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📝 Density and Mass of an Inhomogeneous Lamina (14 MCQs)

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

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What is Density and Mass of an Inhomogeneous Lamina?

Definition:
The mass of a lamina with variable density $\rho(x,y)$ is $M = \iint_R \rho(x,y) \, dA$ .

Example:
If $\rho(x,y) = x^2+y^2$ over the unit disk, $M = \int_0^{2\pi} \int_0^1 r^2 \cdot r \, dr \, d\theta = \frac{\pi}{2}$ .

Reason:
Real materials often have non-uniform density; integrating density over area gives total mass, essential for accurate physical modeling.

Easy

Medium

Hard

📝 All Density and Mass of an Inhomogeneous Lamina MCQs

Q1. A rectangular lamina occupies $0\le x\le 2$ , $0\le y\le 1$ , with density $\rho(x,y)=3+x$ . Which integral correctly represents its mass?

\int_0^2\int_0^1(3+x)\,dy\,dx

✅

\int_0^2\int_0^1(3+y)\,dy\,dx

\int_0^1\int_0^2(3+x)\,dy\,dx

\int_0^2\int_0^1 3x\,dy\,dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For a variable-density lamina, mass is obtained by integrating density over the entire region. The density here depends on $x$ , so the integrand must be $3+x$ . The bounds cover the full rectangle. Option C has incompatible inner-variable bounds because $y$ would run from $0$ to $2$ , while options B and D use incorrect density functions.

Q2. A lamina has density $\rho(x,y)=2x+y$ . Two students claim that the denser half of the lamina must also contain half of its total mass. Which statement best evaluates their claim?

A.It is always true because density is positive.

B.It is true only when the density is constant over the region. ✅

C.It is false because positive density means every subregion has equal mass.

D.It is false because mass depends only on the area of the region.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Equal area does not generally imply equal mass when density varies spatially. A subregion containing points where $\rho(x,y)$ is larger can contribute substantially more mass than another subregion of the same area. Equal masses occur automatically for equal areas only when density is constant or when symmetry and density variation happen to balance.

Q3. A circular lamina is centered at the origin and has density $\rho(x,y)=1+x^2$ . Without evaluating the integral, what can be concluded about the average density of the lamina?

A.It is exactly

1

because the region is centered at the origin.

B.It is greater than

1

because

x^2>0

except along one line. ✅

C.It is less than

1

because positive and negative

x

-values cancel.

D.It is zero because the integral of

x^2

is symmetric.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The density equals $1+x^2$ , and $x^2\ge0$ everywhere. Except on the line $x=0$ , which has zero area, $x^2>0$ . Therefore the density exceeds $1$ over almost the entire lamina, so its area-weighted average density must also be greater than $1$ . Symmetry does not make $x^2$ cancel.

Q4. A triangular lamina is bounded by $x=0$ , $y=0$ , and $x+y=2$ , with density $\rho(x,y)=x+2y$ . Which setup correctly models its mass?

\int_0^2\int_0^{2-x}(x+2y)\,dy\,dx

✅

\int_0^2\int_0^x(x+2y)\,dy\,dx

\int_0^2\int_x^2(x+2y)\,dy\,dx

\int_0^2\int_0^{2}(x+2y)\,dy\,dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For each fixed $x$ between $0$ and $2$ , the line $x+y=2$ gives the upper boundary $y=2-x$ , while the lower boundary is $y=0$ . The density itself is $x+2y$ , so option A correctly combines the geometry of the region with the spatially varying density.

Q5. A manufacturing plate occupies a region $R$ of area $12\text{ cm}^2$ . Its density varies continuously between $4$ and $9\text{ g/cm}^2$ . Which conclusion about its mass $M$ is necessarily valid?

M=78\text{ g}

M

must be between

48\text{ g}

and

108\text{ g}

. ✅

M

must equal

6.5(12)\text{ g}

M

must be less than

48\text{ g}

because density varies.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Since density is everywhere at least $4\text{ g/cm}^2$ and at most $9\text{ g/cm}^2$ , multiplying these bounds by the area gives $4(12)\le M\le9(12)$ . Thus $48\le M\le108$ grams. The exact mass depends on how density is distributed, so the midpoint density cannot automatically be used.

Q6. A lamina occupies $0\le x\le1$ , $0\le y\le2$ , and has density $\rho(x,y)=5-2x+y$ . A student says increasing $y$ cannot increase the total mass because the positive and negative contributions of $y$ should cancel. What is the best response?

A.The student is correct because the interval is symmetric about

y=0

B.The student is correct because

y

integrates to zero over every interval.

C.The student is incorrect because

y

is nonnegative throughout the region, so its contribution increases mass. ✅

D.The student is incorrect because density must decrease whenever

y

increases.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The region has $0\le y\le2$ , so $y$ is never negative. Consequently, the $+y$ term contributes positively throughout the lamina. There is no cancellation between positive and negative values of $y$ . The mistaken reasoning would apply only in a region symmetric about $y=0$ , where positive and negative contributions could cancel.

Q7. A lamina occupies the disk $x^2+y^2\le4$ and has density $\rho(x,y)=6+x$ . A designer wants to estimate its mass using the approximation $M\approx6(\text{area})$ . Is this estimate exact?

A.Yes, because the

x

-term always contributes zero pointwise.

B.No, because

x

is positive everywhere in the disk.

C.Yes, because the integral of

x

over the centered disk is zero. ✅

D.No, because the density is not constant, and symmetry cannot be used in any integral.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The density is not constant, but the mass is still exactly $6$ times the disk's area because the integral of $x$ over a disk centered at the origin is zero by left-right symmetry. Positive $x$ -contributions on one side balance negative $x$ -contributions on the other. Thus the approximation happens to be exact.

Q8. A student computes the mass of a lamina with density $\rho(x,y)=x+y$ by first finding the area of the region and then multiplying by $\rho(1,1)=2$ . Why can this procedure fail?

A.Because density can never be evaluated at a point.

B.Because the density may vary across the region, so a single point value need not equal the average density. ✅

C.Because mass is always independent of density.

D.Because area must always be computed using a triple integral.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For an inhomogeneous lamina, mass is an integral of density over area. Multiplying area by the density at one arbitrary point assumes that this point represents the average density, which is generally unjustified. The procedure is exact only under special circumstances, such as constant density or a point whose density equals the true average.

Q9. A rectangular lamina $0\le x\le2$ , $0\le y\le1$ has density $\rho(x,y)=x^2+2$ . A quality engineer divides the rectangle into two equal-area vertical strips and predicts that the right strip is heavier. Which reasoning is strongest?

A.The strips have equal area, so their masses must be equal.

B.The right strip contains larger

x^2

values throughout, so its average density is larger and its mass is greater. ✅

C.The left strip is heavier because smaller

x

produces smaller density.

D.Both strips have the same mass because the density depends only on

x

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The two strips have equal area, but density is larger at larger $x$ because of the $x^2$ term. Every point in the right strip has a greater $x^2$ value than corresponding points in the left strip. Therefore its average density is larger, and equal area then implies a larger mass.

Q10. A graph of density over a rectangular region shows contour lines labeled $2,4,6,8$ , with the contours becoming larger toward the upper-right corner. Two equal-area subregions are located respectively near the lower-left and upper-right corners. Which comparison is most justified?

A.The lower-left subregion must have greater mass because it is closer to the origin.

B.The two subregions must have equal mass because their areas are equal.

C.The upper-right subregion has greater mass because its density values are systematically larger. ✅

D.No mass comparison is possible without knowing the perimeter of each subregion.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Mass depends on both area and density. Since the two subregions have equal area, their masses can be compared through their average densities. The contour information indicates that density is systematically higher toward the upper-right corner, so the upper-right subregion has the larger average density and therefore greater mass.

Q11. A lamina is symmetric about the $y$ -axis, while its density satisfies $\rho(-x,y)=\rho(x,y)$ . An engineer wants to simplify the mass calculation by integrating only over the right half and doubling the result. Is this valid?

A.No, because mass integrals can never use symmetry.

B.Yes, because both the region and density have reflection symmetry about the

y

-axis. ✅

C.No, because density must be antisymmetric for doubling to work.

D.Yes, but only if the density is constant.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Reflection symmetry applies to the complete mass integrand when both the region and density are unchanged under $x\mapsto-x$ . Here the lamina is symmetric and the density satisfies $\rho(-x,y)=\rho(x,y)$ , so the two halves contribute equally. Therefore integrating over one half and doubling gives the exact mass.

Q12. A rectangular plate has density $\rho(x,y)=1+x+2y$ . Method A integrates directly over the rectangle. Method B first averages the density over the rectangle and multiplies that average by the area. Which statement is correct?

A.Only Method A can ever be correct for variable density.

B.Only Method B can be correct because density must be averaged first.

C.Both methods give the same mass when the average density is computed correctly. ✅

D.The methods differ because averaging density destroys spatial information.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For total mass, integrating density over the region is equivalent to multiplying the region's area by its average density. The average density is itself defined through the integral of $\rho$ divided by area. Therefore Method B does not approximate the mass; when performed correctly, it reproduces Method A exactly.

Q13. Consider a lamina in a region $R$ with density $\rho(x,y)=k(x^2+y^2)$ , where $k>0$ . Suppose the region is changed by scaling every coordinate by a factor of $2$ , producing a geometrically similar lamina. How does the mass scale?

A.It becomes

2

times the original mass.

B.It becomes

4

times the original mass.

C.It becomes

8

times the original mass.

D.It becomes

16

times the original mass. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Under the scaling $x\mapsto2x$ and $y\mapsto2y$ , the area element scales by $4$ , while $x^2+y^2$ scales by $4$ . Therefore the density at corresponding points becomes four times as large, and the area becomes four times as large. Their product causes total mass to scale by $4\cdot4=16$ .

Q14. A lamina occupies a region symmetric about both coordinate axes, and its density is $\rho(x,y)=7+3xy+x^2$ . Which expression gives the correct qualitative simplification of the mass integral?

A.The

3xy

term contributes zero, so only

7+x^2

needs to be integrated. ✅

B.Both

3xy

and

x^2

contribute zero by symmetry.

C.The constant

7

contributes zero because the region is centered at the origin.

D.The entire density integrates to zero because positive and negative coordinates balance.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Because the region is symmetric about either coordinate axis, the function $xy$ changes sign under reflection across one axis, so its integral over the entire region is zero. However, $x^2$ is unchanged by reflection and is nonnegative, so it does not cancel. Thus the mass reduces to the integral of $7+x^2$ over the region.

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