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📝 Area Calculation as a Double Integral (13 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 13 questions available

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What is Area Calculation as a Double Integral?

Definition:
The area of a region $R$ in the xy-plane can be calculated by integrating the constant function $f(x, y) = 1$ over $R$ : $\text{Area} = \iint_R 1 \, dA$ .

Example:
The area of the triangle with vertices (0,0), (1,0), (1,1) is $\int_0^1 \int_0^x 1 \, dy \, dx = \int_0^1 x \, dx = \frac{1}{2}$ .

Reason:
This unifies the concept of area with volume calculation, showing that area is simply the volume under the flat surface $z=1$ , providing a consistent framework for geometric measurements.

Easy

Medium

Hard

📝 All Area Calculation as a Double Integral MCQs

Q1. A region $R$ lies between $y=x^2$ and $y=2x$ . A student wants its area using a double integral. Which setup represents the area correctly without requiring the region to be split?

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

✅

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

\int_0^4\int_{\sqrt{x}}^{2\sqrt{x}}1\,dy\,dx

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

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The area of a planar region is obtained by integrating the constant function $1$ over the entire region. The curves intersect when $x^2=2x$ , giving $x=0$ and $x=2$ . On this interval, $2x$ lies above $x^2$ , so the inner limits must run from $x^2$ to $2x$ .

Q2. Why does the double integral $\iint_R 1\,dA$ measure the area of a region $R$ , even though no explicit geometric area formula appears in the expression?

A.Because every small element

dA

contributes exactly one unit of area density ✅

B.Because

dA

represents the boundary length of

R

C.Because the integrand

1

automatically finds the perimeter

D.Because double integration always produces area regardless of the integrand

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The differential $dA$ represents a small area element, while the integrand $1$ assigns unit density to every point in the region. Summing these infinitesimal contributions over $R$ therefore gives the total area. A different integrand would generally represent a weighted quantity rather than ordinary area.

Q3. A designer models a flower bed by the region $R=\{(x,y):0\le x\le3,\;x\le y\le6-x\}$ . Which expression correctly models the bed's area?

\int_0^3\int_x^{6-x}1\,dy\,dx

✅

\int_0^3\int_{6-x}^{x}1\,dy\,dx

\int_0^6\int_x^{6-x}1\,dx\,dy

\int_0^3\int_x^{6-x}x\,dy\,dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The model directly describes vertical slices: $x$ ranges from $0$ to $3$ , the lower boundary is $y=x$ , and the upper boundary is $y=6-x$ . Since the desired quantity is geometric area rather than a weighted area, the integrand must be $1$ . Thus the first setup exactly represents the stated region.

Q4. A student evaluates $\iint_R 1\,dA$ over a region and obtains $12$ . Another student says the answer cannot be an area because the integrand has no units. Which assessment is most appropriate?

A.The result can represent area because integrating

1

with respect to

dA

sums area elements ✅

B.The result must represent volume because a double integral is two-dimensional

C.The result represents perimeter because

1

has no physical units

D.The calculation is invalid unless the integrand contains

x+y

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The student's objection confuses the integrand with the differential element. In $\iint_R1\,dA$ , the integrand is a unit weighting factor, while $dA$ carries the area element. The integral therefore adds all infinitesimal pieces of area. A numerical result such as $12$ is entirely valid as the region's area when coordinates are measured consistently.

Q5. A park is bounded by $y=4-x^2$ , the $x$ -axis, and the vertical lines $x=-1$ and $x=1$ . Which calculation strategy is most efficient for finding its area?

A.Use

\int_{-1}^{1}\int_0^{4-x^2}1\,dy\,dx

because vertical slices describe the region with one pair of limits ✅

B.Use

\int_{-1}^{1}\int_{4-x^2}^{0}1\,dy\,dx

because the upper curve must always be the lower limit

C.Use

\int_0^4\int_{-1}^{1}1\,dy\,dx

because

y

does not affect the boundary

D.Use

\int_{-1}^{1}\int_0^{4-x^2}(4-x^2)\,dy\,dx

because the height must appear as the integrand

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Vertical slices are especially convenient because for every $x$ from $-1$ to $1$ , the region extends continuously from $y=0$ to $y=4-x^2$ . Therefore one double integral describes the entire region without splitting. The integrand $1$ is essential because the goal is unweighted geometric area.

Q6. A student claims that the area between $y=x$ and $y=x^2$ from $x=0$ to $x=1$ is $\iint_R y\,dA$ . What is the fundamental error in this reasoning?

A.The student is computing a weighted quantity rather than ordinary area ✅

B.The student used

y

instead of

x

as the outer variable

C.The curves cannot enclose any region for

0\le x\le1

D.A double integral cannot be used for a region bounded by two curves

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For ordinary geometric area, each infinitesimal area element must be counted equally, which requires the integrand $1$ . Using $y$ weights different portions of the region differently and therefore calculates a different quantity. The bounds may describe the region correctly, but the integrand does not represent unweighted area.

Q7. Consider the region enclosed by $y=x$ and $y=4-x^2$ . A student writes $\int_{-2}^{1}\int_{x}^{4-x^2}1\,dy\,dx$ . Is this setup correct?

A.Yes, because the curves intersect at

x=-2

and

x=1

, and

4-x^2

is above

x

between them ✅

B.No, because the intersection values are

x=-1

and

x=2

C.No, because

x

must be the inner variable in every double integral

D.Yes, but only if the integrand is changed from

1

x

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Setting $x=4-x^2$ gives $x^2+x-4=0$ , whose roots are not $-2$ and $1$ . Thus the student's setup has incorrect intersection limits even though its general vertical-slice structure is reasonable. The correct $x$ -limits must come from solving the actual intersection equation before constructing the integral.

Q8. A graph shows a closed region whose vertical slice at $x=1$ extends from $y=2$ to $y=5$ , while its vertical slice at $x=3$ extends from $y=1$ to $y=4$ . If the region is described using vertical slices, which statement must be true?

A.The inner limits are determined by the lower and upper

y

-boundaries as functions of

x

✅

B.The inner limits must always be

0

and

1

C.The outer limits must be the

y

-coordinates

2

and

5

D.The integrand must equal the vertical height of each slice

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For a vertical-slice description, $x$ is fixed while $y$ varies from the lower boundary to the upper boundary. The examples show different lower and upper values at different $x$ -positions, confirming that these boundaries generally depend on $x$ . The integrand remains $1$ when calculating ordinary area.

Q9. A rectangular region has $0\le x\le5$ and $2\le y\le7$ . A student calculates its area as $\int_0^5\int_2^7 x\,dy\,dx$ . What does this calculation actually demonstrate?

A.It produces a weighted integral, not the rectangle's ordinary area ✅

B.It still produces the rectangle's area because

x

integrates to

1

C.It calculates the perimeter exactly

D.It is impossible to interpret because

x

is independent of

y

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The rectangle's ordinary area requires $\iint_R1\,dA$ , which equals $5\cdot5=25$ . Replacing $1$ with $x$ gives $\int_0^5\int_2^7x\,dy\,dx$ , which weights each vertical strip according to its $x$ -coordinate. The resulting value is therefore not the geometric area, even though the same region is being integrated.

Q10. A region is described by $0\le y\le2$ and $y^2\le x\le4$ . Which expression correctly calculates its area using horizontal slices?

\int_0^2\int_{y^2}^{4}1\,dx\,dy

✅

\int_0^2\int_{4}^{y^2}1\,dx\,dy

\int_0^4\int_0^2 1\,dx\,dy

\int_0^2\int_{y^2}^{4}y\,dx\,dy

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Horizontal slices fix $y$ and allow $x$ to vary across the region. The given description states that $y$ ranges from $0$ to $2$ , while $x$ extends from the left boundary $x=y^2$ to the right boundary $x=4$ . Since the desired quantity is area, the integrand is $1$ .

Q11. Two methods are proposed for a region bounded by $y=x^2$ , $y=2$ , and $x=0$ . Method I uses vertical slices and requires one integral. Method II uses horizontal slices and also requires one integral. Which conclusion is justified?

A.Both can produce the same area if each method describes exactly the same region ✅

B.The vertical method must be correct because

y=x^2

is written explicitly in terms of

x

C.The horizontal method must be incorrect because

x

is not the dependent variable

D.Only one integration order can ever represent a given planar region

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A planar region can often be described in more than one valid way. Vertical slices may use $x$ as the outer variable, while horizontal slices may use $y$ as the outer variable. If both sets of limits cover precisely the same points and the integrand is $1$ , both double integrals must yield the same geometric area.

Q12. A computer model describes a land parcel by $0\le x\le4$ and $x/2\le y\le4-x$ . Before evaluating its area, what should the analyst check first?

A.That the lower boundary does not exceed the upper boundary throughout the stated interval ✅

B.That the integrand should be

x+y

to account for the parcel's shape

C.That both boundary equations have the same slope

D.That the

x

-interval must always begin at the origin

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A valid region described by vertical slices requires the lower $y$ -boundary to be at or below the upper boundary. Here, $x/2\le4-x$ restricts the usable interval further than the stated $0\le x\le4$ . Checking this consistency before integration prevents calculating negative slice lengths or including points outside the intended parcel.

Q13. Let $R$ be the region inside the circle $x^2+y^2\le9$ . A student argues that using symmetry, $\int_0^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}1\,dy\,dx$ gives only half the area because $x$ is nonnegative. What is the correct evaluation strategy?

A.Double the integral because the displayed region is the right half of the circle ✅

B.The integral already gives the entire circle because the

y

-limits cover both halves

C.Subtract the lower

y

-limit because it is negative

D.Replace the integrand

1

2

and keep the same limits

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The bounds $0\le x\le3$ restrict the region to the right half of the circle, while the $y$ -limits extend from the lower semicircle to the upper semicircle. Therefore the integral represents exactly half of the circular area. By symmetry about the $y$ -axis, multiplying this result by $2$ gives the full area.

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