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📝 Centers of Gravity Using Multiple Integrals (14 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 14 questions available

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What is Centers of Gravity Using Multiple Integrals?

Definition:
The center of gravity $(\bar{x}, \bar{y}, \bar{z})$ of a solid with density $\rho$ is given by moments divided by mass: $\bar{x} = \frac{1}{M} \iiint_E x\rho \, dV$ , etc.

Example:
For a uniform sphere, symmetry implies $\bar{x}=\bar{y}=\bar{z}=0$ if centered at origin.

Reason:
Center of gravity determines balance points and stability, critical in engineering and physics for analyzing structural integrity and motion.

Easy

Medium

Hard

📝 All Centers of Gravity Using Multiple Integrals MCQs

Q1. A uniform lamina occupies a region $R$ in the plane. Which pair of quantities must be divided by the total mass to determine the coordinates of its center of gravity?

A.The first moments about the coordinate axes ✅

B.The area and perimeter of

R

C.The second moments about the coordinate axes

D.The maximum and minimum coordinates of

R

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For a uniform lamina, the center of gravity is determined from first moments divided by total mass. Specifically, the coordinates are obtained from ratios involving $\iint_R x\,dA$ , $\iint_R y\,dA$ , and $\iint_R dA$ . Area or second moments alone do not determine the centroid.

Q2. A lamina is symmetric about the $y$ -axis and has uniform density. Without evaluating any integral, what can be concluded about its center of gravity?

A.Its

y

-coordinate must be zero

B.Its

x

-coordinate must be zero ✅

C.Both coordinates must be zero

D.Its center must lie on the line

y=x

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Reflection across the $y$ -axis pairs every point $(x,y)$ with $(-x,y)$ , producing equal and opposite contributions to the first moment involving $x$ . Therefore, the horizontal coordinate of the center of gravity is zero, while the vertical coordinate depends on the actual shape.

Q3. A uniform lamina occupies a region that is symmetric about the line $y=x$ . Which conclusion follows most directly from this symmetry?

A.The center of gravity must lie on

y=-x

B.The center of gravity must lie on the

x

-axis

C.The center of gravity must lie on

y=x

✅

D.The center of gravity must be at the origin

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A symmetry line of a uniform region also becomes a symmetry line for its mass distribution. Since reflection across $y=x$ interchanges the coordinates without changing the lamina, the center of gravity must remain fixed under that reflection. Hence its coordinates must satisfy $x=y$ .

Q4. Suppose a uniform triangular lamina has vertices $(0,0)$ , $(6,0)$ , and $(0,3)$ . A student claims that its center of gravity is at $(3,1)$ because those coordinates are the averages of the extreme $x$ - and $y$ -values. Which evaluation is most appropriate?

A.The claim is correct because the region is triangular

B.The claim is correct only because the density is uniform

C.The claim is incorrect; the centroid is

(2,1)

✅

D.The claim is incorrect; the centroid is

(3,1.5)

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a triangle, the centroid is located at the average of the three vertex coordinates, not the averages of the extreme coordinate values. Thus $x=(0+6+0)/3=2$ and $y=(0+0+3)/3=1$ . The student's method gives an incorrect horizontal coordinate.

Q5. A uniform lamina lies between $y=x^2$ and $y=4$ , for $-2\le x\le2$ . Which strategy is most efficient for determining the vertical coordinate of its center of gravity?

A.Use symmetry to conclude that the vertical coordinate is zero

B.Integrate

y

over the region and divide by its area ✅

C.Integrate

x

over the region and divide by its area

D.Use only the boundary length to determine the coordinate

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The region is symmetric about the $y$ -axis, so symmetry immediately determines only the horizontal coordinate. The vertical coordinate still requires the first moment involving $y$ . For a uniform lamina, this can be obtained from $\iint_R y\,dA$ divided by the area of the region.

Q6. A rectangular lamina occupies $0\le x\le4$ and $0\le y\le6$ , but its density increases with height according to $\rho(x,y)=1+y$ . Compared with the geometric centroid, where should the center of gravity move?

A.Toward smaller

y

-values

B.Toward larger

y

-values ✅

C.Toward smaller

x

-values only

D.It must remain at the geometric centroid

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Higher values of $y$ have greater density, so more mass is concentrated near the upper part of the rectangle. Consequently, the mass-weighted average of $y$ becomes larger than the geometric centroid's $y$ -coordinate. The horizontal coordinate remains centered because the density does not depend on $x$ .

Q7. A uniform semicircular lamina lies above the $x$ -axis and is centered at the origin. A student argues that its center of gravity must be at the origin because the full circle would have its centroid there. What is the best response?

A.Correct, because every semicircle has its centroid at the origin

B.Correct, because the diameter passes through the origin

C.Incorrect, because the lamina is not symmetric about the

x

-axis and its center lies above the axis ✅

D.Incorrect, because its center lies on the

x

-axis but not at the origin

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The semicircle is symmetric about the $y$ -axis, so its horizontal coordinate is zero. However, it is not symmetric about the $x$ -axis because all its mass lies above that axis. Therefore, the center of gravity must have a positive $y$ -coordinate rather than being located at the origin.

Q8. A designer models a uniform plate as the region between $y=x$ and $y=x^2$ for $0\le x\le1$ . Which reasoning correctly identifies the horizontal coordinate of its center of gravity?

A.It equals

1/2

because the

x

-interval has midpoint

1/2

B.It equals

1/3

because the upper boundary is quadratic

C.It must be found from the area-weighted average of

x

, so the vertical thickness of the region matters ✅

D.It equals

2/3

because the centroid lies closer to the wider end

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The horizontal centroid is not generally the midpoint of the coordinate interval because the region's width varies with $x$ . Here the vertical thickness is $x-x^2$ , so locations with greater thickness contribute more mass. Therefore, $x$ must be weighted by the local area contribution when evaluating the centroid.

Q9. A uniform lamina is bounded by $y=0$ , $x=0$ , and $x+y=6$ . An engineer wants to estimate where its center of gravity lies before performing any calculation. Which statement is most defensible?

A.It must lie at

(3,3)

because the largest coordinates are both 6

B.It must lie at

(2,2)

because the region is triangular with three vertices ✅

C.It must lie at

(1,1)

because most of the region is near the origin

D.It must lie on the line

x=y

because the region is symmetric about that line

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The triangle has vertices $(0,0)$ , $(6,0)$ , and $(0,6)$ . Its centroid is the average of the three vertices, giving $(2,2)$ . The symmetry argument also confirms that the coordinates are equal, but symmetry alone would not determine their numerical value.

Q10. A graph shows a uniform region symmetric about the $y$ -axis, with a narrow lower portion and a much wider upper portion. Which qualitative location is most reasonable for its center of gravity?

A.On the negative

x

-axis

B.On the positive

x

-axis

C.On the

y

-axis, closer to the wider upper portion ✅

D.At the origin regardless of the region's vertical distribution

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Symmetry about the $y$ -axis forces the horizontal coordinate of the center of gravity to be zero. Because the upper portion contains substantially more area and therefore more mass for a uniform lamina, the vertical coordinate is pulled toward that upper region rather than necessarily remaining at the origin.

Q11. Two uniform laminae have equal area. Lamina A is concentrated near $y=1$ , while Lamina B is concentrated near $y=5$ . They are placed together without overlap to form a composite object. Which method best determines the combined center of gravity?

A.Average the two geometric centers without considering their areas

B.Use the sum of their first moments divided by the sum of their masses ✅

C.Use the center of the bounding rectangle

D.Average only their

y

-coordinates because the areas are equal

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The center of gravity of a composite object is obtained by combining the first moments of all components and dividing by total mass. Equal areas and uniform densities make the component masses equal, so a simple average of component centroids would work here, but the general and correct method is the ratio of total first moment to total mass.

Q12. A student computes the centroid of a uniform region using $\bar{x}=\frac{\iint_R x\,dA}{\iint_R dA}$ , but obtains $\bar{x}=7$ even though the entire region lies between $x=1$ and $x=5$ . Which conclusion is strongest?

A.The centroid can lie outside the region

B.The integral formula is invalid for two-dimensional regions

C.There is an error in the setup or evaluation because a weighted average of

x

cannot exceed the largest

x

-value ✅

D.The density must automatically be negative somewhere

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For a uniform region, the centroid coordinate is an area-weighted average of the possible $x$ -values. Since every point satisfies $1\le x\le5$ , the weighted average must also lie between 1 and 5. A result of 7 therefore signals an error in the integration limits, integrand, or algebra.

Q13. A uniform lamina occupies a region symmetric about both coordinate axes but has a circular hole centered at $(2,0)$ . Which qualitative effect does the hole have on the center of gravity of the remaining lamina?

A.The center remains at the origin because the outer boundary is symmetric

B.The center shifts toward the hole because mass was removed there

C.The center shifts away from the hole, toward the opposite side ✅

D.The center moves only vertically

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Removing mass is equivalent to adding a negative mass at the hole's location. Since the removed material is to the right of the origin, the remaining mass has relatively greater influence on the left side. Therefore, the center of gravity shifts away from the removed region, toward the opposite side.

Q14. Consider the uniform region $0\le y\le x\le1$ . A proposed centroid is $(1/2,1/2)$ . Without fully evaluating both double integrals, which observation immediately disproves this proposal?

A.The region is symmetric about the

x

-axis

B.Every point satisfies

y\le x

, so the centroid must satisfy

\bar y\le\bar x

, but equality would require special concentration on

y=x

✅

C.The region has area greater than 1

D.The centroid must always lie on a boundary curve

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The region is the triangular set under $y=x$ and above $y=0$ . Its mass occupies a range of points strictly below the line $y=x$ , except on the boundary, so the average $y$ -coordinate must be strictly less than the average $x$ -coordinate. Thus $(1/2,1/2)$ cannot be the centroid.

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←Previous Topic:Change of Variables in Triple Integrals📝 Centers of Gravity Using Multiple IntegralsDensity and Mass of an Inhomogeneous Lamina:Next Topic→