Multivariable Calculus Quiz: Area In Polar Coordinates
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Area In Polar CoordinatesQuestion 1 of 10

The area enclosed by one petal of the rose curve r=5sin⁡(2θ)r = 5\sin(2\theta) is:

25π8\frac{25\pi}{8}
25π4\frac{25\pi}{4}
25π16\frac{25\pi}{16}
25π2\frac{25\pi}{2}
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Area In Polar Coordinates

Practice Area In Polar Coordinates in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Area In Polar Coordinates, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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Question 1

The area enclosed by one petal of the rose curve r=5sin⁡(2θ)r = 5\sin(2\theta) is:

  1. 25π8\frac{25\pi}{8} (correct answer)
  2. 25π4\frac{25\pi}{4}
  3. 25π16\frac{25\pi}{16}
  4. 25π2\frac{25\pi}{2}
Explanation: One petal occurs when sin⁡(2θ)≥0\sin(2\theta) \geq 0, which happens for 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2} (first petal). The area is A=12∫0π/2(5sin⁡(2θ))2dθ=252∫0π/2sin⁡2(2θ)dθA = \frac{1}{2}\int_0^{\pi/2} (5\sin(2\theta))^2 d\theta = \frac{25}{2}\int_0^{\pi/2} \sin^2(2\theta) d\theta. Using sin⁡2(2θ)=1−cos⁡(4θ)2\sin^2(2\theta) = \frac{1-\cos(4\theta)}{2}: A=254∫0π/2(1−cos⁡(4θ))dθ=254[θ−sin⁡(4θ)4]0π/2=254⋅π2=25π8A = \frac{25}{4}\int_0^{\pi/2} (1-\cos(4\theta)) d\theta = \frac{25}{4}\left[\theta - \frac{\sin(4\theta)}{4}\right]_0^{\pi/2} = \frac{25}{4} \cdot \frac{\pi}{2} = \frac{25\pi}{8}. Choice B forgets the factor of 12\frac{1}{2} in the area formula. Choice C uses incorrect bounds [0,π/4][0, \pi/4]. Choice D compounds both errors from choices B and has an additional factor error.

Question 2

What is the area of a single petal of the rose curve given by the polar equation r=4cos⁡(2θ)r = 4\cos(2\theta)?

  1. 2π2\pi (correct answer)
  2. π\pi
  3. 4π4\pi
  4. 8π8\pi
Explanation: The curve r=4cos⁡(2θ)r = 4\cos(2\theta) is a rose curve with 2n=42n=4 petals. To find the area of one petal, we need to determine the bounds of integration that trace it out once. A petal is traced when rr goes from 00 to a maximum and back to 00. Let's consider the petal symmetric about the polar axis. r=0r=0 when cos⁡(2θ)=0\cos(2\theta)=0, which occurs when 2θ=±π/2,±3π/2,…2\theta = \pm \pi/2, \pm 3\pi/2, \dots. This implies θ=±π/4,±3π/4,…\theta = \pm \pi/4, \pm 3\pi/4, \dots. The petal on the positive polar axis is traced for θ\theta from −π/4-\pi/4 to π/4\pi/4. The area AA is given by: A=12∫−π/4π/4r2dθ=12∫−π/4π/4(4cos⁡(2θ))2dθA = \frac{1}{2} \int_{-\pi/4}^{\pi/4} r^2 d\theta = \frac{1}{2} \int_{-\pi/4}^{\pi/4} (4\cos(2\theta))^2 d\theta A=8∫−π/4π/4cos⁡2(2θ)dθA = 8 \int_{-\pi/4}^{\pi/4} \cos^2(2\theta) d\theta Using the identity cos⁡2(x)=1+cos⁡(2x)2\cos^2(x) = \frac{1+\cos(2x)}{2}, we get: A=8∫−π/4π/41+cos⁡(4θ)2dθ=4[θ+14sin⁡(4θ)]−π/4π/4A = 8 \int_{-\pi/4}^{\pi/4} \frac{1+\cos(4\theta)}{2} d\theta = 4 \left[ \theta + \frac{1}{4}\sin(4\theta) \right]_{-\pi/4}^{\pi/4} A=4[(π4+14sin⁡(π))−(−π4+14sin⁡(−π))]=4(π4−(−π4))=4(π2)=2πA = 4 \left[ (\frac{\pi}{4} + \frac{1}{4}\sin(\pi)) - (-\frac{\pi}{4} + \frac{1}{4}\sin(-\pi)) \right] = 4(\frac{\pi}{4} - (-\frac{\pi}{4})) = 4(\frac{\pi}{2}) = 2\pi

Question 3

Find the area of the region in the first quadrant bounded by the spiral r=θr=\theta, the polar axis, and the line θ=π/2\theta = \pi/2.

  1. π28\frac{\pi^2}{8}
  2. π324\frac{\pi^3}{24}
  3. π216\frac{\pi^2}{16}
  4. π348\frac{\pi^3}{48} (correct answer)
Explanation: When you encounter area problems in polar coordinates, you need to recognize that the area formula is A=12∫αβr2 dθA = \frac{1}{2}\int_{\alpha}^{\beta} r^2 \, d\theta, where the region is bounded by rays at angles α\alpha and β\beta. Here, you're finding the area swept out by the spiral r=θr = \theta from the polar axis (θ=0\theta = 0) to the line θ=π/2\theta = \pi/2. Setting up the integral: A=12∫0π/2θ2 dθA = \frac{1}{2}\int_{0}^{\pi/2} \theta^2 \, d\theta To evaluate this, use the power rule: ∫θ2 dθ=θ33\int \theta^2 \, d\theta = \frac{\theta^3}{3} So: A=12[θ33]0π/2=12⋅13[θ3]0π/2=16(π38−0)=π348A = \frac{1}{2} \left[ \frac{\theta^3}{3} \right]_0^{\pi/2} = \frac{1}{2} \cdot \frac{1}{3} \left[ \theta^3 \right]_0^{\pi/2} = \frac{1}{6} \left( \frac{\pi^3}{8} - 0 \right) = \frac{\pi^3}{48} This confirms answer D is correct. Answer A (π28\frac{\pi^2}{8}) results from incorrectly using r=θr = \theta directly instead of r2=θ2r^2 = \theta^2 in the area formula. Answer B (π324\frac{\pi^3}{24}) comes from forgetting the 12\frac{1}{2} factor in the polar area formula. Answer C (π216\frac{\pi^2}{16}) combines both errors: using rr instead of r2r^2 and making an integration mistake. Remember: polar area always involves r2r^2, and don't forget the 12\frac{1}{2} coefficient. These are the most common mistakes in polar area problems.

Question 4

The area of one petal of the rose curve r=acos⁡(3θ)r = a \cos(3\theta) is equal to π\pi. What is the value of the positive constant aa?

  1. 6\sqrt{6}
  2. 222\sqrt{2}
  3. 232\sqrt{3} (correct answer)
  4. 44
Explanation: The rose curve r=acos⁡(3θ)r = a \cos(3\theta) has 3 petals. One petal is traced as 3θ3\theta goes from −π/2-\pi/2 to π/2\pi/2, which means θ\theta goes from −π/6-\pi/6 to π/6\pi/6. The area AA of one petal is given by: A=12∫−π/6π/6(acos⁡(3θ))2dθ=a22∫−π/6π/6cos⁡2(3θ)dθA = \frac{1}{2} \int_{-\pi/6}^{\pi/6} (a \cos(3\theta))^2 d\theta = \frac{a^2}{2} \int_{-\pi/6}^{\pi/6} \cos^2(3\theta) d\theta Using the identity cos⁡2(x)=1+cos⁡(2x)2\cos^2(x) = \frac{1+\cos(2x)}{2}: A=a22∫−π/6π/61+cos⁡(6θ)2dθ=a24[θ+16sin⁡(6θ)]−π/6π/6A = \frac{a^2}{2} \int_{-\pi/6}^{\pi/6} \frac{1+\cos(6\theta)}{2} d\theta = \frac{a^2}{4} \left[ \theta + \frac{1}{6}\sin(6\theta) \right]_{-\pi/6}^{\pi/6} A=a24[(π6+16sin⁡(π))−(−π6+16sin⁡(−π))]=a24(π6+π6)=a24(2π6)=πa212A = \frac{a^2}{4} \left[ (\frac{\pi}{6} + \frac{1}{6}\sin(\pi)) - (-\frac{\pi}{6} + \frac{1}{6}\sin(-\pi)) \right] = \frac{a^2}{4} (\frac{\pi}{6} + \frac{\pi}{6}) = \frac{a^2}{4} (\frac{2\pi}{6}) = \frac{\pi a^2}{12} We are given that the area is π\pi. So, πa212=π\frac{\pi a^2}{12} = \pi. This simplifies to a2=12a^2 = 12, and since aa is a positive constant, a=12=23a = \sqrt{12} = 2\sqrt{3}.

Question 5

Find the area of the region enclosed by the inner loop of the limaçon r=1−2sin⁡θr = 1 - 2\sin\theta.

  1. π−332\pi - \frac{3\sqrt{3}}{2} (correct answer)
  2. 3π2−332\frac{3\pi}{2} - \frac{3\sqrt{3}}{2}
  3. 2π−332\pi - 3\sqrt{3}
  4. π+332\pi + \frac{3\sqrt{3}}{2}
Explanation: The inner loop of a limaçon is traced when rr is negative. First, we find the values of θ\theta for which r=0r=0: 1−2sin⁡θ=0  ⟹  sin⁡θ=1/21 - 2\sin\theta = 0 \implies \sin\theta = 1/2. In the interval [0,2π][0, 2\pi], this occurs at θ=π/6\theta = \pi/6 and θ=5π/6\theta = 5\pi/6. For θ\theta between π/6\pi/6 and 5π/65\pi/6, sin⁡θ>1/2\sin\theta > 1/2, so r=1−2sin⁡θ<0r = 1-2\sin\theta < 0. This interval traces the inner loop. The area AA is: A=12∫π/65π/6(1−2sin⁡θ)2dθA = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 2\sin\theta)^2 d\theta A=12∫π/65π/6(1−4sin⁡θ+4sin⁡2θ)dθA = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 4\sin\theta + 4\sin^2\theta) d\theta Using the identity sin⁡2θ=1−cos⁡(2θ)2\sin^2\theta = \frac{1-\cos(2\theta)}{2}: A=12∫π/65π/6(1−4sin⁡θ+41−cos⁡(2θ)2)dθ=12∫π/65π/6(3−4sin⁡θ−2cos⁡(2θ))dθA = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 4\sin\theta + 4\frac{1-\cos(2\theta)}{2}) d\theta = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (3 - 4\sin\theta - 2\cos(2\theta)) d\theta A=12[3θ+4cos⁡θ−sin⁡(2θ)]π/65π/6A = \frac{1}{2} [3\theta + 4\cos\theta - \sin(2\theta)]_{\pi/6}^{5\pi/6} A=12[(35π6+4cos⁡(5π6)−sin⁡(5π3))−(3π6+4cos⁡(π6)−sin⁡(π3))]A = \frac{1}{2} [(3\frac{5\pi}{6} + 4\cos(\frac{5\pi}{6}) - \sin(\frac{5\pi}{3})) - (3\frac{\pi}{6} + 4\cos(\frac{\pi}{6}) - \sin(\frac{\pi}{3}))] A=12[(5π2−432−(−32))−(π2+432−32)]=12[(5π2−332)−(π2+332)]=12[2π−33]=π−332A = \frac{1}{2} [(\frac{5\pi}{2} - 4\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})) - (\frac{\pi}{2} + 4\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2})] = \frac{1}{2} [(\frac{5\pi}{2} - \frac{3\sqrt{3}}{2}) - (\frac{\pi}{2} + \frac{3\sqrt{3}}{2})] = \frac{1}{2} [2\pi - 3\sqrt{3}] = \pi - \frac{3\sqrt{3}}{2}

Question 6

Find the area of one loop of the lemniscate given by the equation r2=9cos⁡(2θ)r^2 = 9\cos(2\theta).

  1. 9π2\frac{9\pi}{2}
  2. 99
  3. 9π4\frac{9\pi}{4}
  4. 92\frac{9}{2} (correct answer)
Explanation: When you encounter a lemniscate (figure-eight curve) in polar coordinates, you're working with area calculations that require careful attention to the curve's symmetry and domain. For the lemniscate r2=9cos⁡(2θ)r^2 = 9\cos(2\theta), you first need to determine where the curve exists. Since r2≥0r^2 \geq 0, we need cos⁡(2θ)≥0\cos(2\theta) \geq 0, which occurs when −π4≤θ≤π4-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} for one complete loop. The area formula in polar coordinates is A=12∫αβr2 dθA = \frac{1}{2}\int_{\alpha}^{\beta} r^2 \, d\theta. Since r2=9cos⁡(2θ)r^2 = 9\cos(2\theta), the area of one loop becomes: A=12∫−π/4π/49cos⁡(2θ) dθ=92∫−π/4π/4cos⁡(2θ) dθA = \frac{1}{2}\int_{-\pi/4}^{\pi/4} 9\cos(2\theta) \, d\theta = \frac{9}{2}\int_{-\pi/4}^{\pi/4} \cos(2\theta) \, d\theta Evaluating: A=92[sin⁡(2θ)2]−π/4π/4=94[sin⁡(π/2)−sin⁡(−π/2)]=94[1−(−1)]=92A = \frac{9}{2} \left[\frac{\sin(2\theta)}{2}\right]_{-\pi/4}^{\pi/4} = \frac{9}{4}[\sin(\pi/2) - \sin(-\pi/2)] = \frac{9}{4}[1 - (-1)] = \frac{9}{2} Option A (9π2\frac{9\pi}{2}) incorrectly includes π\pi in the final answer, likely from confusing this with a circular area formula. Option B (99) doubles the correct answer, possibly from calculating both loops. Option C (9π4\frac{9\pi}{4}) combines both errors—including π\pi and using incorrect integration limits. Remember: lemniscates have restricted domains where r2≥0r^2 \geq 0, and the area formula directly uses r2r^2, making the integration straightforward once you identify the correct limits.

Question 7

Find the area of the region defined by the polar inequalities 1≤r≤1+cos⁡θ1 \le r \le 1+\cos\theta.

  1. 2+π22 + \frac{\pi}{2}
  2. 2+π42 + \frac{\pi}{4} (correct answer)
  3. π2\frac{\pi}{2}
  4. 3π2\frac{3\pi}{2}
Explanation: The region is bounded by the circle r=1r=1 and the cardioid r=1+cos⁡θr=1+\cos\theta. The inequality 1≤1+cos⁡θ1 \le 1+\cos\theta implies cos⁡θ≥0\cos\theta \ge 0. This condition holds for θ∈[−π/2,π/2]\theta \in [-\pi/2, \pi/2]. So, these are the bounds for our integration. The area is the difference between the area of the outer curve and the inner curve over this interval. A=12∫−π/2π/2((1+cos⁡θ)2−12)dθA = \frac{1}{2} \int_{-\pi/2}^{\pi/2} ((1+\cos\theta)^2 - 1^2) d\theta By symmetry, we can integrate from 00 to π/2\pi/2 and double the result. A=∫0π/2((1+2cos⁡θ+cos⁡2θ)−1)dθ=∫0π/2(2cos⁡θ+cos⁡2θ)dθA = \int_{0}^{\pi/2} ((1+2\cos\theta+\cos^2\theta) - 1) d\theta = \int_{0}^{\pi/2} (2\cos\theta + \cos^2\theta) d\theta Using the identity cos⁡2θ=1+cos⁡(2θ)2\cos^2\theta = \frac{1+\cos(2\theta)}{2}: A=∫0π/2(2cos⁡θ+12+12cos⁡(2θ))dθA = \int_{0}^{\pi/2} (2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos(2\theta)) d\theta A=[2sin⁡θ+θ2+14sin⁡(2θ)]0π/2A = [2\sin\theta + \frac{\theta}{2} + \frac{1}{4}\sin(2\theta)]_{0}^{\pi/2} A=(2sin⁡(π/2)+π/22+14sin⁡(π))−(0)=2(1)+π4+0=2+π4A = (2\sin(\pi/2) + \frac{\pi/2}{2} + \frac{1}{4}\sin(\pi)) - (0) = 2(1) + \frac{\pi}{4} + 0 = 2 + \frac{\pi}{4} Distractor C results from incorrectly integrating over [0,2π][0, 2\pi] and making an error, which gives the area of the cardioid minus the area of the circle over the whole domain. The integral of 2cos⁡θ2\cos\theta over [0,2π][0, 2\pi] is 0, leaving 12∫02πcos⁡2θdθ=π/2\frac{1}{2} \int_0^{2\pi} \cos^2\theta d\theta = \pi/2.

Question 8

The total area enclosed by the curve r2=9cos⁡(2θ)r^2 = 9\cos(2\theta) is:

  1. 99 (correct answer)
  2. 9π2\frac{9\pi}{2}
  3. 1818
  4. 9π4\frac{9\pi}{4}
Explanation: This is a lemniscate (figure-eight curve). The curve exists only when cos⁡(2θ)≥0\cos(2\theta) \geq 0, which occurs for −π4≤θ≤π4-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} and 3π4≤θ≤5π4\frac{3\pi}{4} \leq \theta \leq \frac{5\pi}{4}. Since r=3cos⁡(2θ)r = 3\sqrt{\cos(2\theta)}, the total area is A=2⋅12∫−π/4π/49cos⁡(2θ)dθ=9∫−π/4π/4cos⁡(2θ)dθ=9[sin⁡(2θ)2]−π/4π/4=9⋅12[sin⁡(π/2)−sin⁡(−π/2)]=9⋅12[1−(−1)]=9A = 2 \cdot \frac{1}{2}\int_{-\pi/4}^{\pi/4} 9\cos(2\theta) d\theta = 9\int_{-\pi/4}^{\pi/4} \cos(2\theta) d\theta = 9\left[\frac{\sin(2\theta)}{2}\right]_{-\pi/4}^{\pi/4} = 9 \cdot \frac{1}{2}[\sin(\pi/2) - \sin(-\pi/2)] = 9 \cdot \frac{1}{2}[1 - (-1)] = 9. Choice B incorrectly uses the standard area formula without accounting for the r2r^2 form. Choice C doubles the answer incorrectly. Choice D uses wrong bounds or formula.

Question 9

Find the area enclosed by the curve r=2+cos⁡(3θ)r = 2 + \cos(3\theta) for 0≤θ≤2π0 \leq \theta \leq 2\pi.

  1. 9π2\frac{9\pi}{2} (correct answer)
  2. 7π2\frac{7\pi}{2}
  3. 5π2\frac{5\pi}{2}
  4. 11π2\frac{11\pi}{2}
Explanation: The area is given by A=12∫02πr2 dθ=12∫02π(2+cos⁡(3θ))2 dθA = \frac{1}{2}\int_0^{2\pi} r^2 \, d\theta = \frac{1}{2}\int_0^{2\pi} (2 + \cos(3\theta))^2 \, d\theta. Expanding: (2+cos⁡(3θ))2=4+4cos⁡(3θ)+cos⁡2(3θ)(2 + \cos(3\theta))^2 = 4 + 4\cos(3\theta) + \cos^2(3\theta). Using cos⁡2(3θ)=1+cos⁡(6θ)2\cos^2(3\theta) = \frac{1 + \cos(6\theta)}{2}, we get A=12∫02π(4+4cos⁡(3θ)+1+cos⁡(6θ)2)dθ=12[4θ+4sin⁡(3θ)3+θ2+sin⁡(6θ)12]02π=12(8π+π)=9π2A = \frac{1}{2}\int_0^{2\pi} \left(4 + 4\cos(3\theta) + \frac{1 + \cos(6\theta)}{2}\right) d\theta = \frac{1}{2}\left[4\theta + \frac{4\sin(3\theta)}{3} + \frac{\theta}{2} + \frac{\sin(6\theta)}{12}\right]_0^{2\pi} = \frac{1}{2}(8\pi + \pi) = \frac{9\pi}{2}. Choice B results from forgetting the cos⁡2\cos^2 term. Choice C comes from using cos⁡2(3θ)=12\cos^2(3\theta) = \frac{1}{2} instead of the correct identity. Choice D adds an extra π\pi term incorrectly.

Question 10

The region bounded by r=1+cos⁡θr = 1 + \cos\theta (cardioid) has area:

  1. 3π2\frac{3\pi}{2} (correct answer)
  2. π\pi
  3. 2π2\pi
  4. π2\frac{\pi}{2}
Explanation: The area is A=12∫02π(1+cos⁡θ)2dθ=12∫02π(1+2cos⁡θ+cos⁡2θ)dθA = \frac{1}{2}\int_0^{2\pi} (1 + \cos\theta)^2 d\theta = \frac{1}{2}\int_0^{2\pi} (1 + 2\cos\theta + \cos^2\theta) d\theta. Using cos⁡2θ=1+cos⁡(2θ)2\cos^2\theta = \frac{1+\cos(2\theta)}{2}: A=12∫02π(1+2cos⁡θ+1+cos⁡(2θ)2)dθ=12∫02π(32+2cos⁡θ+cos⁡(2θ)2)dθA = \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \frac{1+\cos(2\theta)}{2}\right) d\theta = \frac{1}{2}\int_0^{2\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{\cos(2\theta)}{2}\right) d\theta. Evaluating: A=12[3θ2+2sin⁡θ+sin⁡(2θ)4]02π=12⋅3⋅2π2=3π2A = \frac{1}{2}\left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin(2\theta)}{4}\right]_0^{2\pi} = \frac{1}{2} \cdot \frac{3 \cdot 2\pi}{2} = \frac{3\pi}{2}. Choice B forgets the cos⁡2θ\cos^2\theta term. Choice C omits the 12\frac{1}{2} factor. Choice D uses incorrect bounds or makes a computational error.