Calculus 1 Quiz: Cross Sections Triangles And Semicircles
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Cross Sections Triangles And SemicirclesQuestion 1 of 10

The base of a solid is the region enclosed by the graphs of y=xy=x and y=x3y=x^3. Cross-sections perpendicular to the x-axis are semicircles. What is the volume of the solid?

π105\frac{\pi}{105}
π21\frac{\pi}{21}
4π105\frac{4\pi}{105}
2π105\frac{2\pi}{105}
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Calculus 1 Quiz

Calculus 1 Quiz: Cross Sections Triangles And Semicircles

Practice Cross Sections Triangles And Semicircles in Calculus 1 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 Cross Sections Triangles And Semicircles, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

The base of a solid is the region enclosed by the graphs of y=xy=x and y=x3y=x^3. Cross-sections perpendicular to the x-axis are semicircles. What is the volume of the solid?

  1. π105\frac{\pi}{105}
  2. π21\frac{\pi}{21}
  3. 4π105\frac{4\pi}{105}
  4. 2π105\frac{2\pi}{105} (correct answer)
Explanation: This is a solid of revolution problem where you need to set up an integral to find volume using cross-sectional areas. When cross-sections perpendicular to an axis have a known shape, the volume equals the integral of the cross-sectional area over the interval. First, find where the curves intersect by solving x=x3x = x^3, which gives x(1−x2)=0x(1-x^2) = 0, so x=0,1,−1x = 0, 1, -1. Since we want the enclosed region, we integrate from x=0x = 0 to x=1x = 1, where y=xy = x is above y=x3y = x^3. The diameter of each semicircular cross-section is the distance between the curves: x−x3x - x^3. So the radius is x−x32\frac{x - x^3}{2}, and the area of each semicircle is 12πr2=12π(x−x32)2=π(x−x3)28\frac{1}{2}\pi r^2 = \frac{1}{2}\pi\left(\frac{x - x^3}{2}\right)^2 = \frac{\pi(x - x^3)^2}{8}. The volume is: V=∫01π(x−x3)28dx=π8∫01(x2−2x4+x6)dx=π8[x33−2x55+x77]01=π8(13−25+17)=π8⋅4105=2π105V = \int_0^1 \frac{\pi(x - x^3)^2}{8} dx = \frac{\pi}{8}\int_0^1 (x^2 - 2x^4 + x^6) dx = \frac{\pi}{8}\left[\frac{x^3}{3} - \frac{2x^5}{5} + \frac{x^7}{7}\right]_0^1 = \frac{\pi}{8}\left(\frac{1}{3} - \frac{2}{5} + \frac{1}{7}\right) = \frac{\pi}{8} \cdot \frac{4}{105} = \frac{2\pi}{105} Choice A) π105\frac{\pi}{105} likely forgot the factor of 2 in the final calculation. Choice B) π21\frac{\pi}{21} probably used the full circle area instead of semicircle. Choice C) 4π105\frac{4\pi}{105} missed the 18\frac{1}{8} factor when squaring the radius expression. Always carefully track your factors when setting up cross-sectional area formulas—semicircles, quarters of the radius expression, and integration constants can easily be missed.

Question 2

A solid has as its base the region bounded by the curve y=sin⁡(x)y=\sin(x) and the x-axis from x=0x=0 to x=πx=\pi. The cross-sections of the solid perpendicular to the x-axis are isosceles right triangles with their hypotenuse lying on the base. What is the volume of the solid?

  1. π16\frac{\pi}{16}
  2. π8\frac{\pi}{8} (correct answer)
  3. π4\frac{\pi}{4}
  4. π2\frac{\pi}{2}
Explanation: The base of each triangular cross-section has length s(x)=sin⁡(x)s(x) = \sin(x). If this length is the hypotenuse of an isosceles right triangle, its legs have length s/2s/\sqrt{2}. The area of the triangle is A(x)=12⋅(s2)⋅(s2)=14s2A(x) = \frac{1}{2} \cdot (\frac{s}{\sqrt{2}}) \cdot (\frac{s}{\sqrt{2}}) = \frac{1}{4}s^2. Thus, A(x)=14sin⁡2(x)A(x) = \frac{1}{4}\sin^2(x). The volume is V=∫0π14sin⁡2(x)dxV = \int_{0}^{\pi} \frac{1}{4}\sin^2(x) dx. Using the power-reducing identity sin⁡2(x)=1−cos⁡(2x)2\sin^2(x) = \frac{1-\cos(2x)}{2}, we get V=14∫0π1−cos⁡(2x)2dx=18[x−12sin⁡(2x)]0π=18((π−0)−(0−0))=π8V = \frac{1}{4} \int_{0}^{\pi} \frac{1-\cos(2x)}{2} dx = \frac{1}{8} [x - \frac{1}{2}\sin(2x)]_{0}^{\pi} = \frac{1}{8}((\pi - 0) - (0 - 0)) = \frac{\pi}{8}.

Question 3

The base of a solid is the region in the first quadrant bounded by the graph of y=exy=e^x, the x-axis, the y-axis, and the line x=kx=k for some k>0k>0. The cross-sections perpendicular to the x-axis are semicircles. If the volume of the solid is π16(e10−1)\frac{\pi}{16}(e^{10}-1), what is the value of kk?

  1. 2.52.5
  2. 55 (correct answer)
  3. 1010
  4. ln⁡(5)\ln(5)
Explanation: The diameter of a semicircular cross-section at a given xx is s(x)=exs(x) = e^x. The area of the semicircle is A(x)=π8s(x)2=π8(ex)2=π8e2xA(x) = \frac{\pi}{8}s(x)^2 = \frac{\pi}{8}(e^x)^2 = \frac{\pi}{8}e^{2x}. The volume is given by the integral V=∫0kπ8e2xdxV = \int_{0}^{k} \frac{\pi}{8}e^{2x} dx. Evaluating the integral gives V=π8[12e2x]0k=π16[e2k−e0]=π16(e2k−1)V = \frac{\pi}{8} [\frac{1}{2}e^{2x}]_{0}^{k} = \frac{\pi}{16}[e^{2k} - e^0] = \frac{\pi}{16}(e^{2k}-1). We are given that the volume is π16(e10−1)\frac{\pi}{16}(e^{10}-1). Setting the two expressions for the volume equal, we have e2k−1=e10−1e^{2k}-1 = e^{10}-1, which simplifies to e2k=e10e^{2k} = e^{10}. Therefore, 2k=102k=10, and k=5k=5.

Question 4

Let RR be a region in the xy-plane that forms the base of two different solids. For the first solid, the cross-sections perpendicular to the x-axis are isosceles triangles with a height equal to their base, and its volume is VTV_T. For the second solid, the cross-sections are semicircles, and its volume is VSV_S. Which of the following statements must be true for any such region RR?

  1. VT=2πVSV_T = \frac{2}{\pi}V_S
  2. VT=4πVSV_T = \frac{4}{\pi}V_S (correct answer)
  3. VS=4πVTV_S = \frac{4}{\pi}V_T
  4. The relationship between VTV_T and VSV_S depends on the specific shape of the base region RR.
Explanation: Let the length of the cross-section base at a given xx be s(x)s(x). For the first solid, the cross-section is an isosceles triangle with base s(x)s(x) and height s(x)s(x). Its area is AT(x)=12⋅base⋅height=12s(x)2A_T(x) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2}s(x)^2. For the second solid, the cross-section is a semicircle with diameter s(x)s(x). Its area is AS(x)=12π(s(x)2)2=π8s(x)2A_S(x) = \frac{1}{2}\pi (\frac{s(x)}{2})^2 = \frac{\pi}{8}s(x)^2. The ratio of the areas at any xx is constant: AT(x)AS(x)=12s(x)2π8s(x)2=1/2π/8=4π\frac{A_T(x)}{A_S(x)} = \frac{\frac{1}{2}s(x)^2}{\frac{\pi}{8}s(x)^2} = \frac{1/2}{\pi/8} = \frac{4}{\pi}. Since the volumes are the integrals of these area functions over the same interval, the ratio of the volumes is also constant: VTVS=4π\frac{V_T}{V_S} = \frac{4}{\pi}. Therefore, VT=4πVSV_T = \frac{4}{\pi}V_S.

Question 5

The base of a solid is the region in the first quadrant bounded by the ellipse x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1. Cross-sections perpendicular to the y-axis are isosceles triangles with a height that is twice the length of their base. Which of the following integrals represents the volume of the solid?

  1. ∫0389(9−y2)dy\int_{0}^{3} \frac{8}{9}(9-y^2) dy
  2. ∫04916(16−x2)dx\int_{0}^{4} \frac{9}{16}(16-x^2) dx
  3. ∫03169(9−y2)dy\int_{0}^{3} \frac{16}{9}(9-y^2) dy (correct answer)
  4. ∫03329(9−y2)dy\int_{0}^{3} \frac{32}{9}(9-y^2) dy
Explanation: Cross-sections are perpendicular to the y-axis, so we integrate with respect to yy. The bounds are from y=0y=0 to y=3y=3. We need to solve the ellipse equation for xx: x216=1−y29=9−y29  ⟹  x2=169(9−y2)\frac{x^2}{16} = 1 - \frac{y^2}{9} = \frac{9-y^2}{9} \implies x^2 = \frac{16}{9}(9-y^2). Since we are in the first quadrant, x=439−y2x = \frac{4}{3}\sqrt{9-y^2}. This is the length of the base of the triangle, so s(y)=439−y2s(y) = \frac{4}{3}\sqrt{9-y^2}. The height is given as h(y)=2s(y)h(y) = 2s(y). The area of the triangular cross-section is A(y)=12⋅base⋅height=12s(y)(2s(y))=s(y)2A(y) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} s(y) (2s(y)) = s(y)^2. So, A(y)=(439−y2)2=169(9−y2)A(y) = (\frac{4}{3}\sqrt{9-y^2})^2 = \frac{16}{9}(9-y^2). The volume is the integral of the area: V=∫03169(9−y2)dyV = \int_{0}^{3} \frac{16}{9}(9-y^2) dy.

Question 6

The base of a solid is the region enclosed by the upper half of the circle x2+y2=9x^2+y^2=9 and the x-axis. Cross-sections perpendicular to the x-axis are equilateral triangles. What is the volume of the solid?

  1. 36336\sqrt{3}
  2. 18318\sqrt{3}
  3. 939\sqrt{3} (correct answer)
  4. 3636
Explanation: The base is the region defined by y=9−x2y = \sqrt{9-x^2} for xx from −3-3 to 33. At a given xx, the side length of the equilateral triangle is s(x)=ytop−ybottom=9−x2−0=9−x2s(x) = y_{top} - y_{bottom} = \sqrt{9-x^2} - 0 = \sqrt{9-x^2}. The area of an equilateral triangle with side ss is A(x)=34s(x)2=34(9−x2)2=34(9−x2)A(x) = \frac{\sqrt{3}}{4}s(x)^2 = \frac{\sqrt{3}}{4}(\sqrt{9-x^2})^2 = \frac{\sqrt{3}}{4}(9-x^2). The volume is V=∫−3334(9−x2)dxV = \int_{-3}^{3} \frac{\sqrt{3}}{4}(9-x^2) dx. The integrand is an even function, so we can write V=2⋅34∫03(9−x2)dx=32[9x−x33]03=32((27−9)−0)=32(18)=93V = 2 \cdot \frac{\sqrt{3}}{4} \int_{0}^{3} (9-x^2) dx = \frac{\sqrt{3}}{2} [9x - \frac{x^3}{3}]_{0}^{3} = \frac{\sqrt{3}}{2}((27-9)-0) = \frac{\sqrt{3}}{2}(18) = 9\sqrt{3}.

Question 7

The base of a solid, Solid 1, is the region enclosed by the circle x2+y2=4x^2+y^2=4. Its cross-sections perpendicular to the x-axis are equilateral triangles. A second solid, Solid 2, has the same base, but its cross-sections perpendicular to the x-axis are semicircles. What is the ratio of the volume of Solid 1 to the volume of Solid 2?

  1. 23π\frac{2\sqrt{3}}{\pi} (correct answer)
  2. 43π\frac{4\sqrt{3}}{\pi}
  3. 3π\frac{\sqrt{3}}{\pi}
  4. 32π\frac{\sqrt{3}}{2\pi}
Explanation: For any xx in [−2,2][-2, 2], the length of the base of a cross-section is s(x)=4−x2−(−4−x2)=24−x2s(x) = \sqrt{4-x^2} - (-\sqrt{4-x^2}) = 2\sqrt{4-x^2}. For Solid 1, the cross-sectional area is an equilateral triangle, A1(x)=34s(x)2=34(4(4−x2))=3(4−x2)A_1(x) = \frac{\sqrt{3}}{4}s(x)^2 = \frac{\sqrt{3}}{4}(4(4-x^2)) = \sqrt{3}(4-x^2). For Solid 2, the cross-sectional area is a semicircle, A2(x)=π8s(x)2=π8(4(4−x2))=π2(4−x2)A_2(x) = \frac{\pi}{8}s(x)^2 = \frac{\pi}{8}(4(4-x^2)) = \frac{\pi}{2}(4-x^2). The volume is the integral of the area function. The ratio of the volumes is the ratio of the integrals: V1V2=∫−22A1(x)dx∫−22A2(x)dx=∫−223(4−x2)dx∫−22π2(4−x2)dx\frac{V_1}{V_2} = \frac{\int_{-2}^{2} A_1(x) dx}{\int_{-2}^{2} A_2(x) dx} = \frac{\int_{-2}^{2} \sqrt{3}(4-x^2) dx}{\int_{-2}^{2} \frac{\pi}{2}(4-x^2) dx}. Since the integrands differ only by a constant factor, the ratio is 3π/2=23π\frac{\sqrt{3}}{\pi/2} = \frac{2\sqrt{3}}{\pi}.

Question 8

The base of a solid is the region bounded by the graph of y=ln⁡(x)y = \ln(x), the line x=ex=e, and the x-axis. The cross-sections of the solid perpendicular to the y-axis are isosceles right triangles with one leg in the xy-plane. Which integral represents the volume of the solid?

  1. ∫0112(e−ey)2dy\int_{0}^{1} \frac{1}{2} (e - e^y)^2 dy (correct answer)
  2. ∫1e12(ln⁡x)2dx\int_{1}^{e} \frac{1}{2} (\ln x)^2 dx
  3. ∫0114(e−ey)2dy\int_{0}^{1} \frac{1}{4} (e - e^y)^2 dy
  4. ∫0e12(ey)2dy\int_{0}^{e} \frac{1}{2} (e^y)^2 dy
Explanation: Since the cross-sections are perpendicular to the y-axis, we must integrate with respect to yy. We need to express the boundaries in terms of yy. The curve y=ln⁡(x)y=\ln(x) becomes x=eyx=e^y. The line x=ex=e is a right boundary. The x-axis (y=0y=0) is the lower bound for yy. The upper bound for yy is where x=eyx=e^y meets x=ex=e, so e=ey  ⟹  y=1e=e^y \implies y=1. For a given yy between 0 and 1, the length of the base of the triangle is the horizontal distance s(y)=xright−xleft=e−eys(y) = x_{right} - x_{left} = e - e^y. Since the cross-sections are isosceles right triangles with a leg in the base, the area is A(y)=12s(y)2=12(e−ey)2A(y) = \frac{1}{2}s(y)^2 = \frac{1}{2}(e-e^y)^2. The volume is the integral of this area from y=0y=0 to y=1y=1, which is ∫0112(e−ey)2dy\int_{0}^{1} \frac{1}{2} (e - e^y)^2 dy.

Question 9

Let RR be the region enclosed by the graphs of y=4−x2y=4-x^2 and y=x+2y=x+2. The region RR is the base of a solid. For the solid, each cross-section perpendicular to the x-axis is a semicircle. Which of the following integrals gives the volume of the solid?

  1. ∫−21π8(2−x−x2)2dx\int_{-2}^{1} \frac{\pi}{8}(2-x-x^2)^2 dx (correct answer)
  2. ∫−21π2(2−x−x2)2dx\int_{-2}^{1} \frac{\pi}{2}(2-x-x^2)^2 dx
  3. ∫−21π8((4−x2)2−(x+2)2)dx\int_{-2}^{1} \frac{\pi}{8}((4-x^2)^2 - (x+2)^2) dx
  4. ∫−21π8(2−x−x2)dx\int_{-2}^{1} \frac{\pi}{8}(2-x-x^2) dx
Explanation: The points of intersection are found by setting 4−x2=x+24-x^2 = x+2, which gives x2+x−2=0x^2+x-2=0, or (x+2)(x−1)=0(x+2)(x-1)=0. The bounds of integration are x=−2x=-2 and x=1x=1. For a cross-section at a given xx, the diameter of the semicircle is the vertical distance between the curves, s(x)=(4−x2)−(x+2)=2−x−x2s(x) = (4-x^2) - (x+2) = 2-x-x^2. The radius is r(x)=s(x)/2r(x) = s(x)/2. The area of a semicircular cross-section is A(x)=12πr(x)2=12π(s(x)2)2=π8s(x)2=π8(2−x−x2)2A(x) = \frac{1}{2}\pi r(x)^2 = \frac{1}{2}\pi (\frac{s(x)}{2})^2 = \frac{\pi}{8}s(x)^2 = \frac{\pi}{8}(2-x-x^2)^2. The volume is the integral of this area from x=−2x=-2 to x=1x=1, which is ∫−21π8(2−x−x2)2dx\int_{-2}^{1} \frac{\pi}{8}(2-x-x^2)^2 dx.

Question 10

The base of a solid is the region enclosed by the parabola x=y2x=y^2 and the line x=4x=4. Each cross-section perpendicular to the y-axis is an equilateral triangle. What is the volume of the solid?

  1. 838\sqrt{3}
  2. 6435\frac{64\sqrt{3}}{5}
  3. 128315\frac{128\sqrt{3}}{15} (correct answer)
  4. 25615\frac{256}{15}
Explanation: Since cross-sections are perpendicular to the y-axis, we integrate with respect to yy. The region extends from y=−2y=-2 to y=2y=2 (where y2=4y^2=4). At a given yy, the side length of the equilateral triangle is the horizontal distance s(y)=xright−xleft=4−y2s(y) = x_{right} - x_{left} = 4 - y^2. The area of an equilateral triangle with side ss is A=34s2A = \frac{\sqrt{3}}{4}s^2. So, A(y)=34(4−y2)2=34(16−8y2+y4)A(y) = \frac{\sqrt{3}}{4}(4-y^2)^2 = \frac{\sqrt{3}}{4}(16 - 8y^2 + y^4). The volume is V=∫−2234(16−8y2+y4)dyV = \int_{-2}^{2} \frac{\sqrt{3}}{4}(16 - 8y^2 + y^4) dy. By symmetry, V=2⋅34∫02(16−8y2+y4)dy=32[16y−83y3+15y5]02=32(32−643+325)=32(480−320+9615)=32(25615)=128315V = 2 \cdot \frac{\sqrt{3}}{4} \int_{0}^{2} (16 - 8y^2 + y^4) dy = \frac{\sqrt{3}}{2} [16y - \frac{8}{3}y^3 + \frac{1}{5}y^5]_{0}^{2} = \frac{\sqrt{3}}{2} (32 - \frac{64}{3} + \frac{32}{5}) = \frac{\sqrt{3}}{2} (\frac{480-320+96}{15}) = \frac{\sqrt{3}}{2}(\frac{256}{15}) = \frac{128\sqrt{3}}{15}.