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📝 Properties of Double Integrals (14 MCQs)

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

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What is Properties of Double Integrals?

Definition:
Double integrals satisfy linearity and additivity properties: $\iint_R [af(x,y) + bg(x,y)] \, dA = a\iint_R f(x,y) \, dA + b\iint_R g(x,y) \, dA$ and $\iint_{R_1 \cup R_2} f \, dA = \iint_{R_1} f \, dA + \iint_{R_2} f \, dA$ if regions do not overlap.

Example:
If $\iint_R x^2 \, dA = 5$ and $\iint_R y^2 \, dA = 3$ , then $\iint_R (2x^2 - y^2) \, dA = 2(5) - 3 = 7$ .

Reason:
These properties allow us to simplify complex integrals by breaking them into smaller, manageable parts or combining known results, significantly reducing computational effort.

Easy

Medium

Hard

📝 All Properties of Double Integrals MCQs

Q1. Suppose $f(x,y)\geq 0$ on a region $R$ , and $R$ is divided into two non-overlapping subregions $R_1$ and $R_2$ whose union is $R$ . Which relationship must hold when the required integrals exist?

\iint_R f\,dA=\iint_{R_1}f\,dA-\iint_{R_2}f\,dA

\iint_R f\,dA=\iint_{R_1}f\,dA+\iint_{R_2}f\,dA

✅

\iint_R f\,dA=\iint_{R_1\cap R_2}f\,dA

\iint_R f\,dA=\iint_{R_1}f\,dA\iint_{R_2}f\,dA

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The additive property states that an integral over a region can be separated into the sum of integrals over non-overlapping pieces whose union forms the original region. The nonnegative assumption is not actually required for this property, but it makes the interpretation especially intuitive as accumulated volume or mass.

Q2. For a constant $c$ and an integrable function $f$ , which expression correctly describes the effect of multiplying the integrand by $c$ ?

A.The double integral is multiplied by

c

✅

B.The double integral is multiplied by

c^2

C.The constant disappears because integration only depends on the region

D.The double integral is divided by

c

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Double integration is linear with respect to constant scaling. Therefore, multiplying every value of the integrand by $c$ multiplies the accumulated integral by exactly $c$ . The factor does not become $c^2$ , because the scaling applies to function values rather than independently scaling the two coordinate directions.

Q3. A rectangular region $R$ is split into $R_1$ and $R_2$ . The same function $f$ is integrated over both pieces. A student claims that the larger subregion must always produce the larger integral. Which statement best evaluates the claim?

A.It is always true for every integrable function

B.It is true only when

f

is positive and constant

C.It can be false because the values of

f

may be much larger on the smaller region ✅

D.It is false only when

f

changes sign

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Region size alone does not determine the value of a double integral. A smaller region can produce a larger integral if the function has substantially larger values there. The claim becomes reliable under additional conditions such as a nonnegative constant integrand, but not for an arbitrary integrable function.

Q4. Let $R$ be a region and suppose $f(x,y)\leq g(x,y)$ at every point of $R$ . What conclusion follows most directly when both functions are integrable?

\iint_R f\,dA\geq\iint_R g\,dA

\iint_R f\,dA\leq\iint_R g\,dA

✅

C.The two integrals must be equal

D.Their difference must equal the area of

R

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If $f\leq g$ pointwise throughout the same region, then the accumulated contribution from $f$ cannot exceed that from $g$ . Integrating the inequality preserves its direction. Equality occurs only under additional conditions, such as when the functions agree except possibly on a set that does not affect the integral.

Q5. A function satisfies $m\leq f(x,y)\leq M$ throughout a region $R$ having area $A$ . A student wants a quick estimate of the integral without evaluating it exactly. Which bound is justified?

mA\leq\iint_R f\,dA\leq MA

✅

m+A\leq\iint_R f\,dA\leq M+A

mM\leq\iint_R f\,dA\leq A

A/m\leq\iint_R f\,dA\leq A/M

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Every value of $f$ lies between $m$ and $M$ , so the total accumulation over a region of area $A$ must lie between the accumulation obtained from the constant functions $m$ and $M$ . Thus the integral is bounded by $mA$ and $MA$ , provided the stated quantities are appropriate.

Q6. A rectangular plate has uniform surface density $5$ units of mass per unit area and occupies a region of area $18$ . Without evaluating any iterated integral, what is its total mass?

23

units

90

units ✅

180

units

3.6

units

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For uniform surface density, the mass is the integral of the density over the plate. Since the density is the constant $5$ and the area is $18$ , linearity and the constant-factor property give $M=\iint_R5\,dA=5(18)=90$ units. No detailed integration is necessary.

Q7. A model predicts a quantity using $I=\iint_R(f+g)\,dA$ . The analyst already knows $F=\iint_R f\,dA$ and $G=\iint_R g\,dA$ . Which approach gives $I$ without repeating the integration?

A.Compute

I=FG

B.Compute

I=F-G

C.Compute

I=F+G

✅

D.Compute

I=F/G

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The linearity property allows a sum inside the integrand to be separated into a sum of integrals over the same region. Therefore $I=\iint_Rf\,dA+\iint_Rg\,dA=F+G$ . This is useful in modelling because independently computed contributions can be combined without integrating their sum again.

Q8. A student writes $\iint_R 3f(x,y)\,dA=3\iint_R f(x,y)\,dA$ , but another student argues that the factor should become $9$ because there are two variables. Who is correct, and why?

A.The second student, because both

x

and

y

contribute a factor of

3

B.The first student, because the constant multiplies the integrand only once ✅

C.Both are correct depending on the shape of

R

D.Neither is correct because constants cannot be moved outside a double integral

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The factor $3$ multiplies the function value at each point, so linearity pulls out exactly one factor of $3$ . The presence of two integration variables does not create another factor. A factor of $9$ would arise only if the original integrand itself contained a factor $9$ , not merely because two integrations are performed.

Q9. A student argues: If $f(x,y)$ is positive on $R_1$ and negative on $R_2$ , then $\iint_{R_1\cup R_2}f\,dA$ must be positive because $R_1$ and $R_2$ have equal areas. What is the flaw?

A.Equal areas guarantee equal integral magnitudes

B.The sign of the total depends on the magnitudes and distributions of

f

, not only on region areas ✅

C.Negative values cannot occur inside a double integral

D.The integral over the union is always zero

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Equal areas do not imply equal contributions to the integral. If the positive values on $R_1$ have small magnitude while the negative values on $R_2$ have large magnitude, the total can be negative. Additivity permits the two contributions to be combined, but their signs and magnitudes determine the result.

Q10. Consider a graph of a function over two adjacent regions of equal area. Over $R_1$ , the surface lies mostly near height $2$ , while over $R_2$ , it lies mostly near height $6$ . Which comparison is most reasonable if the function remains nonnegative?

A.The integral over

R_1

is likely larger

B.The integral over

R_2

is likely larger ✅

C.The two integrals must be equal because the areas are equal

D.Both integrals must be zero

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A double integral measures accumulated height over area. Since the two regions have equal area but the surface is substantially higher over $R_2$ , its accumulated value is expected to be larger. Equal areas alone would guarantee equal integrals only under stronger conditions, such as matching function behavior across the regions.

Q11. A sensor records $f(x,y)$ over a region $R$ . Due to a calibration error, every recorded value is replaced by $f(x,y)+4$ . If the original integral is $I$ and the area of $R$ is $A$ , what is the corrected integral represented by the new measurements?

I+4

4I

I+4A

✅

I/A+4

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Adding $4$ to the measurement at every point adds a constant function $4$ over the entire region. Therefore the new integral is $\iint_R(f+4)\,dA=I+\iint_R4\,dA=I+4A$ . The correction depends on the area because the calibration offset affects every point.

Q12. Two teams calculate the same integral over $R$ . Team A obtains $12$ . Team B divides $R$ into three non-overlapping regions and obtains contributions $5$ , $4$ , and $3$ . Which conclusion is justified?

A.Team B must be wrong because subdivision changes the integral

B.Team A and Team B agree because the three contributions add to

12

✅

C.Team A must be wrong because subdivision always increases the integral

D.No comparison is possible because double integrals cannot be subdivided

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The additive property provides an independent consistency check. If the three subregions form the original region without overlap except along boundaries, their integrals add to the integral over the whole region. Since $5+4+3=12$ , Team B's calculation is consistent with Team A's result.

Q13. Let $R$ have area $10$ . A function satisfies $-2\leq f(x,y)\leq5$ everywhere on $R$ , and another function $g$ satisfies $g(x,y)=f(x,y)+3$ . Which interval must contain $\iint_R g\,dA$ ?

A.From

-20

50

B.From

10

80

✅

C.From

30

50

D.From

-2

5

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Since $-2\leq f\leq5$ , adding $3$ gives $1\leq g\leq8$ . The integral over a region of area $10$ is therefore bounded by $1(10)$ and $8(10)$ . Hence $10\leq\iint_Rg\,dA\leq80$ , combining order preservation with the constant-function property.

Q14. A function $f$ is nonnegative on a region $R$ , which is split into two parts of equal area. The average value of $f$ on the first part is $3$ , while on the second part it is $7$ . What is the average value of $f$ over all of $R$ ?

A.-3

B.-5 ✅

C.-7

D.-10

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Because the two subregions have equal area, their contributions to the overall average are weighted equally. The total integral is the sum of the two regional integrals, each equal to its average multiplied by the common area. Thus the overall average is the weighted mean $(3+7)/2=5$ , not the sum or either individual average.

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