Multivariable Calculus Quiz: Area And Average Value Double Integrals
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Area And Average Value Double IntegralsQuestion 1 of 7

The region DD is defined by {(x,y):x2+y2≤9,x≥0}\{(x,y): x^2 + y^2 \leq 9, x \geq 0\}. If ∬Df(x,y) dA=18π\iint_D f(x,y) \, dA = 18\pi and the area of DD is 9π2\frac{9\pi}{2}, what is the average value of f(x,y)f(x,y) over region DD?

22
44
92\frac{9}{2}
18π\frac{18}{\pi}
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Area And Average Value Double Integrals

Practice Area And Average Value Double Integrals in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Area And Average Value Double Integrals, 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 region DD is defined by {(x,y):x2+y2≤9,x≥0}\{(x,y): x^2 + y^2 \leq 9, x \geq 0\}. If ∬Df(x,y) dA=18π\iint_D f(x,y) \, dA = 18\pi and the area of DD is 9π2\frac{9\pi}{2}, what is the average value of f(x,y)f(x,y) over region DD?

  1. 22
  2. 44 (correct answer)
  3. 92\frac{9}{2}
  4. 18π\frac{18}{\pi}
Explanation: The average value of a function f(x,y)f(x,y) over a region DD is given by fˉ=1Area(D)∬Df(x,y) dA\bar{f} = \frac{1}{\text{Area}(D)} \iint_D f(x,y) \, dA. We are given that ∬Df(x,y) dA=18π\iint_D f(x,y) \, dA = 18\pi and the area of DD is 9π2\frac{9\pi}{2}. Therefore: fˉ=18π9π/2=18π⋅29π=36π9π=4\bar{f} = \frac{18\pi}{9\pi/2} = \frac{18\pi \cdot 2}{9\pi} = \frac{36\pi}{9\pi} = 4. Note that region DD is a semicircle of radius 3 in the right half-plane, so its area is indeed 12⋅π⋅32=9π2\frac{1}{2} \cdot \pi \cdot 3^2 = \frac{9\pi}{2}, confirming the given information.

Question 2

Consider the region bounded by x=0x = 0, y=0y = 0, and x+2y=6x + 2y = 6 in the first quadrant. If the average value of f(x,y)=x2+y2f(x,y) = x^2 + y^2 over this region is kk, then the average value of g(x,y)=2x2+2y2g(x,y) = 2x^2 + 2y^2 over the same region is:

  1. 2k2k (correct answer)
  2. 4k4k
  3. k+2k + 2
  4. k2k^2
Explanation: This problem tests understanding of the linearity property of average values. The average value of a function ff over a region RR is fˉ=1Area(R)∬Rf(x,y) dA\bar{f} = \frac{1}{\text{Area}(R)} \iint_R f(x,y) \, dA. For the function g(x,y)=2x2+2y2=2f(x,y)g(x,y) = 2x^2 + 2y^2 = 2f(x,y), we have: gˉ=1Area(R)∬Rg(x,y) dA=1Area(R)∬R2f(x,y) dA=2Area(R)∬Rf(x,y) dA=2⋅1Area(R)∬Rf(x,y) dA=2fˉ=2k\bar{g} = \frac{1}{\text{Area}(R)} \iint_R g(x,y) \, dA = \frac{1}{\text{Area}(R)} \iint_R 2f(x,y) \, dA = \frac{2}{\text{Area}(R)} \iint_R f(x,y) \, dA = 2 \cdot \frac{1}{\text{Area}(R)} \iint_R f(x,y) \, dA = 2\bar{f} = 2k. This follows from the linearity property of integrals: ∫cf(x)dx=c∫f(x)dx\int cf(x) dx = c\int f(x) dx. Choice B (4k4k) might be chosen by students who incorrectly think that since we're squaring terms, the average gets squared. Choice C (k+2k + 2) reflects a misunderstanding of how constants affect averages. Choice D (k2k^2) comes from incorrectly squaring the entire average value.

Question 3

The area of the region R={(x,y):0≤x≤2,x2≤y≤2x}R = \{(x,y): 0 \leq x \leq 2, x^2 \leq y \leq 2x\} can be computed as ∫02∫x22xdy dx\int_0^2 \int_{x^2}^{2x} dy \, dx. If we want to express this same area using the order dx dydx \, dy, which of the following integrals is correct?

  1. ∫04∫y/2ydx dy\int_0^4 \int_{y/2}^{\sqrt{y}} dx \, dy
  2. ∫02∫yy/2dx dy+∫24∫y/2ydx dy\int_0^2 \int_{\sqrt{y}}^{y/2} dx \, dy + \int_2^4 \int_{y/2}^{\sqrt{y}} dx \, dy
  3. ∫02∫y/2ydx dy+∫24∫0y/2dx dy\int_0^2 \int_{y/2}^{\sqrt{y}} dx \, dy + \int_2^4 \int_{0}^{y/2} dx \, dy (correct answer)
  4. ∫02∫yy/2dx dy+∫24∫0y/2dx dy\int_0^2 \int_{\sqrt{y}}^{y/2} dx \, dy + \int_2^4 \int_{0}^{y/2} dx \, dy
Explanation: To change the order of integration, we need to determine the range of yy values and express the xx boundaries as functions of yy. The region is bounded by y=x2y = x^2 and y=2xy = 2x for 0≤x≤20 \leq x \leq 2. These curves intersect when x2=2xx^2 = 2x, giving x=0x = 0 and x=2x = 2. The yy values range from 0 to 4 (since y=2xy = 2x reaches y=4y = 4 when x=2x = 2). For 0≤y≤20 \leq y \leq 2: From y=x2y = x^2, we get x=yx = \sqrt{y} (right boundary). From y=2xy = 2x, we get x=y/2x = y/2 (left boundary). So y/2≤x≤yy/2 \leq x \leq \sqrt{y}. For 2≤y≤42 \leq y \leq 4: The curve y=x2y = x^2 no longer bounds the region (since x2>2x^2 > 2 only for x>2≈1.41x > \sqrt{2} \approx 1.41, but the region ends at x=2x = 2). The left boundary is x=0x = 0 and the right boundary is x=y/2x = y/2 (from y=2xy = 2x). So 0≤x≤y/20 \leq x \leq y/2. Therefore: Area=∫02∫y/2ydx dy+∫24∫0y/2dx dy\text{Area} = \int_0^2 \int_{y/2}^{\sqrt{y}} dx \, dy + \int_2^4 \int_0^{y/2} dx \, dy. Choice A uses wrong limits and doesn't split the integral. Choice B has the boundaries reversed in the first integral. Choice D also has reversed boundaries in the first integral.

Question 4

A region RR is bounded by the curves y=x2y = x^2 and y=2xy = 2x. If the average value of the function f(x,y)=xyf(x,y) = xy over region RR is 85\frac{8}{5}, what is the area of region RR?

  1. 43\frac{4}{3} (correct answer)
  2. 83\frac{8}{3}
  3. 165\frac{16}{5}
  4. 3215\frac{32}{15}
Explanation: First, find the intersection points: x2=2xx^2 = 2x gives x=0x = 0 and x=2x = 2. The region is bounded by 0≤x≤20 \leq x \leq 2 and x2≤y≤2xx^2 \leq y \leq 2x. The average value formula is fˉ=1A∬Rf(x,y) dA\bar{f} = \frac{1}{A} \iint_R f(x,y) \, dA, so A=∬Rxy dAfˉA = \frac{\iint_R xy \, dA}{\bar{f}}. Computing ∬Rxy dA=∫02∫x22xxy dy dx=∫02x⋅(2x)2−(x2)22 dx=∫02x⋅4x2−x42 dx=∫024x3−x52 dx=12[x4−x66]02=12(16−646)=1615\iint_R xy \, dA = \int_0^2 \int_{x^2}^{2x} xy \, dy \, dx = \int_0^2 x \cdot \frac{(2x)^2 - (x^2)^2}{2} \, dx = \int_0^2 x \cdot \frac{4x^2 - x^4}{2} \, dx = \int_0^2 \frac{4x^3 - x^5}{2} \, dx = \frac{1}{2}[x^4 - \frac{x^6}{6}]_0^2 = \frac{1}{2}(16 - \frac{64}{6}) = \frac{16}{15}. Therefore, A=16/158/5=1615⋅58=43A = \frac{16/15}{8/5} = \frac{16}{15} \cdot \frac{5}{8} = \frac{4}{3}.

Question 5

Consider the region RR defined by {(x,y):1≤x2+y2≤4,x≥0,y≥0}\{(x,y): 1 \leq x^2 + y^2 \leq 4, x \geq 0, y \geq 0\}. The average value of f(x,y)=x2+y2f(x,y) = x^2 + y^2 over this region is:

  1. 143\frac{14}{3}
  2. 72\frac{7}{2}
  3. 103\frac{10}{3}
  4. 52\frac{5}{2} (correct answer)
Explanation: The region is a quarter annulus in the first quadrant. Using polar coordinates: 1≤r≤21 \leq r \leq 2, 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}, and f(x,y)=r2f(x,y) = r^2. The area is A=∫0π/2∫12r dr dθ=π2⋅[r22]12=π2⋅32=3π4A = \int_0^{\pi/2} \int_1^2 r \, dr \, d\theta = \frac{\pi}{2} \cdot \left[\frac{r^2}{2}\right]_1^2 = \frac{\pi}{2} \cdot \frac{3}{2} = \frac{3\pi}{4}. The integral is ∬Rr2⋅r dr dθ=∫0π/2∫12r3 dr dθ=π2⋅[r44]12=π2⋅154=15π8\iint_R r^2 \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_1^2 r^3 \, dr \, d\theta = \frac{\pi}{2} \cdot \left[\frac{r^4}{4}\right]_1^2 = \frac{\pi}{2} \cdot \frac{15}{4} = \frac{15\pi}{8}. The average value is 15π/83π/4=15π8⋅43π=52\frac{15\pi/8}{3\pi/4} = \frac{15\pi}{8} \cdot \frac{4}{3\pi} = \frac{5}{2}.

Question 6

The area of the region enclosed by the polar curve r=2+cos⁡(3θ)r = 2 + \cos(3\theta) can be expressed as a double integral. Which of the following correctly represents this area?

  1. ∫02π∫02+cos⁡(3θ)r dr dθ\int_0^{2\pi} \int_0^{2+\cos(3\theta)} r \, dr \, d\theta (correct answer)
  2. ∫0π∫02+cos⁡(3θ)r dr dθ\int_0^{\pi} \int_0^{2+\cos(3\theta)} r \, dr \, d\theta
  3. ∫02π∫02+cos⁡(3θ)1 dr dθ\int_0^{2\pi} \int_0^{2+\cos(3\theta)} 1 \, dr \, d\theta
  4. 12∫02π∫02+cos⁡(3θ)r dr dθ\frac{1}{2} \int_0^{2\pi} \int_0^{2+\cos(3\theta)} r \, dr \, d\theta
Explanation: In polar coordinates, the area element is dA=r dr dθdA = r \, dr \, d\theta. For a region bounded by r=f(θ)r = f(\theta), the area is ∬RdA=∫θ1θ2∫0f(θ)r dr dθ\iint_R dA = \int_{\theta_1}^{\theta_2} \int_0^{f(\theta)} r \, dr \, d\theta. The curve r=2+cos⁡(3θ)r = 2 + \cos(3\theta) has period 2π3\frac{2\pi}{3} but traces the complete curve once as θ\theta goes from 00 to 2π2\pi, creating a three-petaled rose-like shape. Since 2+cos⁡(3θ)≥1>02 + \cos(3\theta) \geq 1 > 0 for all θ\theta, the curve doesn't pass through the origin and we integrate from 00 to 2π2\pi to capture the entire region. Choice B uses wrong limits (00 to π\pi would miss part of the curve). Choice C is missing the rr factor needed for the polar area element. Choice D incorrectly includes a factor of 12\frac{1}{2}, which would be used for areas swept by radial lines, not enclosed regions.

Question 7

A lamina occupies the triangular region with vertices at (0,0)(0,0), (4,0)(4,0), and (0,3)(0,3). If the density function is ρ(x,y)=x+y\rho(x,y) = x + y, what is the average density of the lamina?

  1. 73\frac{7}{3} (correct answer)
  2. 83\frac{8}{3}
  3. 52\frac{5}{2}
  4. 72\frac{7}{2}
Explanation: The triangular region has vertices at (0,0)(0,0), (4,0)(4,0), and (0,3)(0,3). The hypotenuse has equation x4+y3=1\frac{x}{4} + \frac{y}{3} = 1, or y=3−3x4y = 3 - \frac{3x}{4}. The region is described by 0≤x≤40 \leq x \leq 4 and 0≤y≤3−3x40 \leq y \leq 3 - \frac{3x}{4}. The average density is ρˉ=∬Rρ(x,y) dAArea\bar{\rho} = \frac{\iint_R \rho(x,y) \, dA}{\text{Area}}. The area of the triangle is 12⋅4⋅3=6\frac{1}{2} \cdot 4 \cdot 3 = 6. Now compute ∬R(x+y) dA=∫04∫03−3x/4(x+y) dy dx=∫04[xy+y22]03−3x/4dx=∫04[x(3−3x4)+(3−3x/4)22]dx=∫04[3x−3x24+(3−3x/4)22]dx\iint_R (x+y) \, dA = \int_0^4 \int_0^{3-3x/4} (x+y) \, dy \, dx = \int_0^4 \left[xy + \frac{y^2}{2}\right]_0^{3-3x/4} dx = \int_0^4 \left[x(3-\frac{3x}{4}) + \frac{(3-3x/4)^2}{2}\right] dx = \int_0^4 \left[3x - \frac{3x^2}{4} + \frac{(3-3x/4)^2}{2}\right] dx. Expanding (3−3x/4)2=9−18x4+9x216=9−9x2+9x216(3-3x/4)^2 = 9 - \frac{18x}{4} + \frac{9x^2}{16} = 9 - \frac{9x}{2} + \frac{9x^2}{16}, so (3−3x/4)22=92−9x4+9x232\frac{(3-3x/4)^2}{2} = \frac{9}{2} - \frac{9x}{4} + \frac{9x^2}{32}. The integrand becomes: 3x−3x24+92−9x4+9x232=92+3x−9x4−3x24+9x232=92+3x4−24x232+9x232=92+3x4−15x2323x - \frac{3x^2}{4} + \frac{9}{2} - \frac{9x}{4} + \frac{9x^2}{32} = \frac{9}{2} + 3x - \frac{9x}{4} - \frac{3x^2}{4} + \frac{9x^2}{32} = \frac{9}{2} + \frac{3x}{4} - \frac{24x^2}{32} + \frac{9x^2}{32} = \frac{9}{2} + \frac{3x}{4} - \frac{15x^2}{32}. Integrating: ∫04[92+3x4−15x232]dx=[9x2+3x28−15x396]04=18+6−15⋅6496=24−10=14\int_0^4 \left[\frac{9}{2} + \frac{3x}{4} - \frac{15x^2}{32}\right] dx = \left[\frac{9x}{2} + \frac{3x^2}{8} - \frac{15x^3}{96}\right]_0^4 = 18 + 6 - \frac{15 \cdot 64}{96} = 24 - 10 = 14. Therefore, ρˉ=146=73\bar{\rho} = \frac{14}{6} = \frac{7}{3}.