Calculus 2 Quiz: Cross Sections Squares And Rectangles
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Cross Sections Squares And RectanglesQuestion 1 of 20

Let R be the region bounded by y=1/x2y = 1/x^2, y=0y=0, x=1x=1, and x=2x=2. A solid has base R and cross-sections perpendicular to the x-axis that are squares. What is the volume of the solid?

12\frac{1}{2}
724\frac{7}{24}
ln⁡2\ln 2
13\frac{1}{3}
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Calculus 2 Quiz

Calculus 2 Quiz: Cross Sections Squares And Rectangles

Practice Cross Sections Squares And Rectangles in Calculus 2 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 Cross Sections Squares And Rectangles, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let R be the region bounded by y=1/x2y = 1/x^2, y=0y=0, x=1x=1, and x=2x=2. A solid has base R and cross-sections perpendicular to the x-axis that are squares. What is the volume of the solid?

  1. 12\frac{1}{2}
  2. 724\frac{7}{24} (correct answer)
  3. ln⁡2\ln 2
  4. 13\frac{1}{3}
Explanation: The limits of integration are from x=1x=1 to x=2x=2. The side of the square cross-section is s(x)=1/x2s(x) = 1/x^2. The area is A(x)=(1/x2)2=1/x4=x−4A(x) = (1/x^2)^2 = 1/x^4 = x^{-4}. The volume is V=∫12x−4dx=[x−3−3]12=[−13x3]12=(−13(23))−(−13(13))=−124+13=−1+824=724V = \int_1^2 x^{-4} dx = [\frac{x^{-3}}{-3}]_1^2 = [-\frac{1}{3x^3}]_1^2 = (-\frac{1}{3(2^3)}) - (-\frac{1}{3(1^3)}) = -\frac{1}{24} + \frac{1}{3} = \frac{-1+8}{24} = \frac{7}{24}.

Question 2

The base of a solid is the region in the first quadrant bounded by y=cos⁡(x)y=\cos(x), y=sin⁡(x)y=\sin(x), and the y-axis. Cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?

  1. π4−12\frac{\pi}{4} - \frac{1}{2} (correct answer)
  2. π2−1\frac{\pi}{2} - 1
  3. 1−221 - \frac{\sqrt{2}}{2}
  4. π4\frac{\pi}{4}
Explanation: The curves intersect when cos⁡(x)=sin⁡(x)\cos(x) = \sin(x), which is at x=π/4x = \pi/4 in the first quadrant. From x=0x=0 to x=π/4x=\pi/4, cos⁡(x)≥sin⁡(x)\cos(x) \ge \sin(x). The side of the square is s(x)=cos⁡(x)−sin⁡(x)s(x) = \cos(x) - \sin(x). The area is A(x)=(cos⁡(x)−sin⁡(x))2=cos⁡2(x)−2sin⁡(x)cos⁡(x)+sin⁡2(x)=1−sin⁡(2x)A(x) = (\cos(x) - \sin(x))^2 = \cos^2(x) - 2\sin(x)\cos(x) + \sin^2(x) = 1 - \sin(2x). The volume is V=∫0π/4(1−sin⁡(2x))dx=[x+12cos⁡(2x)]0π/4=(π4+12cos⁡(π2))−(0+12cos⁡(0))=(π4+0)−(12)=π4−12V = \int_{0}^{\pi/4} (1 - \sin(2x)) dx = [x + \frac{1}{2}\cos(2x)]_{0}^{\pi/4} = (\frac{\pi}{4} + \frac{1}{2}\cos(\frac{\pi}{2})) - (0 + \frac{1}{2}\cos(0)) = (\frac{\pi}{4} + 0) - (\frac{1}{2}) = \frac{\pi}{4} - \frac{1}{2}.

Question 3

The base of a solid is the region enclosed by y=x2y=x^2 and y=ky=k, where k>0k>0. Cross-sections perpendicular to the y-axis are squares. If the volume of the solid is 50, what is the value of kk?

  1. 55 (correct answer)
  2. 5\sqrt{5}
  3. 525\sqrt{2}
  4. 2525
Explanation: For cross-sections perpendicular to the y-axis, we integrate with respect to yy from y=0y=0 to y=ky=k. The curves are x=yx = \sqrt{y} and x=−yx = -\sqrt{y}. The side length of the square is s(y)=y−(−y)=2ys(y) = \sqrt{y} - (-\sqrt{y}) = 2\sqrt{y}. The area is A(y)=(2y)2=4yA(y) = (2\sqrt{y})^2 = 4y. The volume is V=∫0k4ydy=[2y2]0k=2k2V = \int_{0}^{k} 4y dy = [2y^2]_{0}^{k} = 2k^2. Given that V=50V=50, we have 2k2=50⇒k2=25⇒k=52k^2 = 50 \Rightarrow k^2 = 25 \Rightarrow k=5.

Question 4

Let RR be the region in the first quadrant bounded by y=xy=x and y=x2y=x^2. Let VSV_S be the volume of a solid with base RR and square cross-sections perpendicular to the x-axis. Let VRV_R be the volume of a solid with the same base RR but with rectangular cross-sections perpendicular to the x-axis whose height is half their base. What is the ratio VRVS\frac{V_R}{V_S}?

  1. 14\frac{1}{4}
  2. 12\frac{1}{2} (correct answer)
  3. 11
  4. 22
Explanation: For any cross-section at a given xx, the base is s(x)=x−x2s(x) = x - x^2. For the solid with squares, the area is AS(x)=s(x)2=(x−x2)2A_S(x) = s(x)^2 = (x-x^2)^2. The volume is VS=∫01(x−x2)2dxV_S = \int_0^1 (x-x^2)^2 dx. For the solid with rectangles, the area is AR(x)=s(x)⋅h(x)A_R(x) = s(x) \cdot h(x), where h(x)=12s(x)h(x) = \frac{1}{2}s(x). So, AR(x)=s(x)⋅12s(x)=12s(x)2A_R(x) = s(x) \cdot \frac{1}{2}s(x) = \frac{1}{2}s(x)^2. The volume is VR=∫0112(x−x2)2dx=12∫01(x−x2)2dx=12VSV_R = \int_0^1 \frac{1}{2}(x-x^2)^2 dx = \frac{1}{2} \int_0^1 (x-x^2)^2 dx = \frac{1}{2}V_S. Therefore, the ratio VRVS=12\frac{V_R}{V_S} = \frac{1}{2}.

Question 5

The base of a solid is the region bounded by y=xy = \sqrt{x}, x=9x=9, and the x-axis. The cross-sections perpendicular to the x-axis are squares whose diagonals lie in the xy-plane. What is the volume of the solid?

  1. 814\frac{81}{4} (correct answer)
  2. 812\frac{81}{2}
  3. 8181
  4. 2432\frac{243}{2}
Explanation: The length of the diagonal of a square cross-section at xx is d(x)=x−0=xd(x) = \sqrt{x} - 0 = \sqrt{x}. The area of a square with diagonal dd is A=d22A = \frac{d^2}{2}. Therefore, the area of a cross-section is A(x)=(x)22=x2A(x) = \frac{(\sqrt{x})^2}{2} = \frac{x}{2}. The volume is V=∫09x2dx=[x24]09=814V = \int_{0}^{9} \frac{x}{2} dx = [\frac{x^2}{4}]_{0}^{9} = \frac{81}{4}.

Question 6

The base of a solid is the region enclosed by y=exy = e^x, y=e−xy = e^{-x}, and x=1x = 1. Cross sections perpendicular to the x-axis are squares. If the volume is expressed as ∫abf(x)dx\int_{a}^{b} f(x) dx, what is f(1)f(1)?

  1. e2−e−2e^2 - e^{-2}
  2. (e−e−1)2(e - e^{-1})^2 (correct answer)
  3. e2+e−2e^2 + e^{-2}
  4. 2(e−e−1)2(e - e^{-1})
Explanation: The curves y=exy = e^x and y=e−xy = e^{-x} intersect when ex=e−xe^x = e^{-x}, which gives x=0x = 0. For the region from x=0x = 0 to x=1x = 1, the side length of each square cross section is ex−e−xe^x - e^{-x}. The area of the square is (ex−e−x)2(e^x - e^{-x})^2, so f(x)=(ex−e−x)2f(x) = (e^x - e^{-x})^2. Therefore, f(1)=(e1−e−1)2=(e−e−1)2f(1) = (e^1 - e^{-1})^2 = (e - e^{-1})^2. Choice A gives the side length, not the area. Choice C incorrectly adds the exponentials. Choice D gives twice the side length.

Question 7

Consider a solid whose base is bounded by x=y2x = y^2 and x=4x = 4. Cross sections perpendicular to the x-axis are squares. If we set up the volume integral in the form ∫cdg(x)dx\int_{c}^{d} g(x) dx, what is the correct expression for g(x)g(x)?

  1. g(x)=4xg(x) = 4\sqrt{x}
  2. g(x)=2xg(x) = 2\sqrt{x}
  3. g(x)=4xg(x) = 4x (correct answer)
  4. g(x)=xg(x) = \sqrt{x}
Explanation: The parabola x=y2x = y^2 intersects the line x=4x = 4 when y2=4y^2 = 4, so y=±2y = \pm 2. For a given xx-value between 0 and 4, the yy-values range from −x-\sqrt{x} to x\sqrt{x}, giving a total distance of 2x2\sqrt{x}. This is the side length of the square cross section. The area of the square is (2x)2=4x(2\sqrt{x})^2 = 4x, so g(x)=4xg(x) = 4x. Choice A gives twice the distance from the x-axis to one branch. Choice B gives the side length, not the area. Choice D gives half the side length.

Question 8

The base of a solid is the region enclosed by x=y2x = y^2 and x=y+2x = y+2. If the cross-sections perpendicular to the y-axis are squares, which of the following integrals gives the volume of the solid?

  1. ∫−12(y+2−y2)2dy\int_{-1}^{2} (y+2-y^2)^2 dy (correct answer)
  2. ∫−12(y+2−y2)dy\int_{-1}^{2} (y+2-y^2) dy
  3. ∫01(2x)2dx+∫14(x−(x−2))2dx\int_{0}^{1} (2\sqrt{x})^2 dx + \int_{1}^{4} (\sqrt{x} - (x-2))^2 dx
  4. π∫−12((y+2)2−(y2)2)dy\pi \int_{-1}^{2} ((y+2)^2 - (y^2)^2) dy
Explanation: To find the limits, set y2=y+2y^2 = y+2, which yields y2−y−2=0y^2 - y - 2 = 0, or (y−2)(y+1)=0(y-2)(y+1)=0. The limits are y=−1y=-1 to y=2y=2. For cross-sections perpendicular to the y-axis, we integrate with respect to yy. The rightmost curve is x=y+2x = y+2 and the leftmost is x=y2x = y^2. The side length of the square is s(y)=(y+2)−y2s(y) = (y+2) - y^2. The area is A(y)=s(y)2=(y+2−y2)2A(y) = s(y)^2 = (y+2-y^2)^2. The volume is the integral of the area, V=∫−12(y+2−y2)2dyV = \int_{-1}^{2} (y+2-y^2)^2 dy.

Question 9

The base of a solid is the region enclosed by the parabola x=4−y2x = 4 - y^2 and the y-axis. Cross-sections perpendicular to the y-axis are rectangles with a constant height of 5. What is the volume of the solid?

  1. 3232
  2. 51215\frac{512}{15}
  3. 1603\frac{160}{3} (correct answer)
  4. 8080
Explanation: The region is bounded by x=4−y2x=4-y^2 and x=0x=0. The limits of integration are found by setting 4−y2=04-y^2=0, which gives y=±2y=\pm 2. The base of a rectangular cross-section at yy is s(y)=(4−y2)−0=4−y2s(y) = (4-y^2)-0 = 4-y^2. The height is a constant h=5h=5. The area is A(y)=s(y)h=5(4−y2)A(y) = s(y)h = 5(4-y^2). The volume is V=∫−225(4−y2)dy=5[4y−y33]−22=5[(8−83)−(−8+83)]=5[323]=1603V = \int_{-2}^{2} 5(4-y^2) dy = 5[4y - \frac{y^3}{3}]_{-2}^{2} = 5[(8-\frac{8}{3}) - (-8+\frac{8}{3})] = 5[\frac{32}{3}] = \frac{160}{3}.

Question 10

The base of a solid is the region in the first quadrant bounded by y=x2y = x^2, y=4y = 4, and the y-axis. Cross-sections perpendicular to the x-axis are rectangles whose height is equal to xx. What is the volume of the solid?

  1. 44 (correct answer)
  2. 88
  3. 12815\frac{128}{15}
  4. 163\frac{16}{3}
Explanation: The region is bounded by x=0x=0, x=2x=2 (since x2=4x^2=4), y=x2y=x^2 (bottom), and y=4y=4 (top). For cross-sections perpendicular to the x-axis, the base of the rectangle is s(x)=4−x2s(x) = 4 - x^2. The height is given as h(x)=xh(x) = x. The area of the cross-section is A(x)=s(x)h(x)=(4−x2)x=4x−x3A(x) = s(x)h(x) = (4-x^2)x = 4x-x^3. The volume is V=∫02(4x−x3)dx=[2x2−x44]02=2(22)−244=8−4=4V = \int_{0}^{2} (4x-x^3) dx = [2x^2 - \frac{x^4}{4}]_{0}^{2} = 2(2^2) - \frac{2^4}{4} = 8 - 4 = 4.

Question 11

The base of a solid is the region in the first quadrant bounded by y=x3y = x^3 and y=xy = x. The cross-sections perpendicular to the x-axis are rectangles whose height is equal to the y-coordinate of the upper bounding curve. What is the volume of the solid?

  1. 421\frac{4}{21}
  2. 8105\frac{8}{105}
  3. 235\frac{2}{35}
  4. 215\frac{2}{15} (correct answer)
Explanation: This is a volumes by cross-sections problem, where you need to set up an integral based on the geometry of the solid. When cross-sections are perpendicular to the x-axis, you integrate with respect to x, and each cross-section's area becomes your integrand. First, find where the curves intersect by solving x3=xx^3 = x, which gives x(x2−1)=0x(x^2 - 1) = 0, so x=0,1,−1x = 0, 1, -1. Since we're in the first quadrant, the region spans from x=0x = 0 to x=1x = 1. Between these points, y=xy = x is above y=x3y = x^3 (since x>x3x > x^3 when 0<x<10 < x < 1). Each rectangular cross-section has width equal to the distance between the curves: x−x3x - x^3. The height equals the y-coordinate of the upper curve, which is y=xy = x. So each cross-section has area (x−x3)⋅x=x2−x4(x - x^3) \cdot x = x^2 - x^4. The volume is: V=∫01(x2−x4)dx=[x33−x55]01=13−15=5−315=215V = \int_0^1 (x^2 - x^4) dx = \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \frac{1}{3} - \frac{1}{5} = \frac{5-3}{15} = \frac{2}{15} Choice A (421\frac{4}{21}) likely comes from incorrectly using x3x^3 as the height instead of xx. Choice B (8105\frac{8}{105}) might result from errors in setting up the cross-sectional area. Choice C (235\frac{2}{35}) could come from computational mistakes in the integration. Remember: for cross-sections perpendicular to the x-axis, carefully identify the width and height of each cross-section, then integrate the area function over the appropriate interval.

Question 12

Let the base of a solid be the region bounded by the parabola y=4−x2y = 4 - x^2 and the x-axis. The cross-sections perpendicular to the x-axis are rectangles whose height is twice the length of their base. What is the volume of the solid?

  1. 51215\frac{512}{15}
  2. 102415\frac{1024}{15} (correct answer)
  3. 643\frac{64}{3}
  4. 00
Explanation: The base extends from x=−2x=-2 to x=2x=2. The base of a rectangular cross-section at xx is s(x)=(4−x2)−0=4−x2s(x) = (4-x^2) - 0 = 4-x^2. The height is given as h(x)=2s(x)=2(4−x2)h(x) = 2s(x) = 2(4-x^2). The area of the cross-section is A(x)=s(x)h(x)=(4−x2)⋅2(4−x2)=2(4−x2)2A(x) = s(x)h(x) = (4-x^2) \cdot 2(4-x^2) = 2(4-x^2)^2. The volume is V=∫−222(4−x2)2dx=2∫−22(16−8x2+x4)dx=2[16x−8x33+x55]−22=102415V = \int_{-2}^{2} 2(4-x^2)^2 dx = 2\int_{-2}^{2} (16-8x^2+x^4) dx = 2[16x - \frac{8x^3}{3} + \frac{x^5}{5}]_{-2}^{2} = \frac{1024}{15}.

Question 13

A student sets up the integral V=∫04(y)2dyV = \int_{0}^{4} (\sqrt{y})^2 dy to find the volume of a solid. The solid's base is in the xy-plane and its cross-sections are squares. Which of the following statements could describe the solid?

  1. The base is bounded by y=x2y=x^2 and y=4y=4, and cross-sections are perpendicular to the x-axis.
  2. The base is bounded by x=yx=\sqrt{y}, x=0x=0, y=0y=0, and y=4y=4, and cross-sections are perpendicular to the y-axis. (correct answer)
  3. The base is bounded by y=xy=\sqrt{x}, y=0y=0, and x=4x=4, and cross-sections are perpendicular to the y-axis.
  4. The base is bounded by x=y2x=y^2 and x=2x=2, and cross-sections are perpendicular to the y-axis.
Explanation: The integral is with respect to yy, so the cross-sections must be perpendicular to the y-axis. The limits of integration are from y=0y=0 to y=4y=4. The integrand, (y)2(\sqrt{y})^2, represents the area of a square cross-section. This means the side length is s(y)=ys(y) = \sqrt{y}. This side length is a horizontal distance. s(y)=xright−xleft=y−0s(y) = x_{right} - x_{left} = \sqrt{y} - 0. So the base is bounded on the right by x=yx = \sqrt{y} and on the left by x=0x=0 (the y-axis), between y=0y=0 and y=4y=4.

Question 14

The base of a solid is the region bounded by y=1xy = \frac{1}{x}, x=1x=1, x=4x=4, and y=0y=0. Cross-sections perpendicular to the x-axis are rectangles whose height is always 3. What is the volume of the solid?

  1. 3ln⁡43 \ln 4 (correct answer)
  2. 33
  3. −316-\frac{3}{16}
  4. ln⁡64\ln 64
Explanation: The base of a rectangular cross-section at xx is s(x)=1x−0=1xs(x) = \frac{1}{x} - 0 = \frac{1}{x}. The height is given as a constant, h=3h=3. The area of the cross-section is A(x)=s(x)h=1x⋅3=3xA(x) = s(x)h = \frac{1}{x} \cdot 3 = \frac{3}{x}. The volume is V=∫143xdx=3[ln⁡∣x∣]14=3(ln⁡4−ln⁡1)=3ln⁡4V = \int_{1}^{4} \frac{3}{x} dx = 3[\ln|x|]_{1}^{4} = 3(\ln 4 - \ln 1) = 3 \ln 4.

Question 15

The base of a solid is the region in the xy-plane enclosed by the graphs of y=x2y = x^2 and y=x+2y = x+2. The cross-sections of the solid perpendicular to the x-axis are squares. What is the volume of the solid?

  1. 8110\frac{81}{10} (correct answer)
  2. 92\frac{9}{2}
  3. 12815\frac{128}{15}
  4. 725\frac{72}{5}
Explanation: The points of intersection are found by setting x2=x+2x^2 = x+2, which gives x2−x−2=0x^2 - x - 2 = 0, or (x−2)(x+1)=0(x-2)(x+1) = 0. So, the limits of integration are from x=−1x=-1 to x=2x=2. The upper curve is y=x+2y = x+2 and the lower curve is y=x2y = x^2. The side length of a square cross-section at xx is s(x)=(x+2)−x2s(x) = (x+2) - x^2. The area of the cross-section is A(x)=s(x)2=(x+2−x2)2A(x) = s(x)^2 = (x+2-x^2)^2. The volume is the integral of the area: V=∫−12(x+2−x2)2dx=∫−12(x4−2x3−3x2+4x+4)dx=[x55−x42−x3+2x2+4x]−12=8110V = \int_{-1}^{2} (x+2-x^2)^2 dx = \int_{-1}^{2} (x^4 - 2x^3 - 3x^2 + 4x + 4) dx = [\frac{x^5}{5} - \frac{x^4}{2} - x^3 + 2x^2 + 4x]_{-1}^{2} = \frac{81}{10}.

Question 16

A solid has a base bounded by the curves y=x2y = x^2 and y=4y = 4. Each cross section perpendicular to the x-axis is a square whose base lies in the xy-plane. If the volume of the solid is VV, what is the relationship between VV and the integral ∫−22(4−x2)2dx\int_{-2}^{2} (4-x^2)^2 dx?

  1. V=∫−22(4−x2)2dxV = \int_{-2}^{2} (4-x^2)^2 dx (correct answer)
  2. V=2∫−22(4−x2)2dxV = 2\int_{-2}^{2} (4-x^2)^2 dx
  3. V=12∫−22(4−x2)2dxV = \frac{1}{2}\int_{-2}^{2} (4-x^2)^2 dx
  4. V=∫−224−x2dxV = \int_{-2}^{2} \sqrt{4-x^2} dx
Explanation: The base of each square cross section has side length equal to the distance between the curves: 4−x24 - x^2. Since the cross section is a square, its area is (4−x2)2(4-x^2)^2. The volume is found by integrating these areas from x=−2x = -2 to x=2x = 2, giving V=∫−22(4−x2)2dxV = \int_{-2}^{2} (4-x^2)^2 dx. Choice B incorrectly doubles the integral. Choice C incorrectly halves it. Choice D uses the side length instead of the area.

Question 17

A solid is constructed with base region defined by ∣x∣+∣y∣≤2|x| + |y| \leq 2. Cross sections perpendicular to the x-axis are rectangles with height equal to half the base width. What is the volume of this solid?

  1. ∫−22(2−∣x∣)2dx\int_{-2}^{2} (2-|x|)^2 dx (correct answer)
  2. ∫−222(2−∣x∣)2dx\int_{-2}^{2} 2(2-|x|)^2 dx
  3. ∫−22(2−∣x∣)dx\int_{-2}^{2} (2-|x|) dx
  4. ∫−2212(2−∣x∣)2dx\int_{-2}^{2} \frac{1}{2}(2-|x|)^2 dx
Explanation: The region ∣x∣+∣y∣≤2|x| + |y| \leq 2 is a diamond with vertices at (±2,0)(\pm 2, 0) and (0,±2)(0, \pm 2). For a given xx, the yy-values satisfy ∣y∣≤2−∣x∣|y| \leq 2 - |x|, so yy ranges from −(2−∣x∣)-(2-|x|) to (2−∣x∣)(2-|x|). The base width is 2(2−∣x∣)2(2-|x|) and the height is half this: (2−∣x∣)(2-|x|). The cross-sectional area is base × height = 2(2−∣x∣)⋅(2−∣x∣)=(2−∣x∣)22(2-|x|) \cdot (2-|x|) = (2-|x|)^2. Choice B doubles the correct area. Choice C gives base width times height but missing the factor of 2 in base width. Choice D gives half the correct area.

Question 18

A solid is formed with base in the first quadrant bounded by y=sin⁡xy = \sin x, y=cos⁡xy = \cos x, and the y-axis. Cross sections perpendicular to the x-axis are rectangles with height twice the base width. What integral represents the volume?

  1. ∫0π/4(cos⁡x−sin⁡x)dx\int_{0}^{\pi/4} (\cos x - \sin x) dx
  2. ∫0π/42(cos⁡x−sin⁡x)dx\int_{0}^{\pi/4} 2(\cos x - \sin x) dx
  3. ∫0π/42(cos⁡x−sin⁡x)2dx\int_{0}^{\pi/4} 2(\cos x - \sin x)^2 dx (correct answer)
  4. ∫0π/4(cos⁡x−sin⁡x)2dx\int_{0}^{\pi/4} (\cos x - \sin x)^2 dx
Explanation: The curves intersect at x=π/4x = \pi/4 where sin⁡x=cos⁡x\sin x = \cos x. For 0≤x≤π/40 \leq x \leq \pi/4, we have cos⁡x≥sin⁡x\cos x \geq \sin x. The base width of each rectangle is cos⁡x−sin⁡x\cos x - \sin x, and the height is twice this: 2(cos⁡x−sin⁡x)2(\cos x - \sin x). The area of each rectangle is base × height = (cos⁡x−sin⁡x)⋅2(cos⁡x−sin⁡x)=2(cos⁡x−sin⁡x)2(\cos x - \sin x) \cdot 2(\cos x - \sin x) = 2(\cos x - \sin x)^2. Choice A gives the area under one curve. Choice B gives twice the base width, not the rectangle area. Choice D omits the factor of 2 from the height condition.

Question 19

A solid has its base as the region between y=xy = \sqrt{x} and y=x2y = x^2 for 0≤x≤10 \leq x \leq 1. Each cross section perpendicular to the x-axis is a rectangle whose height is 3 times its base. Which expression gives the volume?

  1. ∫01(x−x2)dx\int_{0}^{1} (\sqrt{x} - x^2) dx
  2. 3∫01(x−x2)2dx3\int_{0}^{1} (\sqrt{x} - x^2)^2 dx (correct answer)
  3. ∫013(x−x2)dx\int_{0}^{1} 3(\sqrt{x} - x^2) dx
  4. ∫01(x−x2)2dx\int_{0}^{1} (\sqrt{x} - x^2)^2 dx
Explanation: For 0≤x≤10 \leq x \leq 1, we have x≥x2\sqrt{x} \geq x^2. The base width of each rectangle is x−x2\sqrt{x} - x^2, and the height is 3(x−x2)3(\sqrt{x} - x^2). The area of each rectangle is base × height = (x−x2)⋅3(x−x2)=3(x−x2)2(\sqrt{x} - x^2) \cdot 3(\sqrt{x} - x^2) = 3(\sqrt{x} - x^2)^2. The volume is ∫013(x−x2)2dx=3∫01(x−x2)2dx\int_{0}^{1} 3(\sqrt{x} - x^2)^2 dx = 3\int_{0}^{1} (\sqrt{x} - x^2)^2 dx. Choice A gives the area between the curves, not the cross-sectional area. Choice C gives 3 times the area between curves. Choice D omits the factor of 3 from the height condition.

Question 20

A solid is formed with base region bounded by y=sec⁡xy = \sec x and y=2y = 2 for −π3≤x≤π3-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}. Cross sections perpendicular to the x-axis are rectangles with height equal to 34\frac{3}{4} times the base width. What integral represents the volume?

  1. ∫−π/3π/334(2−sec⁡x)dx\int_{-\pi/3}^{\pi/3} \frac{3}{4}(2 - \sec x) dx
  2. ∫−π/3π/334(sec⁡x−2)2dx\int_{-\pi/3}^{\pi/3} \frac{3}{4}(\sec x - 2)^2 dx
  3. ∫−π/3π/334(sec⁡x−2)dx\int_{-\pi/3}^{\pi/3} \frac{3}{4}(\sec x - 2) dx
  4. ∫−π/3π/334(2−sec⁡x)2dx\int_{-\pi/3}^{\pi/3} \frac{3}{4}(2 - \sec x)^2 dx (correct answer)
Explanation: When you encounter a volume problem with cross-sections, you need to visualize the solid's construction. Here, you're building a solid where each vertical slice at position x is a rectangle, and you integrate these slices to find total volume. First, identify the base width of each rectangular cross-section. The region is bounded above by y=2y = 2 and below by y=sec⁡xy = \sec x. Since sec⁡x≤2\sec x \leq 2 in the given interval, the base width at any x-value is (2−sec⁡x)(2 - \sec x). The problem states that the height equals 34\frac{3}{4} times this base width, so the height is 34(2−sec⁡x)\frac{3}{4}(2 - \sec x). The area of each rectangular cross-section is base × height = (2−sec⁡x)×34(2−sec⁡x)=34(2−sec⁡x)2(2 - \sec x) \times \frac{3}{4}(2 - \sec x) = \frac{3}{4}(2 - \sec x)^2. Integrating from x=−π3x = -\frac{\pi}{3} to x=π3x = \frac{\pi}{3} gives the volume. Option A gives 34(2−sec⁡x)\frac{3}{4}(2 - \sec x) without squaring—this would be the height alone, not the area of the cross-section. Option B incorrectly uses (sec⁡x−2)2(\sec x - 2)^2, which assumes sec⁡x>2\sec x > 2, but in our interval sec⁡x<2\sec x < 2, making this expression negative. Option C uses (sec⁡x−2)(\sec x - 2) without squaring, again missing the area calculation and using the wrong sign. Remember: for volume by cross-sections, always multiply the dimensions to get the cross-sectional area, then integrate. Pay careful attention to which function is on top to avoid sign errors.