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📝 Center of Gravity of an Inhomogeneous Lamina (15 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 15 questions available

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What is Center of Gravity of an Inhomogeneous Lamina?

Definition:
For a lamina with density $\rho(x,y)$ , $\bar{x} = \frac{1}{M} \iint_R x\rho(x,y) \, dA$ and $\bar{y} = \frac{1}{M} \iint_R y\rho(x,y) \, dA$ .

Example:
If density increases with distance from origin, the center of gravity shifts toward heavier regions.

Reason:
Non-uniform density affects balance; calculating the weighted center ensures accurate prediction of rotational behavior and equilibrium.

Easy

Medium

Hard

📝 All Center of Gravity of an Inhomogeneous Lamina MCQs

Q1. A lamina occupies a region $R$ in the $xy$ -plane and has density $\rho(x,y)$ . Which expression correctly identifies the $x$ -coordinate of its center of gravity?

x_c=\frac{\iint_R x\,dA}{\iint_R dA}

x_c=\frac{\iint_R x\rho(x,y)\,dA}{\iint_R \rho(x,y)\,dA}

✅

x_c=\frac{\iint_R \rho(x,y)\,dA}{\iint_R x\rho(x,y)\,dA}

x_c=\iint_R x\rho(x,y)\,dA

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For an inhomogeneous lamina, mass is not proportional to area because density varies from point to point. The first moment about the $y$ -axis is $\iint_R x\rho(x,y)\,dA$ , while total mass is $\iint_R \rho(x,y)\,dA$ . Their ratio therefore gives the mass-weighted average of the $x$ -coordinate.

Q2. Two laminae occupy the same region. Lamina A has constant density, while Lamina B is denser near larger $x$ -values. Without calculating any integral, which conclusion is most reasonable?

A.Both centers must have the same

x

-coordinate

B.Lamina B must have a smaller

x

-coordinate

C.Lamina B must have a larger

x

-coordinate ✅

D.The center of Lamina B must lie at

x=0

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The center of gravity is a mass-weighted average of position. Increasing density on the side with larger $x$ -coordinates gives that side greater influence on the average. Thus the center shifts toward the denser region, so Lamina B has a larger $x$ -coordinate than the uniformly dense lamina, assuming the density change is genuinely concentrated toward larger $x$ .

Q3. A rectangular lamina occupies $0\le x\le2$ , $0\le y\le1$ , with density $\rho(x,y)=1+x$ . Which qualitative statement about its center of gravity is correct?

A.Its

x

-coordinate is less than

1

because the rectangle starts at

x=0

B.Its

x

-coordinate is exactly

1

because the region is symmetric in

x

C.Its

x

-coordinate is greater than

1

because density increases with

x

✅

D.Its

x

-coordinate is greater than

2

because density increases with

x

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Although the geometric midpoint of the rectangle is $x=1$ , the lamina is not uniformly dense. Since $\rho=1+x$ increases as $x$ increases, more mass is concentrated toward the right side. The mass-weighted average of $x$ must therefore move to the right of the geometric midpoint while remaining inside the region.

Q4. A designer models a thin plate over a region $R$ with density $\rho(x,y)=x+y$ , where $x,y\ge0$ . A student claims that because the region is symmetric about $y=x$ , the center must lie on the $x$ -axis. What is the best assessment?

A.Correct, because symmetry always places the center on the nearest coordinate axis

B.Correct, because

x+y

is symmetric

C.Incorrect, because symmetry about

y=x

places the center on the line

y=x

✅

D.Incorrect, because an inhomogeneous lamina can never have a center of gravity

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Symmetry must be considered together with the density function. The region is symmetric under interchange of $x$ and $y$ , and $x+y$ is also unchanged by that interchange. Therefore the mass distribution has the same behavior in both directions, forcing the center of gravity to satisfy $x_c=y_c$ . It need not lie on either coordinate axis.

Q5. A rectangular plate occupies $0\le x\le4$ , $0\le y\le2$ . Its density is $\rho(x,y)=2+x$ . Which setup correctly determines the center of gravity?

x_c=\frac{\int_0^4x\,dx}{\int_0^4dx}

y_c=\frac{\int_0^2y\,dy}{\int_0^2dy}

x_c=\frac{\int_0^4\int_0^2x(2+x)\,dy\,dx}{\int_0^4\int_0^2(2+x)\,dy\,dx}

y_c=\frac{\int_0^4\int_0^2y(2+x)\,dy\,dx}{\int_0^4\int_0^2(2+x)\,dy\,dx}

✅

x_c=\frac{\int_0^4\int_0^2x\,dy\,dx}{\int_0^4\int_0^2(2+x)\,dy\,dx}

y_c=\frac{\int_0^4\int_0^2y\,dy\,dx}{\int_0^4\int_0^2(2+x)\,dy\,dx}

x_c=\frac{\int_0^4\int_0^2(2+x)\,dy\,dx}{\int_0^4\int_0^2x(2+x)\,dy\,dx}

y_c=\frac{\int_0^4\int_0^2(2+x)\,dy\,dx}{\int_0^4\int_0^2y(2+x)\,dy\,dx}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For an inhomogeneous lamina, every infinitesimal area element contributes mass $\rho\,dA$ . Consequently, the first moment for $x_c$ contains $x\rho$ , and the first moment for $y_c$ contains $y\rho$ . Both must be divided by the same total mass integral. Ignoring density in the numerator would incorrectly treat the plate as uniformly dense.

Q6. A thin plate occupies a disk centered at the origin, but its density is $\rho(x,y)=3+x$ . Which prediction is most defensible before evaluating any integrals?

A.The center remains at the origin because the region is circular

B.The center moves toward negative

x

because density is smaller for positive

x

C.The center moves toward positive

x

because density is greater for positive

x

✅

D.The center moves upward because

y

does not occur in the density

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The circular region alone is symmetric about both axes, but the density is not symmetric about the $y$ -axis. Points with positive $x$ receive larger density values than corresponding points with negative $x$ . Their contributions to the total mass are therefore greater, shifting the center of gravity toward positive $x$ . The $y$ -coordinate remains zero by symmetry.

Q7. A rectangular lamina occupies $0\le x\le1$ , $0\le y\le1$ , with density $\rho(x,y)=1+4xy$ . A student argues that $x_c=y_c$ is impossible because the density is not constant. Which response is correct?

A.The student is correct because equality requires constant density

B.The student is correct because the density changes independently in both variables

C.The equality can still hold because both the region and density are unchanged when

x

and

y

are interchanged ✅

D.The equality holds only if total mass is zero

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Uniform density is not required for the coordinates of the center to be equal. Here the square is symmetric under exchanging $x$ and $y$ , and $1+4xy$ is also unchanged by that exchange. Thus the mass distribution has identical first-moment behavior in the two coordinate directions, giving $x_c=y_c$ .

Q8. A student computes the center of gravity using $x_c=\frac{\iint_R x\rho\,dA}{\iint_R\rho\,dA}$ but obtains $x_c=5$ for a lamina entirely contained in $0\le x\le3$ . Which diagnosis is strongest?

A.The result is possible because density can move the center outside the lamina

B.The denominator was probably omitted or the mass integral was evaluated incorrectly ✅

C.The

y

-moment must have been used instead

D.The density must have been negative everywhere

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A center of gravity is a weighted average of the positions occupied by mass. If all mass lies between $x=0$ and $x=3$ , its $x$ -coordinate cannot be $5$ when density is nonnegative. A value outside the physical interval indicates an error in the setup, integration, density handling, or division by total mass.

Q9. A plate lies entirely in the first quadrant. Its density increases continuously as $x$ increases. A student concludes that $x_c$ must equal the largest $x$ -value in the plate because the right edge is the densest. Which criticism is correct?

A.Correct, because maximum density determines the center

B.Incorrect, because

x_c

is a weighted average and generally lies strictly inside the occupied

x

-range ✅

C.Correct, provided the density is increasing

D.Incorrect, because the center must always equal the geometric midpoint

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The center of gravity is determined by the combined contribution of every mass element, not by the location of maximum density alone. Even if density is greatest at the right edge, mass still exists at smaller $x$ -values. For a nondegenerate lamina with nonnegative density, the weighted average normally lies within the occupied range rather than at an extreme endpoint.

Q10. A graph shows a horizontal strip-shaped lamina from $x=0$ to $x=4$ . The shading is darkest near $x=4$ and gradually becomes lighter toward $x=0$ , indicating increasing density toward the right. Which sketch description best represents the expected horizontal position of the center of gravity?

A.At

x=0

, because the left edge contains the lightest material

B.At

x=2

, exactly at the geometric midpoint regardless of density

C.To the right of

x=2

, because the right side contributes more mass ✅

D.At

x=4

, because the right edge has maximum density

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The graph indicates that density increases toward the right. Therefore equal-sized regions on the right contain more mass than corresponding regions on the left. The center of gravity is a mass-weighted average of horizontal positions, so it shifts toward the denser side. It should therefore lie to the right of the geometric midpoint while remaining within the plate.

Q11. Two computational methods are proposed for a lamina with density $\rho(x,y)$ . Method I integrates $x\rho$ and $y\rho$ over the region and divides each result by total mass. Method II first finds the geometric centroid and then multiplies its coordinates by average density. Which method is generally valid?

A.Only Method I, because density variation affects spatial weighting ✅

B.Only Method II, because average density contains all necessary information

C.Both methods are always equivalent

D.Neither method works for an inhomogeneous lamina

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Method I directly accounts for how density varies from point to point and therefore weights each position according to its local mass contribution. Method II generally loses spatial information because an average density does not reveal where the mass is concentrated. The geometric centroid can differ substantially from the center of gravity when density is nonuniform.

Q12. A square plate occupies $0\le x,y\le1$ and has density $\rho(x,y)=1+kx$ , where $k>0$ . As $k$ becomes very large, what happens to the $x$ -coordinate of the center of gravity?

A.It approaches

0

B.It approaches

1/2

C.It approaches

1

✅

D.It becomes negative

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: As $k$ becomes large, the term $kx$ dominates the constant part of the density, so the mass distribution becomes increasingly weighted toward larger $x$ . In the limiting behavior, the density is effectively proportional to $x$ , which places substantially more mass near the right edge. The corresponding weighted average therefore approaches the right-hand boundary $x=1$ .

Q13. A nonnegative-density lamina is entirely contained between the vertical lines $x=1$ and $x=5$ . An engineer calculates $x_c=0.4$ . Before checking the integration details, what is the most decisive conclusion?

A.The value is acceptable because density may be highly concentrated

B.The calculation is incorrect because a mass-weighted average of

x

cannot lie outside the occupied interval ✅

C.The result proves the density is uniform

D.The value can be correct only if

y_c=0.4

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Every mass element of the lamina has an $x$ -coordinate between $1$ and $5$ . With nonnegative density, the center coordinate is a weighted average of these values, so it must also lie between $1$ and $5$ . Therefore $x_c=0.4$ is impossible and signals an error in the mathematical model or computation.

Q14. A lamina occupies a symmetric region about the $y$ -axis. Its density is $\rho(x,y)=a+bx$ , with $a>|b|\max|x|$ so density remains positive. Which statement correctly describes how $b$ affects the center of gravity?

A.Changing

b

has no effect because the region is symmetric

B.Positive

b

shifts the center toward positive

x

, while negative

b

shifts it toward negative

x

✅

C.Positive

b

always shifts the center upward

D.The center leaves the lamina whenever

b\ne0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The symmetric region by itself would place the center on the $y$ -axis, but the density term $bx$ breaks that left-right balance. When $b>0$ , points with positive $x$ have greater density than their reflected counterparts, shifting the center rightward. When $b<0$ , the opposite occurs. The positivity condition ensures the density remains physically meaningful.

Q15. A thin plate occupies the triangular region $x\ge0$ , $y\ge0$ , $x+y\le1$ , with density proportional to $x+y$ . Without fully evaluating both double integrals, which observation provides the strongest route to locating the center?

A.The density is constant along lines

x+y=c

, so the center must lie on

x+y=1

B.The region and density are symmetric under

x\leftrightarrow y

, so

x_c=y_c

, and the increasing density toward

x+y=1

shifts the center toward that boundary ✅

C.The center must be the geometric centroid because density depends only on

x+y

D.The center must lie on the

x

-axis because density is zero at the origin

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The triangle is unchanged when $x$ and $y$ are interchanged, and $x+y$ is also unchanged, so the mass distribution has equal horizontal and vertical behavior. Hence $x_c=y_c$ . Because density increases with $x+y$ , mass is weighted more heavily near the sloping boundary $x+y=1$ , so the center is displaced from the geometric centroid toward that boundary.

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