Statics Quiz: Centroid Via Integration

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Centroid Via IntegrationQuestion 1 of 20

A region is bounded by $y = x^3$ , $x = 2$ , and $y = 0$ . When computing the centroid, a student calculates the area as $A = \int_0^2 x^3 dx = 4$ and the first moment about the y-axis as $M_y = \int_0^2 x \cdot x^3 dx = \frac{32}{5}$ . What error, if any, was made?

No error was made; both calculations are correct for finding the centroid of this region

The area calculation is wrong; it should be

\int_0^8 \frac{\sqrt[3]{y}}{3} dy

to account for horizontal strips

The moment calculation is wrong; it should be

\int_0^2 \frac{x^2}{2} \cdot x^3 dx

for the centroid of each strip

Both calculations assume vertical strips, but the area element

dA

should be

x^3 dx

, not the strip height

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Statics Quiz

Statics Quiz: Centroid Via Integration

Practice Centroid Via Integration in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Centroid Via Integration, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

No error was made; both calculations are correct for finding the centroid of this region (correct answer)
The area calculation is wrong; it should be $\int_0^8 \frac{\sqrt[3]{y}}{3} dy$ to account for horizontal strips
The moment calculation is wrong; it should be $\int_0^2 \frac{x^2}{2} \cdot x^3 dx$ for the centroid of each strip
Both calculations assume vertical strips, but the area element $dA$ should be $x^3 dx$ , not the strip height

Explanation: Both calculations are correct. For the region bounded by

y = x^3

x = 2

, and

y = 0

, using vertical strips: Area

A = \int_0^2 x^3 dx = 4

and

M_y = \int_0^2 x \cdot x^3 dx = \frac{32}{5}

. Choice B suggests an unnecessary alternative method. Choice C incorrectly applies the centroid formula for the strip. Choice D misunderstands that

x^3 dx

is indeed the correct area element for vertical strips of height

x^3

Question 2

For a semicircular region of radius $R$ centered at the origin, the y-coordinate of the centroid is calculated using horizontal strips. Each strip at height $y$ has width $2\sqrt{R^2-y^2}$ . Which expression gives the correct moment about the x-axis?

$\int_0^R y \cdot 2\sqrt{R^2-y^2} \, dy$ (correct answer)
$\int_{-R}^R y \cdot 2\sqrt{R^2-y^2} \, dy$
$\int_0^R y^2 \cdot \sqrt{R^2-y^2} \, dy$
$\int_0^R \sqrt{R^2-y^2} \, dy$
$\int_0^R y \cdot \sqrt{R^2-y^2} \, dy$

Explanation: When calculating centroids using integration, you need to understand that the moment about an axis equals the distance from that axis multiplied by the differential area element. For moments about the x-axis, this distance is the y-coordinate of each strip. The correct approach uses horizontal strips at height

y

from

y = 0

y = R

(covering the upper semicircle). Each strip has width

2\sqrt{R^2-y^2}

and differential height

dy

, giving area

dA = 2\sqrt{R^2-y^2} \, dy

. The moment of this strip about the x-axis is

y \cdot dA = y \cdot 2\sqrt{R^2-y^2} \, dy

. Integrating from 0 to R gives the total moment, making A correct. B is wrong because it integrates from

-R

R

, but the semicircle only exists from

y = 0

y = R

. The lower half-plane contains no area for a semicircle above the x-axis. C uses

y^2

instead of

y

as the moment arm and has an incorrect coefficient. The moment arm for x-axis moments is simply

y

, not

y^2

, and it's missing the factor of 2 in the strip width. D completely omits the moment arm

y

. This integral would give you the area of the semicircle, not the moment about the x-axis. Study tip: Always identify your moment arm first (distance from the axis of interest), then multiply by the differential area element. For x-axis moments, the moment arm is always the y-coordinate of the centroid of each differential element.

Question 3

A region is bounded by $y = x^2$ and $y = 4$ . When calculating the centroid using horizontal strips, each strip at height $y$ extends from $x = -\sqrt{y}$ to $x = \sqrt{y}$ . What is the width function $w(y)$ for integration?

$w(y) = 2\sqrt{y}$ (correct answer)
$w(y) = \sqrt{y}$
$w(y) = y$
$w(y) = 2y$
$w(y) = 4 - y$

Explanation: When calculating centroids using horizontal strips, you need to determine how wide each horizontal slice is at a given height. This width function becomes crucial for setting up your integration. For this region bounded by

y = x^2

and

y = 4

, picture horizontal strips cutting through the area. At any height

y

, the parabola

y = x^2

intersects the horizontal line at two points. Solving

y = x^2

for

x

gives

x = \pm\sqrt{y}

, meaning the strip extends from the left boundary at

x = -\sqrt{y}

to the right boundary at

x = \sqrt{y}

. The width of each strip is simply the distance between these boundaries:

w(y) = \sqrt{y} - (-\sqrt{y}) = \sqrt{y} + \sqrt{y} = 2\sqrt{y}

. This confirms answer A is correct. Looking at the wrong answers: B gives

w(y) = \sqrt{y}

, which represents only half the width—perhaps from confusing the distance from the y-axis to one boundary with the total width. C suggests

w(y) = y

, which might come from incorrectly thinking the width equals the height coordinate. D proposes

w(y) = 2y

, which could result from mistakenly using

y

instead of

\sqrt{y}

for the boundary positions. Remember: when using horizontal strips, always calculate width as "right boundary minus left boundary." Sketch the region and identify where horizontal lines intersect the boundaries—this visual approach prevents algebraic errors and helps you set up the correct width function.

Question 4

For the region bounded by $y = \sin x$ and the x-axis from $x = 0$ to $x = \pi$ , the centroid calculation requires integration. Which integral represents the total area of the region?

$\int_0^\pi \sin x \, dx$ (correct answer)
$\int_0^\pi x \sin x \, dx$
$\int_0^\pi \sin^2 x \, dx$
$\int_0^\pi \frac{\sin^2 x}{2} \, dx$
$\int_0^\pi |\sin x| \, dx$

Explanation: When calculating centroids in statics, you need to find both the total area of the region and the first moments about the coordinate axes. The question specifically asks for the integral representing the total area. For any region bounded by a curve

y = f(x)

and the x-axis between

x = a

and

x = b

, the area is simply

A = \int_a^b f(x) \, dx

. Since our region is bounded by

y = \sin x

and the x-axis from

x = 0

x = \pi

, the area integral is

\int_0^\pi \sin x \, dx

, which is answer A. Let's examine why the other options are incorrect. Option B,

\int_0^\pi x \sin x \, dx

, represents the first moment about the y-axis (

M_y

), which you'd use to find the x-coordinate of the centroid, not the total area. The extra factor of

x

makes this a moment calculation, not an area calculation. Option C,

\int_0^\pi \sin^2 x \, dx

, might appear if you mistakenly think you need to square the function. This would be relevant if you were using the disk method for volumes of revolution, but for simple area under a curve, you don't square the function. Option D,

\int_0^\pi \frac{\sin^2 x}{2} \, dx

, combines the error from option C with an incorrect factor of

\frac{1}{2}

. This resembles formulas from centroid calculations involving

y

-coordinates, but it's not the area integral. Remember: for centroid problems, always start by identifying what each integral represents. Area is always

\int f(x) \, dx

— no extra factors, no squaring the function.

Question 5

A region is defined by $0 \leq x \leq 2$ and $0 \leq y \leq x^2$ . When calculating $\bar{x}$ using double integration in rectangular coordinates, which expression is correct?

$\bar{x} = \frac{\int_0^2 \int_0^{x^2} x \, dy \, dx}{\int_0^2 \int_0^{x^2} 1 \, dy \, dx}$ (correct answer)
$\bar{x} = \frac{\int_0^2 \int_0^{x^2} x^2 \, dy \, dx}{\int_0^2 \int_0^{x^2} x \, dy \, dx}$
$\bar{x} = \frac{\int_0^4 \int_0^{\sqrt{y}} x \, dx \, dy}{\int_0^4 \int_0^{\sqrt{y}} 1 \, dx \, dy}$
$\bar{x} = \frac{\int_0^2 x \cdot x^2 \, dx}{\int_0^2 x^2 \, dx}$
$\bar{x} = \frac{\int_0^2 \int_0^{x^2} y \, dy \, dx}{\int_0^2 \int_0^{x^2} 1 \, dy \, dx}$

Explanation: When finding the centroid of a region using double integration, you need to apply the fundamental formula:

\bar{x} = \frac{\text{first moment about y-axis}}{\text{total area}}

. The first moment about the y-axis requires integrating

x

times the differential area element, while the denominator is simply the area of the region. For the region defined by

0 \leq x \leq 2

and

0 \leq y \leq x^2

, the correct setup integrates over

x

first (from 0 to 2), then

y

(from 0 to

x^2

for each

x

-value). The first moment is

\int_0^2 \int_0^{x^2} x \, dy \, dx

(integrating

x

over the region), and the area is

\int_0^2 \int_0^{x^2} 1 \, dy \, dx

. Answer A correctly applies this formula. Answer B incorrectly uses

x^2

in the numerator instead of

x

—this would give you a different moment calculation, not the x-coordinate of the centroid. Answer C attempts to change the order of integration to

dy \, dx

, but the limits are wrong; when

y

ranges from 0 to 4 and

x

from 0 to

\sqrt{y}

, this describes a different region entirely. Answer D tries to reduce the problem to single integration, but this approach doesn't properly account for the two-dimensional nature of the region. Study tip: Always start centroid problems by clearly identifying your region and sketching it if possible. Then systematically apply

\bar{x} = \frac{\int\int x \, dA}{\int\int dA}

, ensuring your integration limits match the region's boundaries exactly.

Question 6

A triangular region has vertices at $(0,0)$ , $(4,0)$ , and $(2,3)$ . The line from $(0,0)$ to $(2,3)$ has equation $y = \frac{3x}{2}$ , and the line from $(4,0)$ to $(2,3)$ has equation $y = -\frac{3x}{2} + 6$ . Using vertical strips, what is the x-coordinate of the centroid?

$\bar{x} = \frac{3}{2}$
$\bar{x} = \frac{4}{3}$
$\bar{x} = \frac{5}{3}$
$\bar{x} = 2$ (correct answer)
$\bar{x} = \frac{8}{3}$

Explanation: When finding centroids using vertical strips, you're essentially calculating the weighted average x-coordinate of the region. For any area, the x-coordinate of the centroid is

\bar{x} = \frac{\int x \cdot dA}{\int dA}

where the integrals are taken over the entire region. For this triangle with vertices at (0,0), (4,0), and (2,3), you need to integrate from x = 0 to x = 4. At any x-coordinate, the height of the vertical strip extends from y = 0 (the x-axis) to the upper boundary of the triangle. From x = 0 to x = 2, the upper boundary is

y = \frac{3x}{2}

, so the strip height is

\frac{3x}{2}

. From x = 2 to x = 4, the upper boundary is

y = -\frac{3x}{2} + 6

, so the strip height is

-\frac{3x}{2} + 6

. The calculation becomes:

\bar{x} = \frac{\int_0^2 x \cdot \frac{3x}{2} dx + \int_2^4 x \cdot (-\frac{3x}{2} + 6) dx}{\int_0^2 \frac{3x}{2} dx + \int_2^4 (-\frac{3x}{2} + 6) dx}

Working through the integrals gives

\bar{x} = 2

. Choice A (

\frac{3}{2}

) likely comes from only considering the left half of the triangle. Choice B (

\frac{4}{3}

) might result from incorrectly weighting the integration. Choice C (

\frac{5}{3}

) could stem from computational errors in the integration process. Remember: centroids require careful attention to integration limits and the correct height function for each interval. Always verify your boundary equations match the given geometry.

Question 7

A region is bounded by $x = y^2$ and $x = 4$ . Using horizontal strips for centroid calculation, each strip at height $y$ extends from $x = y^2$ to $x = 4$ . What is the correct setup for finding the x-coordinate of the centroid?

$\bar{x} = \frac{\int_{-2}^2 \frac{y^2 + 4}{2} \cdot (4 - y^2) \, dy}{\int_{-2}^2 (4 - y^2) \, dy}$ (correct answer)
$\bar{x} = \frac{\int_{-2}^2 y \cdot (4 - y^2) \, dy}{\int_{-2}^2 (4 - y^2) \, dy}$
$\bar{x} = \frac{\int_0^4 x \cdot 2\sqrt{x} \, dx}{\int_0^4 2\sqrt{x} \, dx}$
$\bar{x} = \frac{\int_{-2}^2 (4 - y^2) \, dy}{\int_{-2}^2 dy}$
$\bar{x} = \frac{\int_{-2}^2 \frac{4 + y^2}{2} \, dy}{\int_{-2}^2 \frac{4 - y^2}{2} \, dy}$

Explanation: When finding centroids using horizontal strips, you need to understand how each strip contributes to the total moment about the y-axis. Each horizontal strip at height

y

acts like a thin rectangle with its own centroid at the midpoint of its width. For this region bounded by

x = y^2

and

x = 4

, each horizontal strip extends from

x = y^2

x = 4

, giving it a width of

(4 - y^2)

. The centroid of each individual strip is located at the midpoint:

\frac{y^2 + 4}{2}

. The moment contribution of each strip is this centroid location times the strip's area (width times differential height

dy

). Answer A correctly sets up

\bar{x} = \frac{\int_{-2}^2 \frac{y^2 + 4}{2} \cdot (4 - y^2) \, dy}{\int_{-2}^2 (4 - y^2) \, dy}

, where the numerator represents the total moment (centroid of each strip times its area) and the denominator is the total area. Answer B incorrectly uses

y

instead of the x-coordinate of each strip's centroid. This confuses the height coordinate with the centroid location. Answer C uses vertical strips instead of the specified horizontal strips—while mathematically equivalent, it doesn't follow the problem's requirement. Answer D uses the wrong numerator entirely, representing area rather than moment. Remember: for horizontal strips, always identify the centroid of each individual strip first, then multiply by the strip's area to get the moment contribution. The x-coordinate of a horizontal strip's centroid is always the midpoint of its width.

Question 8

Consider a region in the first quadrant bounded by the curve $y = \ln x$ , the x-axis, and the line $x = e$ . For centroid calculation using vertical strips, which integral represents the first moment about the x-axis?

$\int_1^e \frac{\ln x}{2} \cdot \ln x \, dx$
$\int_1^e x \cdot \ln x \, dx$
$\int_1^e \frac{(\ln x)^2}{2} \, dx$ (correct answer)
$\int_0^1 e^y \cdot y \, dy$
$\int_1^e \ln x \, dx$

Explanation: When calculating centroids using the method of integration, you need to understand that the first moment about the x-axis involves the distance from the x-axis to the centroid of each differential element. For vertical strips, each strip has height

\ln x

and width

dx

, so its area is

\ln x \, dx

. The key insight is that the centroid of each vertical strip is located at half its height, which is

\frac{\ln x}{2}

above the x-axis. The first moment about the x-axis is the product of this distance and the area of the strip:

\frac{\ln x}{2} \cdot \ln x \, dx = \frac{(\ln x)^2}{2} \, dx

. Integrating from

x = 1

x = e

gives the total first moment. Option A (

\int_1^e \frac{\ln x}{2} \cdot \ln x \, dx

) shows the correct integrand but fails to simplify

\frac{\ln x}{2} \cdot \ln x

\frac{(\ln x)^2}{2}

. While mathematically equivalent to C, it's not in standard form. Option B (

\int_1^e x \cdot \ln x \, dx

) incorrectly uses

x

as the moment arm instead of

\frac{\ln x}{2}

. This would be relevant for moments about the y-axis, not the x-axis. Option D (

\int_0^1 e^y \cdot y \, dy

) represents horizontal strips (using

x = e^y

), but the limits are wrong since

y

ranges from 0 to 1, and the moment arm setup is incorrect. Study tip: Always identify your moment arm correctly—for vertical strips finding

\bar{y}

, use half the strip height as your distance from the x-axis.

Question 9

A parabolic region is bounded by $y = 4 - x^2$ and the x-axis. Using vertical strips, the x-coordinate of the centroid is $\bar{x} = 0$ by symmetry. What is the y-coordinate of the centroid?

$\bar{y} = \frac{8}{5}$
$\bar{y} = \frac{12}{5}$
$\bar{y} = \frac{16}{5}$
$\bar{y} = \frac{32}{15}$ (correct answer)
$\bar{y} = \frac{64}{15}$

Explanation: When finding centroids of regions using integration, you need to calculate both the area and the first moments about the coordinate axes. For the y-coordinate of the centroid, you use the formula

\bar{y} = \frac{M_x}{A}

, where

M_x

is the first moment about the x-axis and

A

is the total area. First, find the area. The parabola

y = 4 - x^2

intersects the x-axis when

4 - x^2 = 0

, so

x = \pm 2

. The area is:

A = \int_{-2}^{2} (4 - x^2) dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \frac{32}{3}

For the first moment about the x-axis using vertical strips, the y-coordinate of each strip's centroid is at the midheight:

\frac{y}{2} = \frac{4-x^2}{2}

. Therefore:

M_x = \int_{-2}^{2} \frac{4-x^2}{2} \cdot (4-x^2) dx = \frac{1}{2}\int_{-2}^{2} (4-x^2)^2 dx

Expanding:

(4-x^2)^2 = 16 - 8x^2 + x^4

M_x = \frac{1}{2}\left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_{-2}^{2} = \frac{256}{15}

Therefore:

\bar{y} = \frac{256/15}{32/3} = \frac{256}{15} \cdot \frac{3}{32} = \frac{32}{15}

Answer D is correct. Answer A (

\frac{8}{5}

) likely comes from using the wrong centroid height for strips. Answer B (

\frac{12}{5}

) might result from calculation errors in the integration. Answer C (

\frac{16}{5}

) could come from forgetting the factor of

\frac{1}{2}

in the moment calculation. Remember: for vertical strips, the centroid of each strip is at its geometric center, which is half the strip's height.

Question 10

A triangular region with vertices at $(0,0)$ , $(a,0)$ , and $(0,b)$ has its centroid calculated using vertical strips. If the strip at position $x$ has height $h(x)$ , what is the correct expression for the y-coordinate of the centroid?

$\bar{y} = \frac{\int_0^a \frac{h(x)}{2} \cdot h(x) \, dx}{\int_0^a h(x) \, dx}$ (correct answer)
$\bar{y} = \frac{\int_0^a h(x) \cdot x \, dx}{\int_0^a h(x) \, dx}$
$\bar{y} = \frac{\int_0^a \frac{h(x)}{2} \cdot dx}{\int_0^a dx}$
$\bar{y} = \frac{\int_0^b y \cdot w(y) \, dy}{\int_0^b w(y) \, dy}$
$\bar{y} = \frac{\int_0^a h(x)^2 \, dx}{2\int_0^a h(x) \, dx}$

Explanation: When finding centroids using integration, you need to remember that the centroid represents the "balance point" of a region. For vertical strips, each strip has its own local centroid at its geometric center, and you're finding the weighted average of all these local centroids. For a vertical strip at position

x

with height

h(x)

, the strip's area is

h(x) \cdot dx

and its local centroid is at

y = \frac{h(x)}{2}

(halfway up the strip). The y-coordinate of the overall centroid follows the fundamental formula:

\bar{y} = \frac{\text{sum of (area × local y-centroid)}}{\text{total area}}

. Answer A correctly applies this principle:

\frac{\int_0^a \frac{h(x)}{2} \cdot h(x) \, dx}{\int_0^a h(x) \, dx}

. The numerator multiplies each strip's area

h(x) \, dx

by its local y-centroid

\frac{h(x)}{2}

, then integrates over all strips. Answer B calculates the x-coordinate of the centroid instead, using

x

as the moment arm rather than

\frac{h(x)}{2}

. Answer C forgets to weight by area—it just averages the local centroids

\frac{h(x)}{2}

without considering that wider strips should have more influence. Answer D uses horizontal strips (integrating over

y

with width

w(y)

) rather than the specified vertical strips. Study tip: Always identify your strip orientation first, then remember that centroids require area-weighting. Each strip contributes to the centroid proportionally to its size, not equally.

Question 11

For a region bounded by $y = x^3$ and $y = x$ in the first quadrant, the curves intersect at $(0,0)$ and $(1,1)$ . When using vertical strips to find $\bar{y}$ , which expression correctly represents the y-coordinate of each strip's centroid?

$\frac{x + x^3}{2}$ (correct answer)
$\frac{x - x^3}{2}$
$\frac{(x + x^3)}{2}$
$x^2$
$\frac{1 + x^2}{2}$

Explanation: When finding centroids using the method of integration, you need to identify the centroid of each differential element. For vertical strips in this region, each strip extends from the lower curve

y = x^3

to the upper curve

y = x

, and the centroid of each rectangular strip lies at its geometric center. The y-coordinate of a strip's centroid is the average of its top and bottom y-coordinates. Since the strip extends from

y = x^3

(bottom) to

y = x

(top), the centroid's y-coordinate is

\frac{x^3 + x}{2} = \frac{x + x^3}{2}

. This is simply the midpoint formula applied vertically. Looking at the wrong answers: Choice B gives

\frac{x - x^3}{2}

, which represents half the height of the strip, not the centroid location. This is a common error where students confuse the strip's dimension with its centroid coordinate. Choice C,

\frac{(x + x^3)}{2}

, is mathematically identical to choice A but written with unnecessary parentheses - if this were listed as a separate option, it would also be correct. Choice D gives

x^2

, which has no geometric relationship to the centroid of a strip bounded by

x

and

x^3

. Study tip: Remember that for any rectangular strip, the centroid coordinates are always the averages of the boundaries. For vertical strips,

\bar{y}_{strip} = \frac{y_{top} + y_{bottom}}{2}

. Don't confuse this with the strip's height, which would be

y_{top} - y_{bottom}

. The correct answer is A.

Question 12

A quarter-circle region in the first quadrant has radius 3. Using the integration method with vertical strips, the setup for finding $\bar{y}$ includes the integral $\int_0^3 \frac{\sqrt{9-x^2}}{2} \cdot \sqrt{9-x^2} \, dx$ . What does the factor $\frac{\sqrt{9-x^2}}{2}$ represent?

The y-coordinate of the centroid of the vertical strip at position x (correct answer)
The total height of the vertical strip at position x
The area of the vertical strip at position x
The x-coordinate of the centroid of the vertical strip at position x
The moment arm for calculating the first moment about the y-axis

Explanation: When finding centroids using integration with vertical strips, you're essentially breaking down a region into thin vertical rectangles and finding where each strip's "center of mass" is located. The key insight is that each strip has its own centroid, and you need to weight these individual centroids by the strip's area. For a quarter-circle with radius 3, a vertical strip at position

x

extends from

y = 0

y = \sqrt{9-x^2}

(the circle's equation). This strip has height

\sqrt{9-x^2}

, and its centroid is located at the midpoint of its height, which is

\frac{\sqrt{9-x^2}}{2}

above the x-axis. This is exactly what the factor

\frac{\sqrt{9-x^2}}{2}

represents - the y-coordinate where the strip's centroid is located. Looking at the wrong answers: B)

\sqrt{9-x^2}

would be the total height, not the factor shown. C) The area of a vertical strip would be

\sqrt{9-x^2} \cdot dx

, involving the differential width. D) The x-coordinate of the strip's centroid would simply be

x

, the strip's horizontal position. The integral structure

\int (\text{strip centroid location}) \times (\text{strip area}) \, dx

is the standard pattern for finding overall centroids. Remember: when using vertical strips, the y-coordinate of each strip's centroid is always at half its height, while the x-coordinate is just the strip's horizontal position.

Question 13

The centroid of a semicircular region of radius $R$ is located at $\bar{y} = \frac{4R}{3\pi}$ from the diameter. If this result is derived using integration with horizontal strips, which integral identity must be used in the calculation?

$\int_0^R \sqrt{R^2 - y^2} \, dy = \frac{\pi R^2}{4}$
$\int_0^R y\sqrt{R^2 - y^2} \, dy = \frac{R^3}{3}$ (correct answer)
$\int_0^R y^2\sqrt{R^2 - y^2} \, dy = \frac{\pi R^4}{16}$
$\int_0^{\pi/2} R^2 \cos^2\theta \, d\theta = \frac{\pi R^2}{8}$
$\int_0^{\pi/2} R^3 \sin\theta \cos\theta \, d\theta = \frac{R^3}{2}$

Explanation: When finding centroids using integration, you need to set up the calculation based on your chosen strip orientation. For a semicircle with horizontal strips, each strip at height

y

has width

2\sqrt{R^2 - y^2}

(extending from

-\sqrt{R^2 - y^2}

+\sqrt{R^2 - y^2}

). The centroid formula requires calculating the first moment about the x-axis:

\bar{y} = \frac{\int y \, dA}{\int dA}

. For horizontal strips,

dA = 2\sqrt{R^2 - y^2} \, dy

, so the first moment becomes

\int_0^R y \cdot 2\sqrt{R^2 - y^2} \, dy = 2\int_0^R y\sqrt{R^2 - y^2} \, dy

. This is exactly twice the integral in option B. Option B is correct because it provides the essential integral

\int_0^R y\sqrt{R^2 - y^2} \, dy = \frac{R^3}{3}

needed to evaluate the first moment. With this result, the first moment equals

2 \cdot \frac{R^3}{3} = \frac{2R^3}{3}

, and dividing by the semicircle's area

\frac{\pi R^2}{2}

gives

\bar{y} = \frac{4R}{3\pi}

. Option A gives the area calculation, not the first moment. Option C involves

y^2

, which would be needed for second moments or moments of inertia, not centroids. Option D uses polar coordinates with

\cos^2\theta

, which doesn't match the horizontal strip approach described. Study tip: When setting up centroid problems, always identify what your differential area element looks like first—this determines which integrals you'll need to evaluate the required moments.

Question 14

A parabolic region is bounded by $y = 4x^2$ and $y = 16$ in the first quadrant. To find the x-coordinate of the centroid using integration, which expression correctly represents the first moment about the y-axis?

$\int_0^2 x(16 - 4x^2) dx$ (correct answer)
$\int_0^{16} \frac{x}{2}\sqrt{\frac{y}{4}} dy$
$\int_0^2 \frac{x^2}{2}(16 - 4x^2) dx$
$\int_0^2 x^2(16 - 4x^2) dx$

Explanation: The first moment about the y-axis is

M_y = \int x \cdot dA

. For a vertical strip of width

dx

, the area element is

dA = (16 - 4x^2)dx

, so

M_y = \int_0^2 x(16 - 4x^2) dx

. Choice B uses horizontal strips incorrectly. Choice C includes an extra factor of

x/2

. Choice D uses

x^2

instead of

x

for the moment arm.

Question 15

A region in polar coordinates is defined by $0 \leq r \leq 2\cos\theta$ and $0 \leq \theta \leq \pi/2$ . This represents a semicircle. The x-coordinate of the centroid requires the integral $\int_0^{\pi/2} \int_0^{2\cos\theta} r\cos\theta \cdot r \, dr \, d\theta$ . After integrating with respect to $r$ , what expression remains?

$\int_0^{\pi/2} \frac{8\cos^4\theta}{3} \, d\theta$ (correct answer)
$\int_0^{\pi/2} \frac{16\cos^3\theta}{3} \, d\theta$
$\int_0^{\pi/2} 4\cos^3\theta \, d\theta$
$\int_0^{\pi/2} 8\cos^4\theta \, d\theta$
$\int_0^{\pi/2} 2\cos^2\theta \, d\theta$

Explanation: When calculating centroids in polar coordinates, you're essentially finding the "balance point" of a region by integrating position-weighted area elements. The integral

\int_0^{\pi/2} \int_0^{2\cos\theta} r\cos\theta \cdot r \, dr \, d\theta

represents the first moment about the y-axis, where

r\cos\theta

is the x-coordinate and

r \, dr \, d\theta

is the area element. To evaluate this double integral, you integrate with respect to

r

first. The integrand is

r^2\cos\theta

, so:

\int_0^{2\cos\theta} r^2\cos\theta \, dr = \cos\theta \int_0^{2\cos\theta} r^2 \, dr = \cos\theta \left[\frac{r^3}{3}\right]_0^{2\cos\theta}

Substituting the limits:

\cos\theta \cdot \frac{(2\cos\theta)^3}{3} = \cos\theta \cdot \frac{8\cos^3\theta}{3} = \frac{8\cos^4\theta}{3}

This gives you

\int_0^{\pi/2} \frac{8\cos^4\theta}{3} \, d\theta

, which is answer A. Answer B has

\cos^3\theta

instead of

\cos^4\theta

, suggesting you forgot to multiply by the

\cos\theta

factor from the x-coordinate. Answer C is missing the factor of

\frac{8}{3}

, indicating an error in evaluating

\frac{(2\cos\theta)^3}{3}

. Answer D has the correct trigonometric part but omits the

\frac{1}{3}

from the power rule integration. Remember: when integrating

r^n

in polar coordinates, don't forget to apply the power rule correctly and carry through all coefficient factors from the original integrand.

Question 16

Consider a region bounded by $y = e^x$ , $x = 0$ , $x = 1$ , and $y = 0$ . If horizontal strips are used instead of vertical strips to find the centroid, what are the correct limits of integration for the y-variable?

From $y = 0$ to $y = e$ (correct answer)
From $y = 1$ to $y = e$
From $y = 0$ to $y = 1$
From $y = e^0$ to $y = e^1$
From $y = \ln(0)$ to $y = \ln(1)$

Explanation: When finding centroids using horizontal strips, you need to identify the range of y-values that the region spans. This requires examining where the region begins and ends vertically. The region is bounded by

y = e^x

x = 0

x = 1

, and

y = 0

. To find the y-limits for horizontal strips, determine the minimum and maximum y-values in the region. The bottom boundary is

y = 0

. For the top boundary, since

y = e^x

and

x

ranges from 0 to 1, you need the maximum value of

e^x

over this interval. Since

e^x

is an increasing function, its maximum occurs at

x = 1

, giving

y = e^1 = e

. Therefore, y ranges from 0 to

e

. Answer A (

y = 0

y = e

) correctly captures this full vertical span of the region. Answer B (

y = 1

y = e

) incorrectly starts at

y = 1

instead of

y = 0

, missing the lower portion of the region since the bottom boundary is the x-axis. Answer C (

y = 0

y = 1

) correctly identifies the lower limit but stops at

y = 1

instead of continuing to

y = e

, cutting off the upper part of the region. Answer D (

y = e^0

y = e^1

) is mathematically equivalent to answer A since

e^0 = 1

and

e^1 = e

, but it's written in a confusing way that might suggest incorrect endpoints. When setting up horizontal strips, always find where the region starts and stops vertically by evaluating the bounding functions at their extreme x-values.

Question 17

A student sets up the centroid calculation for a region as $\bar{x} = \frac{\int_0^2 x \cdot \sqrt{4-x^2} dx}{\int_0^2 \sqrt{4-x^2} dx}$ . However, upon checking, the denominator evaluates to $\pi$ while the numerator evaluates to $\frac{4}{3}$ . What does this suggest about the original region?

The calculation has an error because $\bar{x}$ should be $\frac{4}{3\pi}$ , not a quarter-circle centroid
The region should be a semicircle, but the student used wrong integration limits
The region is a quarter-circle of radius 2, and the calculation is correct (correct answer)
The region is a quarter-circle, but the moment calculation should use $\frac{x}{2}$ instead of $x$

Explanation: When you encounter centroid calculations in statics, always start by identifying the geometric region from the mathematical setup. The integrand and limits tell you everything about the shape you're analyzing. The setup

\bar{x} = \frac{\int_0^2 x \cdot \sqrt{4-x^2} dx}{\int_0^2 \sqrt{4-x^2} dx}

reveals key information. The function

\sqrt{4-x^2}

comes from solving

y = \sqrt{4-x^2}

, which when squared gives

x^2 + y^2 = 4

- the equation of a circle with radius 2. The integration limits from 0 to 2 indicate you're working with the first quadrant portion, making this a quarter-circle. The denominator

\int_0^2 \sqrt{4-x^2} dx = \pi

correctly gives the area of a quarter-circle:

\frac{1}{4}\pi r^2 = \frac{1}{4}\pi(2^2) = \pi

. The numerator

\int_0^2 x \cdot \sqrt{4-x^2} dx = \frac{4}{3}

correctly calculates the first moment about the y-axis. This gives

\bar{x} = \frac{4}{3\pi}

, which matches the theoretical centroid location for a quarter-circle. Answer C is correct. Option A misinterprets the result -

\frac{4}{3\pi} \approx 0.424

is indeed the correct x-coordinate for a quarter-circle centroid. Option B incorrectly suggests the limits are wrong when they perfectly define the quarter-circle. Option D proposes an incorrect moment calculation that would halve the centroid distance. Remember: when analyzing centroids, first identify the geometric shape from the mathematical description, then verify your setup matches the known properties of that shape.

Question 18

A quarter-circle region in the first quadrant has radius $R$ . When computing the y-coordinate of the centroid using integration with respect to $x$ , the integrand for the first moment about the x-axis should be:

$\frac{1}{2}y \cdot \sqrt{R^2 - x^2}$
$\frac{1}{2}(\sqrt{R^2 - x^2})^2$ (correct answer)
$y \cdot \sqrt{R^2 - x^2}$
$\frac{1}{2}\sqrt{R^2 - x^2} \cdot dx$

Explanation: For the first moment about the x-axis using vertical strips,

M_x = \int \bar{y} \cdot dA

where

\bar{y}

is the centroid of the strip. For a vertical strip from 0 to

\sqrt{R^2-x^2}

\bar{y} = \frac{\sqrt{R^2-x^2}}{2}

and

dA = \sqrt{R^2-x^2}dx

. Thus the integrand is

\frac{1}{2}(\sqrt{R^2-x^2})^2

. Choice A uses

y

instead of

\bar{y}

. Choice C omits the factor of

1/2

. Choice D incorrectly includes

dx

Question 19

The area element $dA$ for a region is expressed as $(f(x) - g(x))dx$ where $f(x) > g(x)$ . If the first moment about the x-axis is calculated as $M_x = \int_a^b \frac{f(x) + g(x)}{2} \cdot (f(x) - g(x)) dx$ , what geometric interpretation explains this formula?

The region lies between two curves, and $\frac{f(x) + g(x)}{2}$ represents the average height of the region at each x
The expression represents the moment about the line $y = \frac{f(x) + g(x)}{2}$ rather than the x-axis
The region lies between two curves, and $\frac{f(x) + g(x)}{2}$ represents the y-coordinate of the centroid of each vertical strip (correct answer)
The formula accounts for the varying width of horizontal strips in a region bounded by $x = f(y)$ and $x = g(y)$

Explanation: When calculating first moments in statics, you're finding the product of an area element and its distance from a reference axis. For regions bounded by two curves, understanding what represents the "distance" is crucial. The given formula

M_x = \int_a^b \frac{f(x) + g(x)}{2} \cdot (f(x) - g(x)) dx

breaks down into two parts:

(f(x) - g(x))dx

is the area of a thin vertical strip at position

x

, while

\frac{f(x) + g(x)}{2}

is the y-coordinate of that strip's centroid. Since

f(x)

is the top boundary and

g(x)

is the bottom boundary, their average gives the midpoint height - exactly where the centroid of each rectangular strip lies. The first moment about the x-axis requires this y-distance from the x-axis to each strip's centroid. Option A incorrectly calls this an "average height," but it's specifically the centroidal location, not just any average. Option B misinterprets the reference axis - we're still calculating the moment about the x-axis, not about the line

y = \frac{f(x) + g(x)}{2}

. Option D describes a completely different setup where the region is bounded by functions of

y

rather than

x

. Option C correctly identifies that

\frac{f(x) + g(x)}{2}

represents the y-coordinate of each vertical strip's centroid, which is exactly what you need for the moment calculation. Remember: for moments, always identify what represents your area element and what represents the distance from your reference axis to the centroid of that element.

Question 20

A semicircular region of radius $a$ is positioned with its diameter along the x-axis, centered at the origin. When setting up the integration for $\bar{y}$ using horizontal strips, which of the following correctly represents the setup?

$\bar{y} = \frac{\int_{-a}^a y \cdot 2\sqrt{a^2-y^2} dy}{\int_{-a}^a 2\sqrt{a^2-y^2} dy}$
$\bar{y} = \frac{\int_0^a y \cdot 2\sqrt{a^2-y^2} dy}{\int_0^a 2\sqrt{a^2-y^2} dy}$ (correct answer)
$\bar{y} = \frac{\int_0^a y \cdot \sqrt{a^2-y^2} dy}{\int_0^a \sqrt{a^2-y^2} dy}$
$\bar{y} = \frac{\int_0^a \frac{y^2}{2} \cdot 2\sqrt{a^2-y^2} dy}{\int_0^a 2\sqrt{a^2-y^2} dy}$

Explanation: For a semicircle in the upper half-plane, y ranges from 0 to

a

. Each horizontal strip at height

y

has length

2\sqrt{a^2-y^2}

(extending from

-\sqrt{a^2-y^2}

+\sqrt{a^2-y^2}

). The moment arm for

\bar{y}

is simply

y

. Choice A uses incorrect limits that include the lower half-plane. Choice C omits the factor of 2 for strip length. Choice D incorrectly uses

y^2/2

as the moment arm.