Mass, Density & Center of Mass - Calculus 3

1. A solid occupies the region bounded by $z = 4 - x^2 - y^2$ and $z = 0$ with density $\rho(x,y,z) = z$. What is the total mass of this solid?
   Answer: $\frac{16\pi}{3}$ The solid is bounded below by $z = 0$ and above by $z = 4 - x^2 - y^2$. In cylindrical coordinates, this becomes $0 \leq z \leq 4 - r^2$ with $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$. The mass is $M = \iiint \rho \, dV = \int_0^{2\pi} \int_0^2 \int_0^{4-r^2} z \cdot r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r \left[\frac{z^2}{2}\right]_0^{4-r^2} dr \, d\theta = \int_0^{2\pi} \int_0^2 \frac{r(4-r^2)^2}{2} dr \, d\theta = \int_0^{2\pi} \frac{8}{3} d\theta = \frac{16\pi}{3}$. Choice A results from forgetting the factor of $2\pi$. Choice C results from computational errors in the integration. Choice D results from incorrectly computing the inner integral.
2. A lamina occupies the region $D = \{(x,y): x^2 + y^2 \leq 4, x \geq 0, y \geq 0\}$ with density function $\rho(x,y) = x + y$. Which of the following best approximates the $y$-coordinate of the center of mass?
   Answer: $1.13$ Using polar coordinates with $0 \leq r \leq 2$ and $0 \leq \theta \leq \pi/2$, we have $\rho(r,\theta) = r\cos\theta + r\sin\theta = r(\cos\theta + \sin\theta)$. The total mass is $M = \int_0^{\pi/2} \int_0^2 r(\cos\theta + \sin\theta) \cdot r \, dr \, d\theta = \int_0^{\pi/2} \frac{8}{3}(\cos\theta + \sin\theta) d\theta = \frac{16}{3}$. The first moment about the x-axis is $M_x = \int_0^{\pi/2} \int_0^2 r\sin\theta \cdot r(\cos\theta + \sin\theta) \cdot r \, dr \, d\theta = \int_0^{\pi/2} 4\sin\theta(\cos\theta + \sin\theta) d\theta = \frac{6 + \pi}{2}$. Therefore $\bar{y} = \frac{M_x}{M} = \frac{3(6 + \pi)}{16} \approx 1.13$. The other choices result from computational errors in evaluating the integrals.
3. A thin lamina occupies the region $R = \{(x,y): 0 \leq x \leq 2, 0 \leq y \leq x^2\}$ and has density function $\rho(x,y) = xy + 1$. If the total mass of the lamina is $M$, which expression correctly represents the $x$-coordinate of the center of mass?
   Answer: $\bar{x} = \frac{1}{M} \int_0^2 \int_0^{x^2} x(xy + 1) \, dy \, dx$ The x-coordinate of the center of mass is given by $\bar{x} = \frac{M_y}{M}$ where $M_y = \iint_R x \rho(x,y) \, dA$ is the first moment about the y-axis. Therefore $\bar{x} = \frac{1}{M} \iint_R x \rho(x,y) \, dA = \frac{1}{M} \int_0^2 \int_0^{x^2} x(xy + 1) \, dy \, dx$. Choice B omits the factor of $x$. Choice C incorrectly uses $x^2$ instead of $x$. Choice D incorrectly uses $y$ instead of $x$.
4. A lamina occupies the annular region $1 \leq x^2 + y^2 \leq 4$ with density function $\rho(x,y) = x^2 + y^2$. Due to the symmetry properties of both the region and the density function, what can be concluded about the center of mass?
   Answer: The center of mass lies at the origin $(0,0)$ When you encounter problems involving the center of mass of a lamina, the key insight is recognizing how symmetry can eliminate the need for complex calculations. The center of mass $(\bar{x}, \bar{y})$ is determined by the weighted distribution of mass across the region. The center of mass lies at the origin because both the region and density function exhibit perfect symmetry about both coordinate axes. The annular region $1 \leq x^2 + y^2 \leq 4$ is symmetric about the $x$-axis, $y$-axis, and the origin. The density function $\rho(x,y) = x^2 + y^2$ is also symmetric about both axes since $\rho(-x,y) = \rho(x,-y) = \rho(x,y)$. This means for every point $(x,y)$ with mass $\rho(x,y)$, there are corresponding points $(-x,y)$, $(x,-y)$, and $(-x,-y)$ with identical mass. These symmetric contributions cancel out when calculating $\bar{x}$ and $\bar{y}$, forcing both coordinates to equal zero. Choice A is incorrect because symmetry analysis provides a definitive answer without calculation. Choice B fails because the density function has no preference for positive $x$ values over negative ones. Choice C is wrong because there's no reason the center would favor the first quadrant when all four quadrants are weighted equally by the symmetric density function. Study tip: Always check for symmetry in both the region and density function before diving into integration. If both are symmetric about the same axes, the center of mass lies at the intersection of those axes of symmetry.
5. A thin wire lies along the curve $y = x^2$ from $x = 0$ to $x = 2$ and has linear density $\lambda(x,y) = x + 1$. What is the mass of this wire?
   Answer: $\int_0^2 (x + 1)\sqrt{1 + 4x^2} \, dx$ When you encounter problems involving mass of objects along curves, you're working with line integrals. The key insight is that mass equals density times length, but since density varies along the curve, you need to integrate: $\text{mass} = \int \lambda \, ds$, where $ds$ is the differential arc length element. For a curve given by $y = f(x)$, the arc length element is $ds = \sqrt{1 + (f'(x))^2} \, dx$. Here, $y = x^2$, so $f'(x) = 2x$ and $ds = \sqrt{1 + (2x)^2} \, dx = \sqrt{1 + 4x^2} \, dx$. Since the density is $\lambda(x,y) = x + 1$ and we're moving along the curve from $x = 0$ to $x = 2$, the mass is: $\int_0^2 (x + 1)\sqrt{1 + 4x^2} \, dx$ This confirms answer C is correct. Answer A forgets the arc length element entirely—it treats this like integrating along a straight horizontal line, which would give you the wrong units and incorrect result. Answer B attempts to integrate with respect to $y$, but keeps the density in terms of $x$, creating an inconsistent setup. Answer D incorrectly substitutes the curve equation into the density function, changing $x + 1$ to $x^2 + 1$, which misrepresents the given density. Study tip: For line integral problems, always identify three components: the parametrization (here $x$ from 0 to 2), the function being integrated (density $x + 1$), and the correct differential element ($ds$ for arc length, not just $dx$).
6. A triangular lamina with vertices at $(0,0)$, $(3,0)$, and $(0,4)$ has constant density $\rho = 2$. If the lamina is rotated $90°$ counterclockwise about the origin, what are the new coordinates of its center of mass?
   Answer: $\left(-\frac{4}{3}, 1\right)$ For the original triangle, the center of mass is at the centroid since density is constant: $\bar{x} = \frac{0+3+0}{3} = 1$ and $\bar{y} = \frac{0+0+4}{3} = \frac{4}{3}$. When rotated $90°$ counterclockwise about the origin, the transformation is $(x,y) \mapsto (-y,x)$. Therefore the new center of mass is at $\left(-\frac{4}{3}, 1\right)$. Choice B incorrectly applies the transformation as $(x,y) \mapsto (-x,y)$. Choice C forgets the rotation entirely. Choice D incorrectly applies the transformation as $(x,y) \mapsto (y,x)$.
7. Consider a solid cone with height $h$ and base radius $R$, where the vertex is at the origin and the axis lies along the positive $z$-axis. If the density varies as $\rho(x,y,z) = z$, which expression gives the $z$-coordinate of the center of mass?
   Answer: $\bar{z} = \frac{\int_0^h \int_0^{2\pi} \int_0^{\frac{R(h-z)}{h}} z^2 \cdot r \, dr \, d\theta \, dz}{\int_0^h \int_0^{2\pi} \int_0^{\frac{R(h-z)}{h}} z \cdot r \, dr \, d\theta \, dz}$ When finding the center of mass of a three-dimensional object with variable density, you need to set up the correct formula: $\bar{z} = \frac{\int\int\int z \cdot \rho(x,y,z) \, dV}{\int\int\int \rho(x,y,z) \, dV}$, where the numerator represents the first moment about the xy-plane and the denominator is the total mass. The key insight is correctly describing the cone's geometry. With the vertex at the origin and opening upward along the positive z-axis, at height $z$, the cross-sectional radius decreases linearly from $R$ at the base ($z=h$) to 0 at the vertex ($z=0$). This gives us the relationship: radius = $\frac{R(h-z)}{h}$. Since the density is $\rho(x,y,z) = z$, the numerator becomes $\int_0^h \int_0^{2\pi} \int_0^{\frac{R(h-z)}{h}} z^2 \cdot r \, dr \, d\theta \, dz$ (note $z^2$ from $z \times z$), and the denominator is $\int_0^h \int_0^{2\pi} \int_0^{\frac{R(h-z)}{h}} z \cdot r \, dr \, d\theta \, dz$. This matches option C. Option A is missing the density function entirely in the numerator. Option B uses the wrong radius function $\frac{Rz}{h}$, which would describe a cone opening downward from the base. Option D switches to integrating over $r$ first, but uses an incorrect height relationship. Study tip: Always sketch the region first and identify how the boundaries change with each variable. For cones, pay special attention to whether the vertex is at the top or bottom, as this determines whether the radius increases or decreases with height.

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