📝 Centroid of plane region calculus (33 MCQs)

📖 Explanation: The centroid is the geometric center of a plane region, assuming the lamina is homogeneous (uniform density). For a homogeneous lamina, the centroid coincides with the center of gravity, which is the point where the entire weight of the lamina can be considered to act for purposes of balance. It is a purely geometric property of the shape, independent of the material's mass or density. Option A is incorrect because mass concentration is not the definition, though it's a related concept. Option D is incorrect as it describes the incenter, not the centroid.

📖 Explanation: The first moment of a region R about the y-axis is the integral of the x-coordinate over the region, which is $M_y = \iint_R x \, dA$. This moment measures the distribution of area relative to the y-axis and is used in finding the x-coordinate of the centroid, $\bar{x} = M_y / A$. Option B represents the moment about the x-axis. Options C and D represent higher-order moments (like the polar moment of inertia and second moment about y-axis), which are not used in centroid calculations.

📖 Explanation: The centroid's x-coordinate is the first moment about the y-axis divided by the area. The first moment is $\int_a^b x f(x) \, dx$ and the area is $\int_a^b f(x) \, dx$. Option B is the inverse, which is incorrect. Option C represents the y-coordinate of the centroid for a region under a curve. Option D is a nonsensical combination of these formulas.

📖 Explanation: Symmetry is a powerful tool in locating centroids. If a region is symmetric about the y-axis, the area is evenly distributed on both sides of the axis. Therefore, the first moment about the y-axis, $M_y$, must be zero. Since $\bar{x} = M_y / A$, the x-coordinate of the centroid is zero, meaning the centroid lies on the axis of symmetry. Options B and C are incorrect as they represent specific locations on the y-axis, not the general axis itself.

📖 Explanation: A circle has infinite lines of symmetry (any diameter). The centroid of any region must lie on every axis of symmetry. The only point common to all diameters of a circle is its center. This is a result of the region's geometric symmetry, not its density or area. Option A is a definition, not a reason. Option C is irrelevant here. Pappus's theorem relates volume and centroid location for solids of revolution, but it doesn't explain why the centroid is at the center.

📖 Explanation: The correct formula for the y-coordinate of the centroid for a region between $y = f(x)$ (top) and $y = g(x)$ (bottom) is $\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2} (f(x)^2 - g(x)^2) \, dx$. The student's expression $\frac{1}{2}(f(x)-g(x))^2$ is incorrect. It represents the square of half the height, not the difference of squares. Option A is the integrand for the area. Option B is wrong because the region is symmetric and extends from -2 to 2. Option D is the integrand for $\bar{x}$.

📖 Explanation: Pappus's theorem states that the volume of a solid of revolution generated by rotating a plane region about an external axis is equal to the area of the region times the distance traveled by its centroid. If the axis is the x-axis, the centroid travels a circular path of radius $|\bar{y}|$, so the distance traveled is $2\pi \bar{y}$. Thus, the volume is $V = A \times (2\pi \bar{y}) = 2\pi A \bar{y}$. Option A would be the formula if the axis were the y-axis.

📖 Explanation: The centroid of a triangle is the average of the coordinates of its vertices. The x-coordinate is $(0+2+0)/3 = 2/3$. The y-coordinate is $(0+0+4)/3 = 4/3$. The centroid is therefore (2/3, 4/3). Option B is a common mistake where the coordinates are swapped. Option C is the average of the non-zero coordinates individually, not the average of the points. Option D is a vertex.

📖 Explanation: The area of the region is $A = \int_0^4 \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{2}{3}(8) = \frac{16}{3} \approx 5.333$. The centroid's x-coordinate is $\bar{x} = M_y / A = 12.8 / (16/3) = 12.8 \times 3 / 16 = 2.4$. Option A is a common error where the student might use the wrong formula or miscalculate the area. Option C is a result of using a different point in the curve. Option D is the x-value of the boundary.

📖 Explanation: When revolving a region about the vertical line $x = 5$, the distance from the centroid $(\bar{x} = 3)$ to the axis of revolution is $|5 - 3| = 2$. The distance traveled by the centroid is the circumference of the circle with this radius, which is $2\pi \times 2 = 4\pi$. The volume is $V = A \times \text{distance traveled by centroid} = 10 \times 4\pi = 40\pi$. Wait, the calculation gives 40π, not 120π. Let's re-evaluate. The distance from the centroid to the axis x=5 is 2 units. The distance traveled is 2π*2 = 4π. Volume = 10 * 4π = 40π. I've corrected the final answer. Option C (120π) would be the volume if the distance from the axis was 6, or if the area was 30.

📖 Explanation: The $\bar{y}$ coordinate is the first moment about the x-axis divided by the area. The moment about x-axis for a thin horizontal strip of height $f(x)-g(x)$ at height y is $y \cdot (f(x)-g(x)) \, dx$. Since the height of the strip varies, we integrate, and the center of the strip is at $y = \frac{f(x)+g(x)}{2}$. The moment is $\int_a^b \frac{f(x)+g(x)}{2} (f(x)-g(x)) \, dx = \frac{1}{2} \int_a^b (f(x)^2 - g(x)^2) \, dx$. Option B is the formula for $\bar{x}$. Option C is a common error. Option D is incorrect.

📖 Explanation: In this case, the curves are given as $x = y^2$ and $x = y+2$. To integrate with respect to x, we would have to split the region because the lower boundary changes. However, when integrating with respect to y, the left and right boundaries are clearly defined for all y in the interval, avoiding the need for splitting. This is a key strategic choice in finding centroids. Option B is a consequence, but the primary reason is the avoidance of splitting. Option C is false; you still need to find both coordinates. Option D is incorrect; the centroid is not on the y-axis.

📖 Explanation: The centroid of a semicircle of radius 'r' with its diameter on the x-axis is located at a distance $4r/(3\pi)$ from the diameter. For $r = 2$, this distance is $4(2)/(3\pi) = 8/(3\pi)$. Since the semicircle is symmetric about the y-axis, $\bar{x} = 0$. So the centroid is $(0, 8/(3\pi))$. Option B is the formula for a semicircle of radius 1. Option C is a common incorrect simplification. Option D is the centroid of a quarter circle in the first quadrant.

📖 Explanation: The y-coordinate of the centroid is defined as $\bar{y} = M_x / A$, where $M_x$ is the first moment about the x-axis. We are given $\bar{y} = 4$ and $A = 12$. Therefore, $M_x = \bar{y} \times A = 4 \times 12 = 48$. The units of the first moment are length cubed (e.g., $\text{units}^3$). Option B is the first moment about the y-axis (if $\bar{x}=2$). Option C is a result of using the wrong formula or sign. Option D is a common doubling error.

📖 Explanation: The two parabolas intersect when $x^2 = 8 - x^2$, so $2x^2 = 8$, which gives $x = \pm 2$. The region is bounded between these points. The region is symmetric about the y-axis because the equations are even functions. Therefore, the centroid must lie on the axis of symmetry, which is the y-axis. This means its x-coordinate is 0. This is a classic Medium of symmetry, requiring no integration. Options B, C, and D are the intersection points or their negatives and are not the centroid's x-coordinate.

📖 Explanation: The region between $y = x$ and $y = x^2$ in the first quadrant is not symmetric about y=x because the curves are not symmetric counterparts in that domain; the line $y=x$ is the upper boundary for $0<x<1$. The average y-value will be higher than the average x-value because the region is skewed towards the line $y=x$. The actual centroid is $(1/2, 2/5)$. The student's claim ignores the different distributions of area in the x and y directions. The region is not symmetric about y=x; only the boundaries are reflections.

📖 Explanation: The centroid of a composite region is the weighted average of the centroids of its parts, weighted by their areas. The x-coordinate is $\bar{x} = \frac{A_1 \bar{x}_1 + A_2 \bar{x}_2}{A_1 + A_2} = \frac{12(2) + 6(8)}{12+6} = \frac{24 + 48}{18} = \frac{72}{18} = 4$. Option B (3.5) is the average of the x-coordinates, unweighted. Option C (4.5) might come from incorrectly weighting. Option D (3) is a common arithmetic error.

📖 Explanation: If $\bar{x} = 0$, then the first moment about the y-axis, $M_y = \iint_R x \, dA = 0$. This means the area distribution is balanced about the y-axis. While a common reason for this is symmetry about the y-axis, it is not strictly necessary; a region could have a zero first moment without being symmetric (e.g., a shape with one part on the right and a larger part on the left in such a way that the moments cancel). However, among the options, symmetry about the y-axis is the standard and most likely implication. Option B would imply $\bar{y} = 0$.

📖 Explanation: The y-coordinate of the centroid is $\bar{y} = M_x / A$. If $M_x = 0$ and the area $A$ is non-zero, then $\bar{y} = 0$. This means the centroid lies on the x-axis. While symmetry about the x-axis is a common cause for this, it's not the only one. The conclusion must be about the centroid's location, not a specific type of symmetry. Option A is a special case. Option C would imply $M_y = 0$, not $M_x$.

📖 Explanation: If the centroid is at the origin, then by definition, $\bar{x} = M_y / A = 0$ and $\bar{y} = M_x / A = 0$. Therefore, the first moments about both axes are zero. A common cause is various types of symmetry, but zero moments are the direct mathematical consequence. The region does not have to be symmetric; the moments can cancel in other ways. Option A is a possible but not necessary condition. Option D is a specific shape, not a general conclusion.

📖 Explanation: The region is bounded between $y = x^2$ (top) and $y = x^3$ (bottom) for $0 \le x \le 1$. The formula for the x-coordinate of the centroid is $\bar{x} = \frac{1}{A} \int_a^b x [f(x) - g(x)] \, dx$. Here, $f(x) = x^2$ and $g(x) = x^3$, so the integrand is $x(x^2 - x^3)$. The student's step is a correct Medium of the formula. Option A incorrectly uses the sum of the functions. Option B uses $x^2$, which is incorrect for $\bar{x}$. Option D is wrong because the curves intersect at $x=0$ and $x=1$.

📖 Explanation: Pappus's theorem can be used in reverse here. The volume $V$ generated by revolving the region about the y-axis is $V = 2\pi \bar{x} A$, where $\bar{x}$ is the distance from the centroid to the axis. We are given $V = 8\pi/3$ and $A = 4/3$. Solving for $\bar{x}$: $8\pi/3 = 2\pi \bar{x} (4/3)$. Simplifying, $8\pi/3 = (8\pi/3) \bar{x}$. Therefore, $\bar{x} = 1$. This problem combines the concepts of Pappus's theorem and centroid calculation. Option A is the area of the region. Options B and C are other boundary values.

📖 Explanation: The x-coordinate of the centroid is $\bar{x} = \frac{\int_1^8 x (x^{1/3}) \, dx}{\int_1^8 x^{1/3} \, dx} = \frac{\int_1^8 x^{4/3} \, dx}{\int_1^8 x^{1/3} \, dx}$. The numerator is $\left[ \frac{3}{7} x^{7/3} \right]_1^8 = \frac{3}{7}(128 - 1) = \frac{381}{7}$. The denominator is $\left[ \frac{3}{4} x^{4/3} \right]_1^8 = \frac{3}{4}(16 - 1) = \frac{45}{4}$. The centroid is $\bar{x} = \frac{381/7}{45/4} = \frac{381}{7} \times \frac{4}{45} = \frac{1524}{315} \approx 4.84$. This is closest to 4.5. Option A is lower, C is near the upper bound, and D is outside the interval.

📖 Explanation: The square has its centroid at its center. The semicircle's centroid is located at a distance $4r/(3\pi)$ from its base. If the semicircle is attached to the top of the square, its centroid is above the square's centroid. The composite centroid will be a weighted average of these two centroids. Since the square and semicircle have different areas and their centroids are at different points, the composite centroid will be somewhere between them. It will be closer to the larger area's centroid. The square's area is 4, and the semicircle's area is $\pi/2 \approx 1.57$, so the square has the larger area. Therefore, the composite centroid will be closer to the square's centroid (which is at its center) than to the semicircle's centroid (which is above the top of the square).

📖 Explanation: For a region with a hole or a cutout (a composite region with a missing part), the centroid can be found using the concept of a weighted average with a negative area. The region's centroid $(\bar{x}, \bar{y})$ can be found from $\bar{x} = \frac{A_{\text{large}} \bar{x}_{\text{large}} - A_{\text{small}} \bar{x}_{\text{small}}}{A_{\text{large}} - A_{\text{small}}}$ and similarly for $\bar{y}$. Option A is unnecessarily complex. Option C is incomplete. Pappus's theorem is for volumes of revolution, not for finding centroids.

📖 Explanation: The area is $A = \int_0^\pi \sin x \, dx = 2$. The x-coordinate of the centroid is $\bar{x} = \frac{1}{A} \int_0^\pi x \sin x \, dx$. Using integration by parts, $\int_0^\pi x \sin x \, dx = [ -x\cos x + \sin x ]_0^\pi = \pi$. So $\bar{x} = \pi/2$. The y-coordinate is $\bar{y} = \frac{1}{A} \int_0^\pi \frac{1}{2} \sin^2 x \, dx = \frac{1}{4} \int_0^\pi \sin^2 x \, dx = \frac{1}{4} \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_0^\pi = \frac{\pi}{8}$. The centroid is $(\pi/2, \pi/8)$. Option B uses $1/2$ incorrectly. Option C uses 1. Option D is the centroid of a semicircular region, not this sine wave region.

📖 Explanation: When integrating with respect to y, the x-coordinate of the centroid is $\bar{x} = \frac{1}{A} \int_c^d x_{\text{right}} (x_{\text{right}} - x_{\text{left}}) \, dy$. The horizontal cross-section's center is at x-coordinate $\frac{x_{\text{right}} + x_{\text{left}}}{2}$, but the moment about the y-axis is $x \cdot \text{length}$. Here, $x_{\text{right}} = y+2$ and $x_{\text{left}} = y^2$, so the moment integral is $\int_c^d (y+2) \cdot ((y+2) - y^2) \, dy$. The student's expression $y(y+2 - y^2)$ is a common error where they incorrectly use 'y' as the x-coordinate for the moment, which is wrong. The correct integrand is $(y+2)(y+2 - y^2)$, not $y(y+2 - y^2)$. Options B and C are formulas for $\bar{y}$. Option D is the area integrand.

📖 Explanation: The distance from the centroid $(\bar{x} = 2)$ to the axis of revolution $x = -1$ is $2 - (-1) = 3$. The distance traveled by the centroid is the circumference of the circle with this radius, which is $2\pi \times 3 = 6\pi$. The volume is $V = A \times \text{distance traveled by centroid} = 4 \times 6\pi = 24\pi$. Option B is a result of doubling the radius incorrectly. Option C is the result of using a radius of 3 but forgetting the factor of 2. Option D is the result of using an incorrect area or distance.

📖 Explanation: The formula $\bar{y} = \frac{1}{2A} \int_a^b (f(x))^2 \, dx$ is derived from the method of vertical strips. For a vertical strip at a given x, its height is f(x). The centroid of this strip is at its midpoint, which is at $y = f(x)/2$. The moment of this strip about the x-axis is $y \cdot \text{area} = (f(x)/2) \cdot (f(x) \, dx) = \frac{1}{2} (f(x))^2 \, dx$. Option B would be used for integrating with respect to y. Option C is used in cylindrical shells for volumes, not centroids. Option D is used in disk methods for volumes.

📖 Explanation: The region is bounded between the two parabolas $y = 1 - x^2$ (top) and $y = x^2 - 1$ (bottom). Both functions are even, meaning they are symmetric about the y-axis. Therefore, $\bar{x} = 0$. The region is also symmetric about the x-axis because the top curve is the negative of the bottom curve. If a region is symmetric about the x-axis, its centroid's y-coordinate is zero. Therefore, the centroid is at $(0,0)$. This problem combines two symmetries to find the centroid without any integration. Options B, C, and D are non-zero y-values that would be the centroids of other shapes, like a semicircle or a region above the x-axis.

📖 Explanation: For a triangle, the centroid is located at the intersection of its medians, which are the lines connecting a vertex to the midpoint of the opposite side. The centroid divides each median in a 2:1 ratio, with the longer segment being from the vertex. Therefore, the distance from the base (which is a side) to the centroid is one-third of the height from the opposite vertex. This can be mathematically proven by averaging the vertices' coordinates. Option B is incorrect because the moment about the base is not zero unless the base is the centroid axis. Option C describes the property of a homogeneous lamina, which is true but not the reason. Option D is circular reasoning.

📖 Explanation: Pappus's theorem states that the volume V of the solid of revolution is $V = 2\pi \bar{x} A$, where $\bar{x}$ is the distance from the centroid to the axis of revolution. Here, the axis is the y-axis, so $\bar{x} = 4$. We are given $V = 80\pi$. Solving for A: $80\pi = 2\pi (4) A = 8\pi A$. Thus, $A = 80\pi / 8\pi = 10$. Option A is a result of not multiplying by 2 in the formula. Option C is a doubling error. Option D is a result of using $\bar{x} = 2$.

📖 Explanation: The area is $A = \int_0^1 e^x \, dx = e - 1$. The y-coordinate is $\bar{y} = \frac{1}{A} \int_0^1 \frac{1}{2} (e^x)^2 \, dx = \frac{1}{2(e-1)} \int_0^1 e^{2x} \, dx = \frac{1}{2(e-1)} \left[ \frac{e^{2x}}{2} \right]_0^1 = \frac{1}{4(e-1)} (e^2 - 1) = \frac{(e-1)(e+1)}{4(e-1)} = \frac{e+1}{4}$. This is a direct calculation. However, wait. Let me re-evaluate. The formula is $\bar{y} = \frac{1}{2A} \int (f(x))^2 dx$. So $\bar{y} = \frac{1}{2(e-1)} \left( \frac{e^2 - 1}{2} \right) = \frac{e^2 - 1}{4(e-1)} = \frac{(e-1)(e+1)}{4(e-1)} = \frac{e+1}{4}$. I need to adjust my options. The correct formula gives $\bar{y} = \frac{e+1}{4}$, which is not listed. I'll correct option A to $\frac{e+1}{4(e-1)}$. Let me recalculate: The integral of $e^{2x}$ from 0 to 1 is $(e^2 - 1)/2$. So $\bar{y} = \frac{1}{2(e-1)} \cdot \frac{e^2 - 1}{2} = \frac{e^2 - 1}{4(e-1)} = \frac{(e-1)(e+1)}{4(e-1)} = \frac{e+1}{4}$. So the correct answer is $\frac{e+1}{4}$. I will set option A as $\frac{e+1}{4}$.