\,dA$ and $I_x = \iint y^2(x+y)\,dA$

Recall the definitions: $M_y = \iint x\,\rho\,dA$ (first moment about the $y$-axis), and $I_x = \iint y^2\,\rho\,dA$ (moment of inertia about the $x$-axis). Apply with $\rho = x + y$: Substituting: $M_y = \iint x(x+y)\,dA$ and $I_x = \iint y^2(x+y)\,dA$. Why distractors fail: Option A replaces $y^2$ with $x^2$ in $I_x$, computing $I_y$ instead. Option B uses $y$ in $M_y$, which would give $M_x$. Option D uses $y$ instead of $y^2$ in $I_x$, confusing first moment with second moment.

Answer: The square of the distance from each point to the $x$-axis

Understand moment of inertia: Moment of inertia measures resistance to rotation about an axis. It is computed by integrating $r^2 \rho\,dA$ where $r$ is the perpendicular distance from the axis of rotation. Why the correct answer works: For $I_x$ (rotation about the $x$-axis), the perpendicular distance from a point $(x,y)$ to the $x$-axis is $|y|$, so $r^2 = y^2$. Why distractors fail: Option A confuses a geometric distance with a density component. Option C describes the distance to the $y$-axis, which would be $x^2$ and corresponds to $I_y$. Option D has no connection to the physical meaning of moment of inertia.

Answer: No, $I_y$ should use $x^2$ in the integrand because $x$ measures the distance to the $y$-axis.

Recall the moment of inertia definitions: By definition, $I_y = \iint x^2 \rho\,dA$ and $I_x = \iint y^2 \rho\,dA$. The subscript indicates the axis of rotation, and the integrand uses the square of the distance to that axis. Why the correct answer works: The distance from a point $(x,y)$ to the $y$-axis is $|x|$, so $I_y = \iint x^2 \rho\,dA$. The student incorrectly used $y^2$. Why distractors fail: Option A has the formula backwards. Option C uses $(x^2+y^2)$, which gives the polar moment of inertia $I_0 = I_x + I_y$, not $I_y$ alone. Option D incorrectly suggests the formula depends on whether $\rho$ is constant.

Answer: $\iiint_E \rho(x,y,z)\,dV$

Recall the 3D mass formula: The mass of a solid with variable density is $m = \iiint_E \rho(x,y,z)\,dV$. Why the correct answer works: Option B is the direct triple integral of density over the volume, which gives total mass. Why distractors fail: Option A uses a double integral ($dA$), which is for laminae, not solids. Option C includes a distance-squared factor, which would compute a second moment (moment of inertia). Option D integrates the gradient of $\rho$, which is a vector quantity and unrelated to mass.

Answer: Spherical coordinates with $\rho_{\text{sph}} \cos\phi$ as the density and limits $0 \le \phi \le \pi/2$

Analyze the region and density: The region is a hemisphere, and the density $\rho = z$ depends only on the vertical coordinate. In spherical coordinates, $z = \rho_{\text{sph}}\cos\phi$, and the volume element is $\rho_{\text{sph}}^2 \sin\phi\,d\rho_{\text{sph}}\,d\phi\,d\theta$. Why spherical is most efficient: In spherical coordinates, all limits become constants: $0 \le \rho_{\text{sph}} \le R$, $0 \le \phi \le \pi/2$, $0 \le \theta \le 2\pi$. This avoids square root bounds entirely, making the integral easiest to evaluate. Use symmetry for center of mass: By symmetry of the hemisphere and the density (both symmetric about the $z$-axis), $\bar{x} = \bar{y} = 0$. Only $\bar{z}$ requires a nontrivial computation. Why distractors fail: Option A works but involves messy square-root bounds. Option B (cylindrical) is viable but still requires $\sqrt{R^2 - r^2}$ as a limit. Option D incorrectly reduces a 3D solid to a 2D integral.

Answer: $15$

Find the area of the triangle: The triangle with vertices $(0,0)$, $(2,0)$, $(0,3)$ has base $2$ and height $3$, so its area is $A = \frac{1}{2}(2)(3) = 3$. Compute the mass: For constant density, $m = \rho \cdot A = 5 \cdot 3 = 15$. Why distractors fail: Option A ($30$) uses $\rho \cdot \text{base} \cdot \text{height}$ without the $\frac{1}{2}$ factor. Option C ($6$) is the area times 2, forgetting to multiply by $\rho$. Option D ($25$) likely results from an incorrect area calculation.

Answer: $\bar{x} = \frac{1}{m}\iint x\,\rho\,dA$, $\bar{y} = \frac{1}{m}\iint y\,\rho\,dA$

Recall the center of mass formulas: By definition, $\bar{x} = M_y/m$ where $M_y = \iint x\,\rho\,dA$, and $\bar{y} = M_x/m$ where $M_x = \iint y\,\rho\,dA$. Why the correct answer works: Option A matches these definitions exactly: $\bar{x}$ uses the integral of $x\rho$ and $\bar{y}$ uses the integral of $y\rho$, both divided by $m$. Why distractors fail: Option B swaps the roles of $x$ and $y$, a common confusion between $M_x$ and $M_y$. Option C omits the division by $m$, giving moments rather than coordinates. Option D uses second moments ($x^2$ and $y^2$), which relate to moments of inertia, not center of mass.

Answer: Toward larger $x$ values (right)

Analyze the density distribution: Since $\rho(x,y) = x$, the density increases as $x$ increases. More mass is concentrated on the right side of the rectangle. Effect on center of mass: The center of mass shifts toward the region of higher density. With density increasing in $x$, $\bar{x}$ moves to the right compared to the uniform case ($\bar{x}_{\text{uniform}} = 1$). Why distractors fail: Option A claims the opposite direction. Option C is wrong because $\rho$ does not depend on $y$, so $\bar{y}$ remains at the midpoint. Option D ignores that non-uniform density shifts the center of mass.

Answer: $\frac{2}{3}$

Compute $M_y$: $M_y = \int_0^1 \int_0^1 x \cdot 6xy\,dy\,dx = \int_0^1 \int_0^1 6x^2y\,dy\,dx$. Evaluate: Inner: $\int_0^1 6x^2 y\,dy = 3x^2$. Outer: $\int_0^1 3x^2\,dx = 1$. So $M_y = 1$. Find $\bar{x}$: $\bar{x} = M_y / m = 1 / (3/2) = 2/3$. Why distractors fail: Option A ($\frac{1}{2}$) assumes uniform density, for which the center of a square is at $\frac{1}{2}$. Option B ($\frac{1}{3}$) confuses $M_x$ and $M_y$ or makes an integration error. Option D ($\frac{3}{4}$) results from an incorrect moment calculation.

Answer: At the origin $(0, 0, 0)$

Use symmetry: The sphere is symmetric about all three coordinate planes, and $\rho = k(x^2+y^2+z^2) = kr^2$ is a radial function, symmetric about the origin. Why the correct answer works: Because both the region and the density function are symmetric with respect to each coordinate plane, all first moments vanish and $(\bar{x}, \bar{y}, \bar{z}) = (0,0,0)$. Why distractors fail: Option A places the center of mass off-center with no reason from the symmetry. Option C shifts the center along the $z$-axis, which contradicts the radial symmetry of $\rho$. Option D is wrong because symmetry arguments determine the center of mass regardless of the value of $k$.

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

Convert to polar coordinates: In polar coordinates, $\rho(x,y) = \sqrt{x^2+y^2} = r$, and the area element is $dA = r\,dr\,d\theta$. The disk $x^2+y^2 \le 4$ becomes $0 \le r \le 2$. Form the mass integral: $m = \iint \rho\,dA = \int_0^{2\pi}\int_0^2 r \cdot r\,dr\,d\theta = \int_0^{2\pi}\int_0^2 r^2\,dr\,d\theta$. Why distractors fail: Option A has the integrand $r$, which omits the density (it computes area). Option C uses the upper limit $r = 4$ instead of $r = 2$ (confusing $r^2 \le 4$ with $r \le 4$). Option D has $r^3$, which would arise if density were $r^2$ instead of $r$.