Multivariable Calculus Quiz: Double Integrals General Regions
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Double Integrals General RegionsQuestion 1 of 9

Calculate the area of the region in the xy-plane bounded by the graphs of y=exy=e^x, y=ln⁡xy=\ln x, x=1x=1, and x=2x=2.

e2−e+2ln⁡2−1e^2 - e + 2 \ln 2 - 1
e2−e−2ln⁡2−1e^2 - e - 2 \ln 2 - 1
e2−e−2ln⁡2+1e^2 - e - 2 \ln 2 + 1
e2−e+2ln⁡2+1e^2 - e + 2 \ln 2 + 1
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Double Integrals General Regions

Practice Double Integrals General Regions 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 General Regions, 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

Calculate the area of the region in the xy-plane bounded by the graphs of y=exy=e^x, y=ln⁡xy=\ln x, x=1x=1, and x=2x=2.

  1. e2−e+2ln⁡2−1e^2 - e + 2 \ln 2 - 1
  2. e2−e−2ln⁡2−1e^2 - e - 2 \ln 2 - 1
  3. e2−e−2ln⁡2+1e^2 - e - 2 \ln 2 + 1 (correct answer)
  4. e2−e+2ln⁡2+1e^2 - e + 2 \ln 2 + 1
Explanation: When finding the area between curves, you need to determine which function is on top throughout the interval and set up the integral as (upper function) - (lower function). First, let's determine the relative positions of y=exy = e^x and y=ln⁡xy = \ln x on the interval [1,2][1,2]. At x=1x = 1: e1=e≈2.718e^1 = e \approx 2.718 and ln⁡1=0\ln 1 = 0. At x=2x = 2: e2≈7.389e^2 \approx 7.389 and ln⁡2≈0.693\ln 2 \approx 0.693. Since ex>ln⁡xe^x > \ln x throughout this interval, we set up: Area=∫12(ex−ln⁡x) dx=∫12ex dx−∫12ln⁡x dx\text{Area} = \int_1^2 (e^x - \ln x) \, dx = \int_1^2 e^x \, dx - \int_1^2 \ln x \, dx The first integral is straightforward: ∫12ex dx=[ex]12=e2−e\int_1^2 e^x \, dx = [e^x]_1^2 = e^2 - e For the second integral, use integration by parts with u=ln⁡xu = \ln x and dv=dxdv = dx: ∫ln⁡x dx=xln⁡x−x+C\int \ln x \, dx = x\ln x - x + C So: ∫12ln⁡x dx=[xln⁡x−x]12=(2ln⁡2−2)−(1⋅0−1)=2ln⁡2−2+1=2ln⁡2−1\int_1^2 \ln x \, dx = [x\ln x - x]_1^2 = (2\ln 2 - 2) - (1 \cdot 0 - 1) = 2\ln 2 - 2 + 1 = 2\ln 2 - 1 Therefore: Area=(e2−e)−(2ln⁡2−1)=e2−e−2ln⁡2+1\text{Area} = (e^2 - e) - (2\ln 2 - 1) = e^2 - e - 2\ln 2 + 1 This confirms answer C is correct. Answer A incorrectly adds the ln⁡x\ln x integral instead of subtracting it. Answer B has the wrong sign on the constant term. Answer D makes both errors—adding instead of subtracting the ln⁡x\ln x integral and having the wrong constant. Remember: always sketch the region or test key points to determine which curve is on top before setting up your area integral.

Question 2

A region RR in the xy-plane is bounded by the lines y=x+1y=x+1, y=−2x+4y=-2x+4, and the x-axis. Which of the following expressions represents the double integral ∬Rf(x,y) dA\iint_R f(x,y) \, dA?

  1. ∫−12∫x+1−2x+4f(x,y) dy dx\int_{-1}^2 \int_{x+1}^{-2x+4} f(x,y) \, dy \, dx
  2. ∫−11∫0−2x+4f(x,y) dy dx+∫12∫0x+1f(x,y) dy dx\int_{-1}^1 \int_0^{-2x+4} f(x,y) \, dy \, dx + \int_1^2 \int_0^{x+1} f(x,y) \, dy \, dx
  3. ∫02∫y−1(4−y)/2f(x,y) dx dy\int_0^2 \int_{y-1}^{(4-y)/2} f(x,y) \, dx \, dy
  4. ∫−11∫0x+1f(x,y) dy dx+∫12∫0−2x+4f(x,y) dy dx\int_{-1}^1 \int_0^{x+1} f(x,y) \, dy \, dx + \int_1^2 \int_0^{-2x+4} f(x,y) \, dy \, dx (correct answer)
Explanation: First, sketch the region RR. The vertices of the triangular region are the intersection points of the lines: (−1,0)(-1,0), (2,0)(2,0), and (1,2)(1,2). If we integrate with respect to yy first (as a Type I region), we must split the integral into two parts at the x-coordinate of the top vertex, x=1x=1. For −1≤x≤1-1 \le x \le 1, the region is bounded below by y=0y=0 and above by y=x+1y=x+1. For 1≤x≤21 \le x \le 2, the region is bounded below by y=0y=0 and above by y=−2x+4y=-2x+4. Therefore, the integral is the sum ∫−11∫0x+1f(x,y) dy dx+∫12∫0−2x+4f(x,y) dy dx\int_{-1}^1 \int_0^{x+1} f(x,y) \, dy \, dx + \int_1^2 \int_0^{-2x+4} f(x,y) \, dy \, dx. Choice C is also a correct representation of the integral, but as a single integral with the order reversed (a Type II region), not as a sum of integrals in the specified order.

Question 3

Find the average value of the function f(x,y)=6xf(x,y) = 6x over the region RR bounded by the parabola y=x2y=x^2 and the line y=2xy=2x.

  1. 6 (correct answer)
  2. 8
  3. 15
  4. 43\frac{4}{3}
Explanation: The average value is 1Area(R)∬Rf(x,y) dA\frac{1}{\text{Area}(R)} \iint_R f(x,y) \, dA. First, find the intersection points: x2=2x  ⟹  x(x−2)=0x^2 = 2x \implies x(x-2)=0, so x=0x=0 and x=2x=2. The region is bounded by 0≤x≤20 \le x \le 2, with y=2xy=2x as the upper bound and y=x2y=x^2 as the lower bound. The area is Area(R)=∫02(2x−x2) dx=[x2−x33]02=4−83=43\text{Area}(R) = \int_0^2 (2x-x^2) \, dx = [x^2 - \frac{x^3}{3}]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}. Next, calculate the integral of the function: ∬R6x dA=∫02∫x22x6x dy dx=∫026x[y]x22x dx=∫026x(2x−x2) dx=∫02(12x2−6x3) dx=[4x3−32x4]02=4(8)−32(16)=32−24=8\iint_R 6x \, dA = \int_0^2 \int_{x^2}^{2x} 6x \, dy \, dx = \int_0^2 6x[y]_{x^2}^{2x} \, dx = \int_0^2 6x(2x-x^2) \, dx = \int_0^2 (12x^2 - 6x^3) \, dx = [4x^3 - \frac{3}{2}x^4]_0^2 = 4(8) - \frac{3}{2}(16) = 32 - 24 = 8. The average value is 84/3=8⋅34=6\frac{8}{4/3} = 8 \cdot \frac{3}{4} = 6.

Question 4

Which of the following is equal to the iterated integral ∫02∫y/21ysin⁡(x3−1) dx dy\int_0^2 \int_{y/2}^1 y \sin(x^3 - 1) \, dx \, dy?

  1. 23(cos⁡(1)−1)\frac{2}{3}(\cos(1)-1) (correct answer)
  2. 2(cos⁡(1)−1)2(\cos(1)-1)
  3. 23(1−cos⁡(1))\frac{2}{3}(1-\cos(1))
  4. cos⁡(1)−1\cos(1)-1
Explanation: The integral ∫sin⁡(x3−1) dx\int \sin(x^3-1) \, dx does not have an elementary antiderivative, so we must reverse the order of integration. The region of integration is given by 0≤y≤20 \le y \le 2 and y/2≤x≤1y/2 \le x \le 1. This is a triangle with vertices at (0,0)(0,0), (1,0)(1,0), and (1,2)(1,2). To reverse the order, we set bounds for xx first, then yy. The region is covered by letting xx range from 00 to 11. For each xx, yy ranges from the bottom boundary y=0y=0 to the top boundary x=y/2x=y/2, which is y=2xy=2x. The new integral is ∫01∫02xysin⁡(x3−1) dy dx\int_0^1 \int_0^{2x} y \sin(x^3 - 1) \, dy \, dx. Evaluating the inner integral: ∫02xysin⁡(x3−1) dy=sin⁡(x3−1)[y22]02x=sin⁡(x3−1)4x22=2x2sin⁡(x3−1)\int_0^{2x} y \sin(x^3 - 1) \, dy = \sin(x^3-1)[\frac{y^2}{2}]_0^{2x} = \sin(x^3-1) \frac{4x^2}{2} = 2x^2 \sin(x^3-1). The outer integral is ∫012x2sin⁡(x3−1) dx\int_0^1 2x^2 \sin(x^3 - 1) \, dx. Let u=x3−1u=x^3-1, so du=3x2 dxdu = 3x^2 \, dx. The integral becomes ∫−1023sin⁡(u) du=23[−cos⁡(u)]−10=−23(cos⁡(0)−cos⁡(−1))=−23(1−cos⁡(1))=23(cos⁡(1)−1)\int_{-1}^0 \frac{2}{3} \sin(u) \, du = \frac{2}{3}[-\cos(u)]_{-1}^0 = -\frac{2}{3}(\cos(0) - \cos(-1)) = -\frac{2}{3}(1 - \cos(1)) = \frac{2}{3}(\cos(1) - 1).

Question 5

The sum of iterated integrals ∫01∫0xf(x,y) dy dx+∫12∫02−xf(x,y) dy dx\int_0^1 \int_0^x f(x,y) \, dy \, dx + \int_1^2 \int_0^{2-x} f(x,y) \, dy \, dx is equal to which of the following single iterated integrals?

  1. ∫02∫0yf(x,y) dx dy\int_0^2 \int_0^{y} f(x,y) \, dx \, dy
  2. ∫01∫y2−yf(x,y) dx dy\int_0^1 \int_y^{2-y} f(x,y) \, dx \, dy (correct answer)
  3. ∫01∫2−yyf(x,y) dx dy\int_0^1 \int_{2-y}^y f(x,y) \, dx \, dy
  4. ∫02∫02−yf(x,y) dx dy\int_0^2 \int_0^{2-y} f(x,y) \, dx \, dy
Explanation: When you encounter a sum of iterated integrals like this, you're being asked to combine two regions of integration and express them as a single integral with reversed order of integration. The key is to visualize the regions geometrically and find their union. Let's analyze what regions these integrals represent. The first integral ∫01∫0xf(x,y) dy dx\int_0^1 \int_0^x f(x,y) \, dy \, dx describes the region where 0≤x≤10 \leq x \leq 1 and 0≤y≤x0 \leq y \leq x. The second integral ∫12∫02−xf(x,y) dy dx\int_1^2 \int_0^{2-x} f(x,y) \, dy \, dx covers 1≤x≤21 \leq x \leq 2 and 0≤y≤2−x0 \leq y \leq 2-x. To reverse the order of integration, we need to express this combined region in terms of yy first, then xx. Looking at both regions together, yy ranges from 0 to 1. For a fixed yy in [0,1][0,1], the xx-values span from where the line y=2−xy = 2-x intersects (giving x=2−yx = 2-y) to where y=xy = x intersects (giving x=yx = y). Therefore, xx ranges from 2−y2-y to yy. This gives us ∫01∫2−yyf(x,y) dx dy\int_0^1 \int_{2-y}^y f(x,y) \, dx \, dy, which is choice B. Choice A has incorrect bounds that don't match our region. Choice C has the xx-limits reversed, which would give a negative area. Choice D extends yy to 2 instead of 1, covering too large a region. Study tip: Always sketch the region of integration first. Converting between orders of integration becomes much clearer when you can visualize the geometric boundaries.

Question 6

Evaluate ∬R(x+y)2 dA\iint_R (x+y)^2 \, dA where RR is the region in the xy-plane defined by ∣x∣+∣y∣≤1|x|+|y| \le 1.

  1. 13\frac{1}{3}
  2. 23\frac{2}{3} (correct answer)
  3. 1
  4. 14\frac{1}{4}
Explanation: When evaluating double integrals over regions defined by absolute value inequalities like ∣x∣+∣y∣≤1|x| + |y| \leq 1, you need to recognize the geometry and exploit symmetry. This region forms a diamond (or square rotated 45°) with vertices at (1,0)(1,0), (0,1)(0,1), (−1,0)(-1,0), and (0,−1)(0,-1). Since both the region and integrand (x+y)2(x+y)^2 have useful symmetry properties, we can simplify the calculation. The region is symmetric about both axes, and (x+y)2(x+y)^2 is an even function with respect to the transformation (x,y)↦(−x,−y)(x,y) \mapsto (-x,-y). We can evaluate this by breaking the diamond into four triangular regions or by using the substitution u=x+yu = x+y, v=x−yv = x-y. Using the latter approach with Jacobian 12\frac{1}{2}, the region transforms to ∣u∣+∣v∣≤2|u| + |v| \leq 2, and our integral becomes: 12∬∣u∣+∣v∣≤2u2 du dv\frac{1}{2}\iint_{|u|+|v| \leq 2} u^2 \, du \, dv By symmetry about the vv-axis and direct integration, this evaluates to 23\frac{2}{3}. Answer (B) 23\frac{2}{3} is correct. Answer (A) 13\frac{1}{3} likely results from forgetting the Jacobian factor or miscalculating the limits. Answer (C) 1 might come from incorrectly treating this as the area of the region rather than the weighted integral. Answer (D) 14\frac{1}{4} could result from errors in setting up the integration limits or confusing this with a simpler integral. Strategy tip: When you see absolute value constraints defining regions, immediately sketch the region and look for symmetries that can simplify your calculation before diving into integration.

Question 7

Evaluate the iterated integral ∫08∫y32ex4 dx dy\int_0^8 \int_{\sqrt[3]{y}}^2 e^{x^4} \, dx \, dy.

  1. (e16−1)(e^{16}-1)
  2. 14(e8−1)\frac{1}{4}(e^8-1)
  3. 14e16\frac{1}{4}e^{16}
  4. 14(e16−1)\frac{1}{4}(e^{16}-1) (correct answer)
Explanation: The inner integral with respect to xx, ∫ex4 dx\int e^{x^4} \, dx, cannot be evaluated in terms of elementary functions. This suggests reversing the order of integration. The region of integration is defined by 0≤y≤80 \le y \le 8 and y3≤x≤2\sqrt[3]{y} \le x \le 2. This can be rewritten as y=x3y = x^3, x=2x=2, and y=0y=0. To reverse the order, we express the limits for yy in terms of xx. The region is bounded by xx from 00 to 22, and for each xx, yy ranges from 00 to x3x^3. The new integral is ∫02∫0x3ex4 dy dx\int_0^2 \int_0^{x^3} e^{x^4} \, dy \, dx. The inner integral is ∫0x3ex4 dy=[yex4]0x3=x3ex4\int_0^{x^3} e^{x^4} \, dy = [y e^{x^4}]_0^{x^3} = x^3 e^{x^4}. The outer integral is ∫02x3ex4 dx\int_0^2 x^3 e^{x^4} \, dx. Using the substitution u=x4u = x^4, we have du=4x3 dxdu = 4x^3 \, dx, so x3 dx=du/4x^3 \, dx = du/4. The limits become u(0)=0u(0)=0 and u(2)=16u(2)=16. The integral becomes ∫016eudu4=14[eu]016=14(e16−e0)=14(e16−1)\int_0^{16} e^u \frac{du}{4} = \frac{1}{4}[e^u]_0^{16} = \frac{1}{4}(e^{16}-e^0) = \frac{1}{4}(e^{16}-1).

Question 8

Evaluate ∬Ry dA\iint_R y \, dA where RR is the region bounded by the semicircle y=9−x2y=\sqrt{9-x^2} and the x-axis.

  1. 9π2\frac{9\pi}{2}
  2. 0
  3. 18 (correct answer)
  4. 4π\frac{4}{\pi}
Explanation: When you encounter a double integral over a region defined by geometric boundaries, start by visualizing the region and choosing appropriate integration limits. Here, the semicircle y=9−x2y = \sqrt{9-x^2} with the x-axis forms the upper half of a circle with radius 3 centered at the origin. To evaluate ∬Ry dA\iint_R y \, dA, set up the integral with bounds: xx ranges from −3-3 to 33, and for each xx, yy ranges from 00 to 9−x2\sqrt{9-x^2}. This gives us: ∫−33∫09−x2y dy dx\int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} y \, dy \, dx First, integrate with respect to yy: ∫09−x2y dy=[y22]09−x2=9−x22\int_{0}^{\sqrt{9-x^2}} y \, dy = \left[\frac{y^2}{2}\right]_{0}^{\sqrt{9-x^2}} = \frac{9-x^2}{2} Now integrate with respect to xx: ∫−339−x22 dx=12[9x−x33]−33=12[(27−9)−(−27+9)]=12[36]=18\int_{-3}^{3} \frac{9-x^2}{2} \, dx = \frac{1}{2}\left[9x - \frac{x^3}{3}\right]_{-3}^{3} = \frac{1}{2}[(27-9) - (-27+9)] = \frac{1}{2}[36] = 18 Answer choice A (9π2\frac{9\pi}{2}) likely comes from confusing this with an area calculation or using polar coordinates incorrectly. Answer choice B (0) would result from incorrectly thinking the integral is zero due to symmetry—but yy is always positive in this region. Answer choice D (4π\frac{4}{\pi}) appears unrelated to this geometry. Remember: when integrating yy over a region where y≥0y \geq 0, the result represents the first moment about the x-axis, not just area. Always check that your bounds match the actual geometric region.

Question 9

Find the volume of the solid in the first octant (x≥0,y≥0,z≥0x \ge 0, y \ge 0, z \ge 0) bounded by the cylinder y2+z2=4y^2 + z^2 = 4, the plane x+y=2x+y=2, and the coordinate planes.

  1. 2π−832\pi - \frac{8}{3} (correct answer)
  2. 43\frac{4}{3}
  3. 2π−1632\pi - \frac{16}{3}
  4. 2π+832\pi + \frac{8}{3}
Explanation: The volume is given by the double integral of the height function z=f(x,y)z=f(x,y) over a region RR in the xy-plane. From the cylinder equation, z=4−y2z = \sqrt{4-y^2} (since z≥0z \ge 0). The region RR is bounded by x+y=2x+y=2 and the coordinate planes x=0,y=0x=0, y=0. This is a triangle with vertices (0,0),(2,0),(0,2)(0,0), (2,0), (0,2). The volume is V=∬R4−y2 dAV = \iint_R \sqrt{4-y^2} \, dA. It is easier to integrate with respect to xx first. The limits are 0≤y≤20 \le y \le 2 and 0≤x≤2−y0 \le x \le 2-y. The integral is V=∫02∫02−y4−y2 dx dyV = \int_0^2 \int_0^{2-y} \sqrt{4-y^2} \, dx \, dy. The inner integral is ∫02−y4−y2 dx=(2−y)4−y2\int_0^{2-y} \sqrt{4-y^2} \, dx = (2-y)\sqrt{4-y^2}. The outer integral is ∫02(2−y)4−y2 dy=∫0224−y2 dy−∫02y4−y2 dy\int_0^2 (2-y)\sqrt{4-y^2} \, dy = \int_0^2 2\sqrt{4-y^2} \, dy - \int_0^2 y\sqrt{4-y^2} \, dy. The first term, 2∫024−y2 dy2\int_0^2 \sqrt{4-y^2} \, dy, is twice the area of a quarter-circle of radius 2, which is 2(14π(22))=2π2(\frac{1}{4}\pi(2^2)) = 2\pi. The second term can be solved with u-substitution (u=4−y2u=4-y^2), yielding 83\frac{8}{3}. Thus, the volume is 2π−832\pi - \frac{8}{3}.