Multivariable Calculus Quiz: Double Integrals In Polar
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Double Integrals In PolarQuestion 1 of 17

The probability density function for the location of a particle on the xyxy-plane is given by f(x,y)=Ce−(x2+y2)f(x,y) = C e^{-(x^2+y^2)} for all (x,y)(x,y). What is the probability that the particle is located within the disk x2+y2≤4x^2+y^2 \le 4?

1−e−21 - e^{-2}
e−4e^{-4}
π(1−e−4)\pi(1 - e^{-4})
1−e−41 - e^{-4}
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Double Integrals In Polar

Practice Double Integrals In Polar 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 Double Integrals In Polar, 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 probability density function for the location of a particle on the xyxy-plane is given by f(x,y)=Ce−(x2+y2)f(x,y) = C e^{-(x^2+y^2)} for all (x,y)(x,y). What is the probability that the particle is located within the disk x2+y2≤4x^2+y^2 \le 4?

  1. 1−e−21 - e^{-2}
  2. e−4e^{-4}
  3. π(1−e−4)\pi(1 - e^{-4})
  4. 1−e−41 - e^{-4} (correct answer)
Explanation: When you encounter a probability density function over a region, you need to integrate the function over that region to find the probability. However, since this is a probability density function, you must first ensure it's properly normalized. To find the probability that the particle lies within the disk x2+y2≤4x^2 + y^2 \leq 4, you need to evaluate ∫∫x2+y2≤4f(x,y) dA\int\int_{x^2+y^2 \leq 4} f(x,y) \, dA. Since the function and region both involve x2+y2x^2 + y^2, convert to polar coordinates where x2+y2=r2x^2 + y^2 = r^2, dA=r dr dθdA = r \, dr \, d\theta, and the disk becomes 0≤r≤20 \leq r \leq 2, 0≤θ≤2π0 \leq \theta \leq 2\pi. The integral becomes: ∫02π∫02Ce−r2⋅r dr dθ=2πC∫02re−r2 dr\int_0^{2\pi} \int_0^2 Ce^{-r^2} \cdot r \, dr \, d\theta = 2\pi C \int_0^2 re^{-r^2} \, dr Using substitution u=−r2u = -r^2, du=−2r drdu = -2r \, dr: 2πC∫02re−r2 dr=πC∫0−4eu du=πC(1−e−4)2\pi C \int_0^2 re^{-r^2} \, dr = \pi C \int_0^{-4} e^u \, du = \pi C(1 - e^{-4}) Since f(x,y)f(x,y) is a probability density function, it must integrate to 1 over all space. By similar calculation over the entire plane (with limits r:0→∞r: 0 \to \infty), we get C=1πC = \frac{1}{\pi}. Therefore, the probability is π⋅1π⋅(1−e−4)=1−e−4\pi \cdot \frac{1}{\pi} \cdot (1 - e^{-4}) = 1 - e^{-4}, which is answer D. Answer A uses the wrong exponent (-2 instead of -4). Answer B gives e−4e^{-4} instead of 1−e−41 - e^{-4}. Answer C fails to account for the normalization constant. Key strategy: Always check if the density function is normalized, and use polar coordinates when dealing with circular regions involving x2+y2x^2 + y^2.

Question 2

Consider the double integral ∬R(x2+y2) dA\iint_R (x^2 + y^2) \, dA where RR is the region bounded by the circles x2+y2=1x^2 + y^2 = 1 and x2+y2=9x^2 + y^2 = 9, and the rays θ=π6\theta = \frac{\pi}{6} and θ=π3\theta = \frac{\pi}{3}. After converting to polar coordinates, which of the following represents the correct setup?

  1. ∫π/6π/3∫13r2⋅r dr dθ\int_{\pi/6}^{\pi/3} \int_1^3 r^2 \cdot r \, dr \, d\theta (correct answer)
  2. ∫π/6π/3∫19r2⋅r dr dθ\int_{\pi/6}^{\pi/3} \int_1^9 r^2 \cdot r \, dr \, d\theta
  3. ∫13∫π/6π/3r2⋅r dθ dr\int_1^3 \int_{\pi/6}^{\pi/3} r^2 \cdot r \, d\theta \, dr
  4. ∫π/6π/3∫13(rcos⁡θ)2+(rsin⁡θ)2 dr dθ\int_{\pi/6}^{\pi/3} \int_1^3 (r\cos\theta)^2 + (r\sin\theta)^2 \, dr \, d\theta
Explanation: In polar coordinates, x2+y2=r2x^2 + y^2 = r^2 and dA=r dr dθdA = r \, dr \, d\theta. The region has 1≤r≤31 \leq r \leq 3 (not 9, since r=x2+y2r = \sqrt{x^2 + y^2}) and π6≤θ≤π3\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}. The integrand becomes r2⋅r=r3r^2 \cdot r = r^3. Choice B incorrectly uses r=9r = 9 instead of r=3r = 3. Choice C has the correct limits but wrong order of integration for the standard setup. Choice D doesn't simplify the integrand and omits the Jacobian rr.

Question 3

The integral ∫0a∫0a2−x2f(x2+y2) dy dx\int_0^a \int_0^{\sqrt{a^2-x^2}} f(x^2+y^2) \, dy \, dx is converted to polar coordinates. If the resulting integral is ∫0α∫0af(r2)r dr dθ\int_0^{\alpha} \int_0^a f(r^2) r \, dr \, d\theta, what is α\alpha?

  1. π4\frac{\pi}{4}
  2. π2\frac{\pi}{2} (correct answer)
  3. π\pi
  4. 2π2\pi
Explanation: The original region is described by 0≤x≤a0 \leq x \leq a and 0≤y≤a2−x20 \leq y \leq \sqrt{a^2-x^2}. The constraint y≤a2−x2y \leq \sqrt{a^2-x^2} means y2≤a2−x2y^2 \leq a^2-x^2, so x2+y2≤a2x^2+y^2 \leq a^2. Combined with x≥0x \geq 0 and y≥0y \geq 0, this describes a quarter-circle of radius aa in the first quadrant. In polar coordinates, this corresponds to 0≤r≤a0 \leq r \leq a and 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}. Therefore, α=π2\alpha = \frac{\pi}{2}. Choice A would represent only an eighth of the circle. Choice C would be a semicircle. Choice D would be the full circle.

Question 4

Consider evaluating ∫01∫01−y2e−(x2+y2) dx dy\int_0^{1} \int_0^{\sqrt{1-y^2}} e^{-(x^2+y^2)} \, dx \, dy using polar coordinates. After the conversion and evaluation of the inner integral, what expression must be integrated with respect to θ\theta?

  1. 12(1−e−1)cos⁡θ\frac{1}{2}(1-e^{-1})\cos\theta
  2. 12(e−1−1)\frac{1}{2}(e^{-1}-1)
  3. 12(1−e−1)\frac{1}{2}(1-e^{-1}) (correct answer)
  4. 12(1−e−1)sin⁡θ\frac{1}{2}(1-e^{-1})\sin\theta
Explanation: When you encounter a double integral with x2+y2x^2 + y^2 in the integrand and a region that suggests circular boundaries, converting to polar coordinates is usually the most efficient approach. The region of integration here is bounded by x=1−y2x = \sqrt{1-y^2}, which describes a quarter-circle of radius 1 in the first quadrant. In polar coordinates, we have x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, and dx dy=r dr dθdx\,dy = r\,dr\,d\theta. The region becomes 0≤r≤10 \leq r \leq 1 and 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}. The integral transforms to: ∫0π/2∫01e−r2⋅r dr dθ\int_0^{\pi/2} \int_0^{1} e^{-r^2} \cdot r \, dr \, d\theta Evaluating the inner integral with respect to rr: Let u=−r2u = -r^2, so du=−2r drdu = -2r\,dr, giving us r dr=−12dur\,dr = -\frac{1}{2}du. When r=0r = 0, u=0u = 0; when r=1r = 1, u=−1u = -1. This yields: ∫01re−r2 dr=−12∫0−1eu du=−12[eu]0−1=−12(e−1−1)=12(1−e−1)\int_0^{1} re^{-r^2} \, dr = -\frac{1}{2}\int_0^{-1} e^u \, du = -\frac{1}{2}[e^u]_0^{-1} = -\frac{1}{2}(e^{-1} - 1) = \frac{1}{2}(1 - e^{-1}) This expression is independent of θ\theta, confirming answer C. Answer A incorrectly includes cos⁡θ\cos\theta, D incorrectly includes sin⁡θ\sin\theta, and B has the wrong sign (e−1−1e^{-1} - 1 instead of 1−e−11 - e^{-1}). Strategy tip: When the inner integral in polar coordinates yields a constant (no θ\theta dependence), that constant gets integrated over the θ\theta range. Always check whether your result depends on the angular variable.

Question 5

Consider the integral ∫−22∫−4−x24−x2x2+y2 dy dx\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2+y^2} \, dy \, dx. After converting to polar coordinates and evaluating, what is the result?

  1. 4π3\frac{4\pi}{3}
  2. 8π3\frac{8\pi}{3}
  3. 32π3\frac{32\pi}{3}
  4. 16π3\frac{16\pi}{3} (correct answer)
Explanation: When you encounter a double integral with limits that describe a circle and an integrand involving x2+y2\sqrt{x^2+y^2}, polar coordinates are your best tool. The region −2≤x≤2-2 \leq x \leq 2 and −4−x2≤y≤4−x2-\sqrt{4-x^2} \leq y \leq \sqrt{4-x^2} describes a disk of radius 2 centered at the origin, since x2+y2≤4x^2 + y^2 \leq 4. Converting to polar coordinates: x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, and dx dy=r dr dθdx\,dy = r\,dr\,d\theta. The integrand x2+y2\sqrt{x^2+y^2} becomes simply rr. The region becomes 0≤r≤20 \leq r \leq 2 and 0≤θ≤2π0 \leq \theta \leq 2\pi. The integral transforms to: ∫02π∫02r⋅r dr dθ=∫02π∫02r2 dr dθ\int_0^{2\pi} \int_0^2 r \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r^2 \, dr \, d\theta Evaluating: ∫02π[r33]02dθ=∫02π83dθ=83⋅2π=16π3\int_0^{2\pi} \left[\frac{r^3}{3}\right]_0^2 d\theta = \int_0^{2\pi} \frac{8}{3} d\theta = \frac{8}{3} \cdot 2\pi = \frac{16\pi}{3} Looking at the wrong answers: Choice A (4π3\frac{4\pi}{3}) results from forgetting the extra rr from the Jacobian. Choice B (8π3\frac{8\pi}{3}) comes from integrating over only half the circle (00 to π\pi instead of 00 to 2π2\pi). Choice C (32π3\frac{32\pi}{3}) occurs if you incorrectly use r3r^3 as your integrand instead of r2r^2. The answer is D. Key strategy: Always remember that converting to polar coordinates requires the Jacobian factor rr, making your integrand r⋅x2+y2=r2r \cdot \sqrt{x^2+y^2} = r^2, not just rr.

Question 6

A student sets up the polar integral ∫02π∫0sin⁡θr3 dr dθ\int_0^{2\pi} \int_0^{\sin\theta} r^3 \, dr \, d\theta but realizes the limits are incorrect because sin⁡θ\sin\theta becomes negative. What is the correct way to handle this integral?

  1. 2∫0π∫0sin⁡θr3 dr dθ2\int_0^{\pi} \int_0^{\sin\theta} r^3 \, dr \, d\theta by symmetry
  2. ∫02π∫0∣sin⁡θ∣r3 dr dθ\int_0^{2\pi} \int_0^{|\sin\theta|} r^3 \, dr \, d\theta using absolute value
  3. ∫0π∫0sin⁡θr3 dr dθ+∫π2π∫0−sin⁡θr3 dr dθ\int_0^{\pi} \int_0^{\sin\theta} r^3 \, dr \, d\theta + \int_{\pi}^{2\pi} \int_0^{-\sin\theta} r^3 \, dr \, d\theta
  4. ∫0π∫0sin⁡θr3 dr dθ\int_0^{\pi} \int_0^{\sin\theta} r^3 \, dr \, d\theta since sin⁡θ≥0\sin\theta \geq 0 on [0,π][0,\pi] (correct answer)
Explanation: When working with polar integrals, you must ensure that your radial limits make physical sense. The radius rr represents a distance, so it must always be non-negative. This is the key insight for handling bounds that involve trigonometric functions. The correct approach is D: ∫0π∫0sin⁡θr3 dr dθ\int_0^{\pi} \int_0^{\sin\theta} r^3 \, dr \, d\theta. Since sin⁡θ≥0\sin\theta \geq 0 for θ∈[0,π]\theta \in [0,\pi], this integral has valid bounds throughout the entire domain. The region being integrated is likely the upper half of a circle or rose petal where rr ranges from 0 to sin⁡θ\sin\theta. A is incorrect because doubling the integral assumes the integrand has the same sign over both halves of the region, which isn't necessarily true for r3r^3. The symmetry argument fails here. B uses absolute value, which seems logical but actually changes the geometry of the region. When sin⁡θ<0\sin\theta < 0, using ∣sin⁡θ∣|\sin\theta| means you're integrating over a different region than originally intended, potentially double-counting areas. C attempts to split the integral but makes an error in the second part. Using −sin⁡θ-\sin\theta as an upper limit when sin⁡θ<0\sin\theta < 0 does give a positive bound, but this approach unnecessarily complicates the problem and risks computational errors. Study tip: When you encounter negative bounds in polar coordinates, first check if restricting the angular domain to where the bound is naturally positive solves your problem. This is often the intended approach and avoids the complications of absolute values or integral splitting.

Question 7

To evaluate ∬Dxy dA\iint_D xy \, dA where DD is the region in the first quadrant bounded by x2+y2=1x^2+y^2=1, x2+y2=9x^2+y^2=9, y=xy=x, and y=x3y=x\sqrt{3}, a student converts to polar coordinates. What is the correct setup?

  1. ∫π/4π/3∫13r2cos⁡θsin⁡θ dr dθ\int_{\pi/4}^{\pi/3} \int_1^3 r^2\cos\theta\sin\theta \, dr \, d\theta
  2. ∫π/6π/4∫13r3cos⁡θsin⁡θ dr dθ\int_{\pi/6}^{\pi/4} \int_1^3 r^3\cos\theta\sin\theta \, dr \, d\theta
  3. ∫π/4π/3∫13r3cos⁡θsin⁡θ dr dθ\int_{\pi/4}^{\pi/3} \int_1^3 r^3\cos\theta\sin\theta \, dr \, d\theta (correct answer)
  4. ∫π/3π/4∫13r3cos⁡θsin⁡θ dr dθ\int_{\pi/3}^{\pi/4} \int_1^3 r^3\cos\theta\sin\theta \, dr \, d\theta
Explanation: When converting double integrals to polar coordinates, you need to carefully transform both the integrand and the region of integration. The key steps are: convert x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta, include the Jacobian rr, and determine the correct bounds. For the integrand xyxy, substituting gives xy=rcos⁡θ⋅rsin⁡θ=r2cos⁡θsin⁡θxy = r\cos\theta \cdot r\sin\theta = r^2\cos\theta\sin\theta. Including the Jacobian rr from the coordinate transformation, the integrand becomes r3cos⁡θsin⁡θr^3\cos\theta\sin\theta. Now for the region bounds. The circles x2+y2=1x^2 + y^2 = 1 and x2+y2=9x^2 + y^2 = 9 correspond to r=1r = 1 and r=3r = 3. For the angular bounds, convert the lines: y=xy = x gives tan⁡θ=1\tan\theta = 1, so θ=π/4\theta = \pi/4. The line y=x3y = x\sqrt{3} gives tan⁡θ=3\tan\theta = \sqrt{3}, so θ=π/3\theta = \pi/3. Since we're in the first quadrant moving counterclockwise from the steeper angle to the less steep angle, θ\theta goes from π/4\pi/4 to π/3\pi/3. Choice C correctly has r3cos⁡θsin⁡θr^3\cos\theta\sin\theta with bounds π/4\pi/4 to π/3\pi/3. Choice A uses r2r^2 instead of r3r^3, missing the Jacobian factor. Choice B has the wrong angular bounds (π/6\pi/6 to π/4\pi/4). Choice D reverses the order of the angular limits, which would give a negative result. Remember: always include the Jacobian rr when converting to polar coordinates, and sketch the region to verify your angle bounds match the geometry.

Question 8

Find the volume of the solid lying under the surface z=x2+y2z = \sqrt{x^2+y^2} and above the region RR in the xyxy-plane enclosed by one leaf of the rose r=4sin⁡(2θ)r=4\sin(2\theta) in the first quadrant.

  1. 2π2\pi
  2. 649\frac{64}{9}
  3. 1289\frac{128}{9} (correct answer)
  4. 2569\frac{256}{9}
Explanation: The surface is z=rz=r. The region RR corresponding to the leaf in the first quadrant is traced for 0≤θ≤π/20 \le \theta \le \pi/2. The volume is given by the double integral of zz over RR. The area element is dA=r dr dθdA = r \, dr \, d\theta. V=∬Rz dA=∫0π/2∫04sin⁡(2θ)r⋅r dr dθ=∫0π/2∫04sin⁡(2θ)r2 dr dθV = \iint_R z \, dA = \int_0^{\pi/2} \int_0^{4\sin(2\theta)} r \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_0^{4\sin(2\theta)} r^2 \, dr \, d\theta The inner integral is ∫04sin⁡(2θ)r2 dr=[r33]04sin⁡(2θ)=643sin⁡3(2θ)\int_0^{4\sin(2\theta)} r^2 \, dr = [\frac{r^3}{3}]_0^{4\sin(2\theta)} = \frac{64}{3}\sin^3(2\theta). The outer integral is: V=643∫0π/2sin⁡3(2θ) dθV = \frac{64}{3} \int_0^{\pi/2} \sin^3(2\theta) \, d\theta Let u=2θu=2\theta, du=2dθdu=2d\theta. The integral becomes 643∫0πsin⁡3(u)du2=323∫0π(1−cos⁡2u)sin⁡u du\frac{64}{3} \int_0^{\pi} \sin^3(u) \frac{du}{2} = \frac{32}{3} \int_0^{\pi} (1-\cos^2u)\sin u \, du. Let v=cos⁡u,dv=−sin⁡u duv=\cos u, dv=-\sin u \, du. V=323∫1−1(1−v2)(−dv)=323∫−11(1−v2) dv=323[v−v33]−11=323[(1−13)−(−1+13)]=323[23−(−23)]=323⋅43=1289V = \frac{32}{3} \int_{1}^{-1} (1-v^2) (-dv) = \frac{32}{3} \int_{-1}^{1} (1-v^2) \, dv = \frac{32}{3} [v-\frac{v^3}{3}]_{-1}^1 = \frac{32}{3} [(1-\frac{1}{3}) - (-1+\frac{1}{3})] = \frac{32}{3} [\frac{2}{3} - (-\frac{2}{3})] = \frac{32}{3} \cdot \frac{4}{3} = \frac{128}{9}

Question 9

Evaluate the improper integral ∬R1(1+x2+y2)2 dA\iint_R \frac{1}{(1+x^2+y^2)^2} \, dA where RR is the first quadrant of the xyxy-plane.

  1. π8\frac{\pi}{8}
  2. π4\frac{\pi}{4} (correct answer)
  3. π28\frac{\pi^2}{8}
  4. The integral diverges.
Explanation: In polar coordinates, the first quadrant is described by 0≤r<∞0 \le r < \infty and 0≤θ≤π/20 \le \theta \le \pi/2. The integrand becomes 1(1+r2)2\frac{1}{(1+r^2)^2}, and dA=r dr dθdA = r \, dr \, d\theta. The integral is: I=∫0π/2∫0∞1(1+r2)2r dr dθI = \int_0^{\pi/2} \int_0^{\infty} \frac{1}{(1+r^2)^2} r \, dr \, d\theta First, evaluate the inner improper integral: ∫0∞r(1+r2)2 dr\int_0^{\infty} \frac{r}{(1+r^2)^2} \, dr. Use the substitution u=1+r2u = 1+r^2, so du=2r drdu = 2r \, dr. As r→0r \to 0, u→1u \to 1. As r→∞r \to \infty, u→∞u \to \infty. ∫1∞1u2du2=12[−1u]1∞=12(lim⁡b→∞−1b−(−11))=12(0+1)=12\int_1^{\infty} \frac{1}{u^2} \frac{du}{2} = \frac{1}{2} \left[ -\frac{1}{u} \right]_1^{\infty} = \frac{1}{2} \left( \lim_{b \to \infty} -\frac{1}{b} - (-\frac{1}{1}) \right) = \frac{1}{2}(0+1) = \frac{1}{2} Now, evaluate the outer integral: I=∫0π/212 dθ=12[θ]0π/2=12⋅π2=π4I = \int_0^{\pi/2} \frac{1}{2} \, d\theta = \frac{1}{2} [\theta]_0^{\pi/2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}

Question 10

What is the average value of the function f(x,y)=yf(x,y) = y over the region DD which is the upper half of the disk x2+y2≤4x^2+y^2 \le 4?

  1. 2π\frac{2}{\pi}
  2. 43π\frac{4}{3\pi}
  3. 83π\frac{8}{3\pi} (correct answer)
  4. 163\frac{16}{3}
Explanation: The average value of a function ff over a region DD is given by 1Area(D)∬Df(x,y) dA\frac{1}{\text{Area}(D)} \iint_D f(x,y) \, dA. The region DD is a semicircle of radius 2, so its area is 12π(22)=2π\frac{1}{2}\pi(2^2) = 2\pi. In polar coordinates, the region is 0≤r≤20 \le r \le 2, 0≤θ≤π0 \le \theta \le \pi. The function is f(x,y)=y=rsin⁡θf(x,y) = y = r\sin\theta. We need to compute the integral: ∬Dy dA=∫0π∫02(rsin⁡θ)⋅r dr dθ=∫0πsin⁡θ(∫02r2 dr)dθ\iint_D y \, dA = \int_0^{\pi} \int_0^2 (r\sin\theta) \cdot r \, dr \, d\theta = \int_0^{\pi} \sin\theta \left( \int_0^2 r^2 \, dr \right) d\theta The inner integral is ∫02r2 dr=[r33]02=83\int_0^2 r^2 \, dr = [\frac{r^3}{3}]_0^2 = \frac{8}{3}. The outer integral is ∫0π83sin⁡θ dθ=83[−cos⁡θ]0π=83(−(−1)−(−1))=83(2)=163\int_0^{\pi} \frac{8}{3}\sin\theta \, d\theta = \frac{8}{3}[-\cos\theta]_0^{\pi} = \frac{8}{3}(-(-1) - (-1)) = \frac{8}{3}(2) = \frac{16}{3}. The average value is 12π⋅163=83π\frac{1}{2\pi} \cdot \frac{16}{3} = \frac{8}{3\pi}.

Question 11

Let RR be the region in the xyxy-plane bounded by the limaçon r=2+sin⁡θr=2+\sin\theta. Evaluate the integral ∬Rx dA\iint_R x \, dA.

  1. 00 (correct answer)
  2. 9π2\frac{9\pi}{2}
  3. 3π2\frac{3\pi}{2}
  4. 4π4\pi
Explanation: The region of integration RR is bounded by r=2+sin⁡θr=2+\sin\theta. Since sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta, the region is symmetric with respect to the yy-axis. The integrand is f(x,y)=xf(x,y) = x. This is an odd function with respect to xx, meaning f(−x,y)=−f(x,y)f(-x, y) = -f(x, y). When integrating an odd function with respect to xx over a region symmetric with respect to the yy-axis, the result is 0. Explicitly, the integral is: ∬Rx dA=∫02π∫02+sin⁡θ(rcos⁡θ)⋅r dr dθ=∫02πcos⁡θ[r33]02+sin⁡θdθ=13∫02π(2+sin⁡θ)3cos⁡θ dθ\iint_R x \, dA = \int_0^{2\pi} \int_0^{2+\sin\theta} (r\cos\theta) \cdot r \, dr \, d\theta = \int_0^{2\pi} \cos\theta \left[ \frac{r^3}{3} \right]_0^{2+\sin\theta} d\theta = \frac{1}{3} \int_0^{2\pi} (2+\sin\theta)^3 \cos\theta \, d\theta Let u=2+sin⁡θu = 2+\sin\theta, so du=cos⁡θ dθdu = \cos\theta \, d\theta. When θ=0\theta=0, u=2u=2. When θ=2π\theta=2\pi, u=2u=2. The integral becomes 13∫22u3 du=0\frac{1}{3} \int_2^2 u^3 \, du = 0.

Question 12

An iterated integral is given by I=∫01∫01−(x−1)21x2+y2 dy dxI = \int_{0}^{1} \int_{0}^{\sqrt{1-(x-1)^2}} \frac{1}{\sqrt{x^2+y^2}} \, dy \, dx. Which polar integral is equivalent to II?

  1. ∫0π/2∫01 dr dθ\int_{0}^{\pi/2} \int_0^1 \, dr \, d\theta
  2. ∫0π/2∫02sin⁡θ dr dθ\int_{0}^{\pi/2} \int_0^{2\sin\theta} \, dr \, d\theta
  3. ∫0π/4∫02cos⁡θr dr dθ\int_{0}^{\pi/4} \int_0^{2\cos\theta} r \, dr \, d\theta
  4. ∫0π/2∫02cos⁡θ dr dθ\int_{0}^{\pi/2} \int_0^{2\cos\theta} \, dr \, d\theta (correct answer)
Explanation: When converting double integrals from Cartesian to polar coordinates, you need to identify the region of integration and transform both the limits and the integrand appropriately. First, let's analyze the region. The limits show 0≤x≤10 \leq x \leq 1 and 0≤y≤1−(x−1)20 \leq y \leq \sqrt{1-(x-1)^2}. The upper boundary y=1−(x−1)2y = \sqrt{1-(x-1)^2} can be rewritten as y2=1−(x−1)2y^2 = 1-(x-1)^2, or (x−1)2+y2=1(x-1)^2 + y^2 = 1. This is a circle with center at (1,0)(1,0) and radius 1. Since we're taking the positive square root and xx goes from 0 to 1, we're integrating over the upper semicircle from the y-axis to the rightmost point. In polar coordinates, this circle has equation r=2cos⁡θr = 2\cos\theta. To see why, substitute x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta into (x−1)2+y2=1(x-1)^2 + y^2 = 1: (rcos⁡θ−1)2+(rsin⁡θ)2=1(r\cos\theta - 1)^2 + (r\sin\theta)^2 = 1, which simplifies to r=2cos⁡θr = 2\cos\theta. The angular limits are 0≤θ≤π/20 \leq \theta \leq \pi/2 for the upper semicircle. The integrand 1x2+y2=1r\frac{1}{\sqrt{x^2+y^2}} = \frac{1}{r} in polar coordinates, and we multiply by the Jacobian rr, giving us 1r⋅r=1\frac{1}{r} \cdot r = 1. Answer D correctly gives ∫0π/2∫02cos⁡θ dr dθ\int_{0}^{\pi/2} \int_0^{2\cos\theta} \, dr \, d\theta. Answer A uses incorrect radial limits (should be 2cos⁡θ2\cos\theta, not 1). Answer B uses 2sin⁡θ2\sin\theta, which describes a different circle. Answer C has wrong angular limits and includes an extra rr factor. Study tip: Always sketch the region first, then identify key geometric features like circle centers and radii to determine the correct polar equation.

Question 13

Which of the following double integrals represents the area of the region in the first quadrant that lies inside the circle r=4sin⁡θr=4\sin\theta but outside the circle r=2r=2?

  1. ∫0π/6∫24sin⁡θr dr dθ\int_{0}^{\pi/6} \int_2^{4\sin\theta} r \, dr \, d\theta
  2. ∫0π/2∫24sin⁡θr dr dθ\int_{0}^{\pi/2} \int_2^{4\sin\theta} r \, dr \, d\theta
  3. ∫π/6π/2∫4sin⁡θ2r dr dθ\int_{\pi/6}^{\pi/2} \int_{4\sin\theta}^2 r \, dr \, d\theta
  4. ∫π/6π/2∫24sin⁡θr dr dθ\int_{\pi/6}^{\pi/2} \int_2^{4\sin\theta} r \, dr \, d\theta (correct answer)
Explanation: When setting up double integrals in polar coordinates to find areas between curves, you need to carefully determine both the angular limits and the radial limits by analyzing where the curves intersect and which curve forms the outer boundary. First, let's understand our region. The circle r=4sin⁡θr = 4\sin\theta can be rewritten in Cartesian form as x2+(y−2)2=4x^2 + (y-2)^2 = 4, which is a circle centered at (0,2)(0,2) with radius 2. The circle r=2r = 2 is centered at the origin with radius 2. These circles intersect where 2=4sin⁡θ2 = 4\sin\theta, giving us sin⁡θ=12\sin\theta = \frac{1}{2}, so θ=π6\theta = \frac{\pi}{6}. For the angular limits, our region exists from θ=π6\theta = \frac{\pi}{6} (the intersection point) to θ=π2\theta = \frac{\pi}{2} (where the upper circle reaches its maximum). For radial limits, we integrate from the inner circle r=2r = 2 to the outer circle r=4sin⁡θr = 4\sin\theta. The area element in polar coordinates is r dr dθr \, dr \, d\theta. Choice A uses incorrect angular limits starting from 0 instead of π6\frac{\pi}{6}. Choice B also starts from 0, missing the intersection point entirely. Choice C has the radial limits reversed, integrating from the outer curve to the inner curve, which would give a negative result. Choice D correctly captures the region with θ\theta from π6\frac{\pi}{6} to π2\frac{\pi}{2} and rr from 2 to 4sin⁡θ4\sin\theta. Study tip: Always find intersection points first to determine your integration limits, and remember that in polar coordinates, you typically integrate from inner to outer curve for the radial component.

Question 14

Let RR be the smaller region bounded by the circle r=2r=2, the line y=xy=x, and the yy-axis in the upper-half plane. Which of the following integrals correctly represents the area of RR?

  1. ∫0π/4∫02r dr dθ\int_{0}^{\pi/4} \int_0^2 r \, dr \, d\theta
  2. ∫π/4π/2∫02r dr dθ\int_{\pi/4}^{\pi/2} \int_0^2 r \, dr \, d\theta (correct answer)
  3. ∫0π/4∫02sec⁡θr dr dθ\int_{0}^{\pi/4} \int_0^{2\sec\theta} r \, dr \, d\theta
  4. ∫π/4π/2∫02csc⁡θr dr dθ\int_{\pi/4}^{\pi/2} \int_0^{2\csc\theta} r \, dr \, d\theta
Explanation: The region RR is in the upper-half plane (y≥0y \ge 0). It is bounded by the line y=xy=x, the yy-axis (x=0x=0), and the circle r=2r=2 (x2+y2=4x^2+y^2=4). The yy-axis corresponds to θ=π/2\theta = \pi/2. The line y=xy=x corresponds to θ=π/4\theta = \pi/4. The region is a sector of the circle r=2r=2. The angle θ\theta ranges from the line y=xy=x to the yy-axis, so π/4≤θ≤π/2\pi/4 \le \theta \le \pi/2. The radius rr ranges from the origin to the circle, so 0≤r≤20 \le r \le 2. The area is given by ∬RdA=∬Rr dr dθ\iint_R dA = \iint_R r \, dr \, d\theta. Therefore, the correct setup is ∫π/4π/2∫02r dr dθ\int_{\pi/4}^{\pi/2} \int_0^2 r \, dr \, d\theta.

Question 15

Find the area of the region that lies inside the circle r=3cos⁡θr=3\cos\theta and outside the cardioid r=1+cos⁡θr=1+\cos\theta.

  1. π2\frac{\pi}{2}
  2. π\pi (correct answer)
  3. 2π2\pi
  4. π−334\pi - \frac{3\sqrt{3}}{4}
Explanation: First, find the intersection points of the two curves: 3cos⁡θ=1+cos⁡θ  ⟹  2cos⁡θ=1  ⟹  cos⁡θ=1/23\cos\theta = 1+\cos\theta \implies 2\cos\theta = 1 \implies \cos\theta = 1/2. This occurs at θ=−π/3\theta = -\pi/3 and θ=π/3\theta = \pi/3. The region is symmetric about the polar axis, so we can integrate from 00 to π/3\pi/3 and multiply by 2. The area is given by A=12∫αβ(router2−rinner2) dθA = \frac{1}{2} \int_{\alpha}^{\beta} (r_{outer}^2 - r_{inner}^2) \, d\theta. In the interval [−π/3,π/3][-\pi/3, \pi/3], 3cos⁡θ≥1+cos⁡θ3\cos\theta \ge 1+\cos\theta. Thus, router=3cos⁡θr_{outer} = 3\cos\theta and rinner=1+cos⁡θr_{inner} = 1+\cos\theta. A=2⋅12∫0π/3[(3cos⁡θ)2−(1+cos⁡θ)2] dθ=∫0π/3[9cos⁡2θ−(1+2cos⁡θ+cos⁡2θ)] dθA = 2 \cdot \frac{1}{2} \int_0^{\pi/3} [(3\cos\theta)^2 - (1+\cos\theta)^2] \, d\theta = \int_0^{\pi/3} [9\cos^2\theta - (1+2\cos\theta+\cos^2\theta)] \, d\theta =∫0π/3(8cos⁡2θ−2cos⁡θ−1) dθ= \int_0^{\pi/3} (8\cos^2\theta - 2\cos\theta - 1) \, d\theta Using cos⁡2θ=1+cos⁡(2θ)2\cos^2\theta = \frac{1+\cos(2\theta)}{2}: =∫0π/3(81+cos⁡(2θ)2−2cos⁡θ−1) dθ=∫0π/3(3+4cos⁡(2θ)−2cos⁡θ) dθ= \int_0^{\pi/3} (8\frac{1+\cos(2\theta)}{2} - 2\cos\theta - 1) \, d\theta = \int_0^{\pi/3} (3 + 4\cos(2\theta) - 2\cos\theta) \, d\theta =[3θ+2sin⁡(2θ)−2sin⁡θ]0π/3=(3(π3)+2sin⁡(2π3)−2sin⁡(π3))−0=π+2(32)−2(32)=π= [3\theta + 2\sin(2\theta) - 2\sin\theta]_0^{\pi/3} = (3(\frac{\pi}{3}) + 2\sin(\frac{2\pi}{3}) - 2\sin(\frac{\pi}{3})) - 0 = \pi + 2(\frac{\sqrt{3}}{2}) - 2(\frac{\sqrt{3}}{2}) = \pi

Question 16

Consider the integral I=∫0a∫0xx2+y2 dy dxI = \int_0^a \int_0^x \sqrt{x^2+y^2} \, dy \, dx. After converting to polar coordinates, what is the value of the inner integral with respect to rr?

  1. a3sec⁡3θ3\frac{a^3 \sec^3\theta}{3} (correct answer)
  2. a3csc⁡3θ3\frac{a^3 \csc^3\theta}{3}
  3. a3sec⁡2θ3\frac{a^3 \sec^2\theta}{3}
  4. a3tan⁡θsec⁡2θ3\frac{a^3 \tan\theta \sec^2\theta}{3}
Explanation: The region of integration is a triangle in the first quadrant with vertices at (0,0)(0,0), (a,0)(a,0), and (a,a)(a,a). The bounds are 0≤x≤a0 \le x \le a and 0≤y≤x0 \le y \le x. The line y=0y=0 is the polar axis, θ=0\theta=0. The line y=xy=x corresponds to θ=π/4\theta = \pi/4. The vertical line x=ax=a is rcos⁡θ=ar\cos\theta = a, or r=asec⁡θr = a\sec\theta. So, the region in polar coordinates is 0≤θ≤π/40 \le \theta \le \pi/4 and 0≤r≤asec⁡θ0 \le r \le a\sec\theta. The integrand is x2+y2=r\sqrt{x^2+y^2} = r, and dA=r dr dθdA = r \, dr \, d\theta. The integral becomes: I=∫0π/4∫0asec⁡θr⋅r dr dθ=∫0π/4(∫0asec⁡θr2 dr)dθI = \int_0^{\pi/4} \int_0^{a\sec\theta} r \cdot r \, dr \, d\theta = \int_0^{\pi/4} \left( \int_0^{a\sec\theta} r^2 \, dr \right) d\theta The question asks for the value of the inner integral. ∫0asec⁡θr2 dr=[r33]0asec⁡θ=(asec⁡θ)33−0=a3sec⁡3θ3\int_0^{a\sec\theta} r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^{a\sec\theta} = \frac{(a\sec\theta)^3}{3} - 0 = \frac{a^3 \sec^3\theta}{3}

Question 17

Evaluate the definite integral ∫02∫04−x2(x2+y2)3/2 dy dx\int_0^2 \int_0^{\sqrt{4-x^2}} (x^2+y^2)^{3/2} \, dy \, dx.

  1. 2π2\pi
  2. 4π3\frac{4\pi}{3}
  3. 16π5\frac{16\pi}{5} (correct answer)
  4. 32π5\frac{32\pi}{5}
Explanation: The region of integration is described by 0≤x≤20 \le x \le 2 and 0≤y≤4−x20 \le y \le \sqrt{4-x^2}, which is a quarter-circle of radius 2 in the first quadrant. In polar coordinates, this region is 0≤r≤20 \le r \le 2 and 0≤θ≤π/20 \le \theta \le \pi/2. The integrand (x2+y2)3/2(x^2+y^2)^{3/2} becomes (r2)3/2=r3(r^2)^{3/2} = r^3. The area element dy dxdy \, dx becomes r dr dθr \, dr \, d\theta. The integral is transformed to: ∫0π/2∫02(r3)⋅r dr dθ=∫0π/2∫02r4 dr dθ\int_0^{\pi/2} \int_0^2 (r^3) \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_0^2 r^4 \, dr \, d\theta Evaluating the inner integral: ∫02r4 dr=[r55]02=325\int_0^2 r^4 \, dr = \left[ \frac{r^5}{5} \right]_0^2 = \frac{32}{5} Evaluating the outer integral: ∫0π/2325 dθ=325[θ]0π/2=325⋅π2=16π5\int_0^{\pi/2} \frac{32}{5} \, d\theta = \frac{32}{5} [\theta]_0^{\pi/2} = \frac{32}{5} \cdot \frac{\pi}{2} = \frac{16\pi}{5}