← Back to 15. Multiple Integrals Calculus

📝 Double Integrals in Polar Coordinates (14 MCQs)

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

←Previous Topic:Area Calculation as a Double Integral📝 Double Integrals in Polar CoordinatesDouble integrals in Simple Polar Regions:Next Topic→

What is Double Integrals in Polar Coordinates?

Definition:
In polar coordinates, $x = r\cos\theta$ , $y = r\sin\theta$ , and the area element becomes $dA = r \, dr \, d\theta$ . The double integral transforms to $\iint_R f(r\cos\theta, r\sin\theta) \, r \, dr \, d\theta$ .

Example:
To integrate $f(x,y) = x^2+y^2$ over the unit disk, we use $\int_0^{2\pi} \int_0^1 (r^2) \, r \, dr \, d\theta = \int_0^{2\pi} d\theta \int_0^1 r^3 \, dr = 2\pi \cdot \frac{1}{4} = \frac{\pi}{2}$ .

Reason:
Polar coordinates simplify integrals over circular or radially symmetric regions, avoiding complex square root boundaries that appear in Cartesian coordinates.

Easy

Medium

Hard

📝 All Double Integrals in Polar Coordinates MCQs

Q1. A region is described by $0\leq r\leq 3$ and $0\leq\theta\leq\pi/2$ . Which integral correctly represents the area of this region?

\int_0^{\pi/2}\int_0^3 r\,dr\,d\theta

✅

\int_0^3\int_0^{\pi/2} r\,d\theta\,dr

\int_0^{\pi/2}\int_0^3 dr\,d\theta

\int_0^3\int_0^{\pi/2} \theta\,d\theta\,dr

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: In polar coordinates, the differential area is $dA=r\,dr\,d\theta$ . The bounds describe a quarter-circle of radius $3$ , so the radial variable must range from $0$ to $3$ , while the angle ranges from $0$ to $\pi/2$ . Including the factor $r$ is essential because polar coordinates stretch area according to distance from the origin.

Q2. Why does a double integral over a polar-coordinate region require the factor $r$ in $dA=r\,dr\,d\theta$ , rather than simply using $dr\,d\theta$ ?

A.Because

r

represents the height of the region

B.Because an angular change

d\theta

corresponds to an arc length approximately

r\,d\theta

✅

C.Because

r

converts radians into degrees

D.Because

dr

already represents an area change

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A small polar element is approximately a rectangle whose radial width is $dr$ and whose tangential length is $r\,d\theta$ . Multiplying these dimensions gives $r\,dr\,d\theta$ . Omitting $r$ treats equal changes in angle as having equal physical width at every radius, which is geometrically incorrect.

Q3. A region is bounded by the circles $r=1$ and $r=4$ , with $0\leq\theta\leq\pi$ . Which description best identifies the region and its area integral?

A.A semicircular disk with

\int_0^\pi\int_0^4 r\,dr\,d\theta

B.An upper half-annulus with

\int_0^\pi\int_1^4 r\,dr\,d\theta

✅

C.A full annulus with

\int_0^{2\pi}\int_1^4 r\,dr\,d\theta

D.A quarter-annulus with

\int_0^{\pi/2}\int_1^4 r\,dr\,d\theta

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The radial bounds $1\leq r\leq4$ exclude the central disk of radius $1$ , while $0\leq\theta\leq\pi$ selects the upper half-plane. Therefore the region is a semicircular annulus. The integral must contain the polar area factor $r$ , giving $\int_0^\pi\int_1^4 r\,dr\,d\theta$ .

Q4. A student writes $\int_0^{2\pi}\int_0^2 r^2\,dr\,d\theta$ for the area of a disk of radius $2$ . What is the most important error in this setup?

A.The angular bounds should be

0\leq\theta\leq\pi

B.The radial bounds should begin at

1

C.The integrand should be

r

, not

r^2

✅

D.The angular variable must be integrated first

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For area, the integrand comes only from the Jacobian $dA=r\,dr\,d\theta$ , so the correct integrand is $r$ . The extra $r$ in $r^2$ changes the quantity being calculated and would produce a value different from the actual geometric area. The bounds themselves correctly describe the disk.

Q5. A circular region satisfies $x^2+y^2\leq 9$ , but only points in the first quadrant are included. Which polar-coordinate integral evaluates $\iint_R (x^2+y^2)\,dA$ ?

\int_0^{\pi/2}\int_0^3 r^2\,dr\,d\theta

\int_0^{\pi/2}\int_0^3 r^3\,dr\,d\theta

✅

\int_0^3\int_0^{\pi/2} r^2\,d\theta\,dr

\int_0^{\pi/2}\int_0^9 r\,dr\,d\theta

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Since $x^2+y^2=r^2$ , the function becomes $r^2$ in polar coordinates. The differential area contributes another factor $r$ , producing $r^3\,dr\,d\theta$ . The first-quadrant restriction gives $0\leq\theta\leq\pi/2$ , and the circle of radius $3$ gives $0\leq r\leq3$ .

Q6. A designer wants to integrate over the region inside $r=2\cos\theta$ and above the $x$ -axis. Which bounds correctly describe the region without including points below the axis?

0\leq\theta\leq\pi,\ 0\leq r\leq2\cos\theta

-\pi/2\leq\theta\leq\pi/2,\ 0\leq r\leq2\cos\theta

0\leq\theta\leq\pi/2,\ 0\leq r\leq2\cos\theta

✅

0\leq\theta\leq\pi,\ 0\leq r\leq2\sin\theta

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The curve $r=2\cos\theta$ represents a circle centered at $(1,0)$ with radius $1$ . The portion above the $x$ -axis corresponds to $0\leq\theta\leq\pi/2$ . Within those angles, $r$ runs from the origin to the circle, so $0\leq r\leq2\cos\theta$ .

Q7. A region lies between the curves $r=2$ and $r=4\cos\theta$ , where both curves intersect for $0\leq\theta\leq\pi/3$ . If the region is intended to contain points closer to the origin than the outer boundary, which radial description is appropriate?

2\leq r\leq4\cos\theta

✅

4\cos\theta\leq r\leq2

0\leq r\leq2

for all angles

0\leq r\leq4\cos\theta

for all angles

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: At a fixed angle in the stated range, the circle $r=2$ forms the inner boundary while $r=4\cos\theta$ forms the outer boundary. Thus the radial coordinate must begin at $2$ and end at $4\cos\theta$ . The angle restriction is important because the outer boundary must remain at least as large as the inner boundary.

Q8. A student claims that the integral $\int_0^{2\pi}\int_0^1 r\,dr\,d\theta$ represents the area of a unit circle but evaluates it as $1$ . Which reasoning best identifies the mistake?

A.The factor

r

should be removed

B.The radial upper limit should be

2

C.The student ignored the contribution from the full angular sweep ✅

D.The order of integration must be reversed

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The integral is correctly constructed for the unit disk. Evaluating the radial part gives $1/2$ , and the angular interval contributes $2\pi$ , producing $\pi$ . The value $1$ results from considering only the radial contribution and neglecting the full $0\leq\theta\leq2\pi$ angular range.

Q9. A graph shows a region bounded by the origin, the positive $x$ -axis, and the curve $r=3\sin\theta$ . Which setup correctly captures the entire region shown?

\int_0^\pi\int_0^{3\sin\theta} r\,dr\,d\theta

✅

\int_0^{\pi/2}\int_0^{3\sin\theta} r\,dr\,d\theta

\int_0^{2\pi}\int_0^{3\sin\theta} r\,dr\,d\theta

\int_0^\pi\int_0^3 \sin\theta\,dr\,d\theta

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The curve $r=3\sin\theta$ is a circle centered on the $y$ -axis, and the interval $0\leq\theta\leq\pi$ traces the entire circle while keeping $r\geq0$ . The radial coordinate extends from the origin to the curve. Therefore the first setup represents the complete region exactly.

Q10. A circular region centered at the origin is used to model heat distribution, with temperature $T(x,y)=x^2+y^2$ . If the radius is $5$ , which expression gives the total modeled heat contribution when density is proportional to $T$ ?

\int_0^{2\pi}\int_0^5 r\,dr\,d\theta

\int_0^{2\pi}\int_0^5 r^2\,dr\,d\theta

\int_0^{5}\int_0^{2\pi} r^2\,d\theta\,dr

\int_0^{2\pi}\int_0^5 r^3\,dr\,d\theta

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The temperature function becomes $T=r^2$ because $x^2+y^2=r^2$ . To obtain the total contribution over area, this function must be multiplied by $dA=r\,dr\,d\theta$ . Therefore the resulting integrand is $r^3$ , with $0\leq r\leq5$ and $0\leq\theta\leq2\pi$ .

Q11. Consider the integral $\int_0^{\pi/2}\int_0^{2} (1+r^2)r\,dr\,d\theta$ . Which interpretation is most accurate?

A.It computes only the area of a quarter-disk

B.It integrates a radial-dependent quantity over a quarter-disk ✅

C.It computes the circumference of a quarter-circle

D.It describes a full disk with a constant integrand

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The bounds describe a quarter-disk because the radius runs from $0$ to $2$ and the angle spans $90^\circ$ . The factor $1+r^2$ is an additional function being integrated over the region, while $r$ is the area Jacobian. Therefore the integral represents the accumulation of a radial-dependent quantity over that quarter-disk.

Q12. Two methods are proposed for evaluating the integral of $x^2+y^2$ over the disk $x^2+y^2\leq4$ . Method A uses Cartesian coordinates and Method B uses polar coordinates. Which statement best compares them?

A.Method A is invalid because circles cannot be integrated in Cartesian coordinates

B.Method B is generally simpler because both the region and integrand become radial ✅

C.Both methods must produce different values because their variables differ

D.Method B is invalid because

x^2+y^2

cannot be transformed

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Both coordinate systems can represent the same double integral, so they must give the same mathematical result when set up correctly. However, polar coordinates are more efficient here because the circular boundary becomes $0\leq r\leq2$ , while $x^2+y^2$ becomes $r^2$ . This greatly simplifies the setup.

Q13. A student uses $0\leq\theta\leq2\pi$ and $0\leq r\leq1+\cos\theta$ to integrate over a cardioid-like region. Another student argues that only $0\leq\theta\leq\pi$ is necessary because the curve is symmetric about the $x$ -axis. Which conclusion is correct for integrating over the entire region?

A.Only

0\leq\theta\leq\pi

is correct because symmetry always halves the integral

B.The full

0\leq\theta\leq2\pi

range is valid and naturally covers the entire region ✅

C.Neither range is valid because polar coordinates cannot represent the curve

D.The angle must range from

-\pi/2

\pi/2

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Symmetry does not automatically mean that the angular interval can be shortened unless the integral is explicitly doubled and the integrand and region permit that reduction. The interval $0\leq\theta\leq2\pi$ directly traces the entire region. Using only half the interval without compensating would generally omit part of the region.

Q14. For the region inside $r=2$ and outside $r=2\cos\theta$ , determine the most suitable angular strategy for covering the region once without unnecessary duplication.

A.Use

0\leq\theta\leq2\pi

with

2\cos\theta\leq r\leq2

B.Use

0\leq\theta\leq\pi/2

with

2\cos\theta\leq r\leq2

, then double the result

C.Use

-\pi/2\leq\theta\leq\pi/2

with

2\cos\theta\leq r\leq2

✅

D.Use

0\leq\theta\leq\pi

with

0\leq r\leq2

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The condition outside $r=2\cos\theta$ requires $r\geq2\cos\theta$ , while remaining inside $r=2$ . For angles where $\cos\theta$ is negative, the inner expression is negative and therefore imposes no additional restriction on nonnegative $r$ . The interval $-\pi/2\leq\theta\leq\pi/2$ handles the boundary-changing portion directly, while other angles require recognizing the full geometric structure rather than blindly applying one radial inequality.

🔗 Related Topics (MCQs)

📝 Double Integrals in Calculus MCQs 📝 Volume under surface double integral MCQs 📝 Double integral definition Riemann sums MCQs 📝 How to evaluate double integrals MCQs 📝 Fubini's theorem for double integrals MCQs

←Previous Topic:Area Calculation as a Double Integral📝 Double Integrals in Polar CoordinatesDouble integrals in Simple Polar Regions:Next Topic→