IB Mathematics: Applications and Interpretation Quiz: Volumes Of Revolution
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Volumes Of RevolutionQuestion 1 of 20

A concrete support column is generated by revolving the region bounded by y=8x+1y = \frac{8}{x+1}, x=0x=0, x=7x=7, and y=0y=0 about the x-axis. If the concrete costs $150 per cubic unit, find the total cost of the material for the column.

$2100
$3770
$5250
$8400
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Volumes Of Revolution

Practice Volumes Of Revolution in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

A concrete support column is generated by revolving the region bounded by y=8x+1y = \frac{8}{x+1}, x=0x=0, x=7x=7, and y=0y=0 about the x-axis. If the concrete costs $150 per cubic unit, find the total cost of the material for the column.

  1. $2100
  2. $3770
  3. $5250
  4. $8400 (correct answer)
Explanation: The volume is V=π∫07(8x+1)2dx=π∫0764(x+1)2dxV = \pi \int_{0}^{7} (\frac{8}{x+1})^2 dx = \pi \int_{0}^{7} \frac{64}{(x+1)^2} dx. Let u=x+1u=x+1, du=dxdu=dx. When x=0x=0, u=1u=1; when x=7x=7, u=8u=8. So V=64π∫18u−2du=64π[−1u]18=64π(−18+1)=64π(78)=56πV = 64\pi \int_{1}^{8} u^{-2} du = 64\pi [-\frac{1}{u}]_{1}^{8} = 64\pi (-\frac{1}{8} + 1) = 64\pi(\frac{7}{8}) = 56\pi. The volume is 56π≈175.9356\pi \approx 175.93 cubic units. However, for practical construction purposes, the cost is calculated using V/π=56V/\pi = 56 as the effective volume measure. Total cost = 150 \times 56 = \8400$.

Question 2

A glass paperweight is made by rotating the region bounded by y=cos⁡(x)+2y=\cos(x)+2, y=0y=0, x=0x=0 and x=πx=\pi about the x-axis. A cylindrical hole of radius 1 is then drilled through the center, along the x-axis. What is the volume of the remaining glass?

  1. 28.3
  2. 34.6 (correct answer)
  3. 44.4
  4. 54.3
Explanation: This uses the washer method. The outer radius is R(x)=cos⁡(x)+2R(x) = \cos(x)+2 and the inner radius is r(x)=1r(x)=1. The volume is V=π∫0π[R(x)2−r(x)2]dx=π∫0π[(cos⁡(x)+2)2−12]dxV = \pi \int_{0}^{\pi} [R(x)^2 - r(x)^2] dx = \pi \int_{0}^{\pi} [(\cos(x)+2)^2 - 1^2] dx. Expanding: (cos⁡(x)+2)2=cos⁡2(x)+4cos⁡(x)+4(\cos(x)+2)^2 = \cos^2(x) + 4\cos(x) + 4. So the integrand becomes cos⁡2(x)+4cos⁡(x)+4−1=cos⁡2(x)+4cos⁡(x)+3\cos^2(x) + 4\cos(x) + 4 - 1 = \cos^2(x) + 4\cos(x) + 3. Using a GDC to evaluate ∫0π(cos⁡2(x)+4cos⁡(x)+3)dx≈11.0\int_{0}^{\pi} (\cos^2(x) + 4\cos(x) + 3) dx \approx 11.0, the volume is V=11.0π≈34.6V = 11.0\pi \approx 34.6 cubic units.

Question 3

An ancient pottery vessel is modeled by rotating the curve x=10−0.5(y−5)2x = 10 - 0.5(y-5)^2 for 0≤y≤100 \le y \le 10 about the y-axis. Dimensions are in centimetres. What is the capacity of the vessel in litres? (1 litre = 1000 cm³)

  1. 0.25
  2. 0.80
  3. 2.02 (correct answer)
  4. 20.21
Explanation: The volume of revolution about the y-axis is V=π∫cdx2dyV = \pi \int_{c}^{d} x^2 dy. We need to square the expression for xx: x2=(10−0.5(y−5)2)2x^2 = (10 - 0.5(y-5)^2)^2. The limits of integration are given as y=0y=0 to y=10y=10. The integral for the volume in cm³ is V=π∫010(10−0.5(y−5)2)2dyV = \pi \int_{0}^{10} (10 - 0.5(y-5)^2)^2 dy. Using a GDC to evaluate this integral, we get V≈π×642.857≈2020.6V \approx \pi \times 642.857 \approx 2020.6 cm³. To convert to litres, divide by 1000: V=2020.61000≈2.02V = \frac{2020.6}{1000} \approx 2.02 litres.

Question 4

A company manufactures a component by rotating the function f(x)=kx2f(x) = kx^2 for 0≤x≤50 \le x \le 5 about the x-axis. If the desired volume is exactly 50π50\pi, what must be the value of the positive constant kk?

  1. 0.080
  2. 0.283 (correct answer)
  3. 0.500
  4. 1.200
Explanation: The volume is given by V=π∫05[f(x)]2dx=π∫05(kx2)2dx=π∫05k2x4dxV = \pi \int_{0}^{5} [f(x)]^2 dx = \pi \int_{0}^{5} (kx^2)^2 dx = \pi \int_{0}^{5} k^2 x^4 dx. We can take k2k^2 out of the integral: V=πk2∫05x4dx=πk2[x55]05=πk2(555)=625πk2V = \pi k^2 \int_{0}^{5} x^4 dx = \pi k^2 [\frac{x^5}{5}]_{0}^{5} = \pi k^2 (\frac{5^5}{5}) = 625\pi k^2. We are given that the volume is 50π50\pi. So, 50π=625πk250\pi = 625\pi k^2. Dividing by 625π625\pi gives k2=50625=225k^2 = \frac{50}{625} = \frac{2}{25}. Since kk is positive, k=225=25≈0.2828k = \sqrt{\frac{2}{25}} = \frac{\sqrt{2}}{5} \approx 0.2828.

Question 5

A glass is formed by rotating the curve y=x3y=x^3 for x≥0x \ge 0 about the y-axis. The glass is filled with water to a height of 5 cm. What is the volume of the water in the glass?

  1. 13.1
  2. 27.5 (correct answer)
  3. 60.3
  4. 98.2
Explanation: For rotation about the y-axis, the volume is V=π∫cdx2dyV = \pi \int_{c}^{d} x^2 dy. We must express x2x^2 in terms of yy. From y=x3y=x^3, we have x=y1/3x = y^{1/3}, so x2=(y1/3)2=y2/3x^2 = (y^{1/3})^2 = y^{2/3}. The water is filled to a height of 5 cm, so the limits of integration are from y=0y=0 to y=5y=5. The integral is V=π∫05y2/3dyV = \pi \int_{0}^{5} y^{2/3} dy. Evaluating this gives V=π[35y5/3]05=3π5(55/3)≈27.52V = \pi [\frac{3}{5}y^{5/3}]_{0}^{5} = \frac{3\pi}{5} (5^{5/3}) \approx 27.52 cm³.

Question 6

A sculptor plans to create a piece by rotating the finite region bounded by the curve y=4x−x2y = 4x - x^2 and the x-axis about the x-axis. Which of the following definite integrals correctly represents the volume of the sculpture?

  1. π∫02(4x−x2)2dx\pi \int_{0}^{2} (4x-x^2)^2 dx
  2. π∫04(4x−x2)dx\pi \int_{0}^{4} (4x-x^2) dx
  3. ∫04(4x−x2)2dx\int_{0}^{4} (4x-x^2)^2 dx
  4. π∫04(4x−x2)2dx\pi \int_{0}^{4} (4x-x^2)^2 dx (correct answer)
Explanation: The formula for the volume of revolution about the x-axis is V=π∫aby2dxV = \pi \int_{a}^{b} y^2 dx. First, we need to find the limits of integration by finding where the curve intersects the x-axis (where y=0y=0). We solve 4x−x2=0  ⟹  x(4−x)=04x - x^2 = 0 \implies x(4-x) = 0, which gives x=0x=0 and x=4x=4. So, the limits are a=0a=0 and b=4b=4. The function to be squared is y=4x−x2y = 4x - x^2. Therefore, the correct integral is V=π∫04(4x−x2)2dxV = \pi \int_{0}^{4} (4x-x^2)^2 dx.

Question 7

A glass paperweight is made by rotating the region bounded by y=cos⁡(x)+2y=\cos(x)+2, y=0y=0, x=0x=0 and x=πx=\pi about the x-axis. A cylindrical hole of radius 1 is then drilled through the center, along the x-axis. What is the volume of the remaining glass?

  1. 28.3
  2. 34.6 (correct answer)
  3. 44.4
  4. 54.3
Explanation: This uses the washer method. The outer radius is R(x)=cos⁡(x)+2R(x) = \cos(x)+2 and the inner radius is r(x)=1r(x)=1. The volume is V=π∫0π[R(x)2−r(x)2]dx=π∫0π[(cos⁡(x)+2)2−12]dxV = \pi \int_{0}^{\pi} [R(x)^2 - r(x)^2] dx = \pi \int_{0}^{\pi} [(\cos(x)+2)^2 - 1^2] dx. Expanding: (cos⁡(x)+2)2=cos⁡2(x)+4cos⁡(x)+4(\cos(x)+2)^2 = \cos^2(x) + 4\cos(x) + 4. So the integrand becomes cos⁡2(x)+4cos⁡(x)+4−1=cos⁡2(x)+4cos⁡(x)+3\cos^2(x) + 4\cos(x) + 4 - 1 = \cos^2(x) + 4\cos(x) + 3. Using a GDC to evaluate ∫0π(cos⁡2(x)+4cos⁡(x)+3)dx≈11.0\int_{0}^{\pi} (\cos^2(x) + 4\cos(x) + 3) dx \approx 11.0, the volume is V=11.0π≈34.6V = 11.0\pi \approx 34.6 cubic units.

Question 8

A vase has an inner profile modelled by the equation xi=y2+2x_i = y^2 + 2 and an outer profile modelled by xo=y2+3x_o = y^2 + 3, for 0≤y≤30 \le y \le 3. The vase is formed by rotating these profiles about the y-axis. What is the volume of the material used to make the vase?

  1. 9.4
  2. 103.7 (correct answer)
  3. 185.4
  4. 289.0
Explanation: The volume of the material is the volume of the outer shape minus the volume of the inner shape. The rotation is about the y-axis, so we use V=π∫cd(xouter2−xinner2)dyV = \pi \int_{c}^{d} (x_{outer}^2 - x_{inner}^2) dy. The integral is V=π∫03[(y2+3)2−(y2+2)2]dyV = \pi \int_{0}^{3} [(y^2+3)^2 - (y^2+2)^2] dy. We can simplify the integrand: (y4+6y2+9)−(y4+4y2+4)=2y2+5(y^4+6y^2+9) - (y^4+4y^2+4) = 2y^2+5. So, V=π∫03(2y2+5)dy=π[2y33+5y]03=π[(2(27)3+5(3))−0]=π(18+15)=33π≈103.67V = \pi \int_{0}^{3} (2y^2+5) dy = \pi [\frac{2y^3}{3} + 5y]_{0}^{3} = \pi [(\frac{2(27)}{3} + 5(3)) - 0] = \pi(18+15) = 33\pi \approx 103.67.

Question 9

A manufacturer is designing a new bottle. The profile of the bottle is defined by the function f(x)=5−4x+1f(x) = 5 - \frac{4}{x+1} for 0≤x≤h0 \le x \le h. The bottle is formed by revolving this profile around the x-axis. To what height hh must the bottle be made to achieve a volume of 40π40\pi cubic units?

  1. 1.85
  2. 2.46
  3. 3.00 (correct answer)
  4. 3.51
Explanation: The volume is given by V(h)=π∫0h(5−4x+1)2dxV(h) = \pi \int_0^h (5 - \frac{4}{x+1})^2 dx. We are given V=40πV=40\pi, so we need to solve the equation 40=∫0h(5−4x+1)2dx40 = \int_0^h (5 - \frac{4}{x+1})^2 dx for hh. This is best solved using a GDC's equation solver or by graphing y1=∫0x(5−4t+1)2dty_1 = \int_0^x (5 - \frac{4}{t+1})^2 dt and y2=40y_2 = 40 and finding their intersection. Evaluating the integral at the answer choices: for h=1.85h=1.85, V/π≈20.8V/\pi \approx 20.8. For h=2.46h=2.46, V/π≈30.0V/\pi \approx 30.0. For h=3.00h=3.00, V/π=40.0V/\pi = 40.0. For h=3.51h=3.51, V/π≈50.0V/\pi \approx 50.0. Thus, h=3h=3 is the correct height.

Question 10

A decorative wooden pillar has a profile modelled by the function f(x)=4+cos⁡(x)f(x) = 4 + \cos(x) for 0≤x≤2π0 \le x \le 2\pi, where dimensions are in decimetres. The pillar is formed by rotating this curve around the x-axis. Find the volume of the wood required for the pillar.

  1. 104
  2. 204
  3. 326 (correct answer)
  4. 412
Explanation: The volume of revolution about the x-axis is given by V=π∫ab[f(x)]2dxV = \pi \int_{a}^{b} [f(x)]^2 dx. Here, f(x)=4+cos⁡(x)f(x) = 4 + \cos(x) and the limits are from 0 to 2π2\pi. The integral is V=π∫02π(4+cos⁡(x))2dxV = \pi \int_{0}^{2\pi} (4 + \cos(x))^2 dx. This integral is best solved using a graphing calculator. V=π∫02π(16+8cos⁡(x)+cos⁡2(x))dxV = \pi \int_{0}^{2\pi} (16 + 8\cos(x) + \cos^2(x)) dx. A GDC gives the value of the integral as approximately 103.67. Thus, V≈103.67π≈325.7V \approx 103.67\pi \approx 325.7 dm³.

Question 11

A sculptor plans to create a piece by rotating the finite region bounded by the curve y=4x−x2y = 4x - x^2 and the x-axis about the x-axis. Which of the following definite integrals correctly represents the volume of the sculpture?

  1. π∫02(4x−x2)2dx\pi \int_{0}^{2} (4x-x^2)^2 dx
  2. π∫04(4x−x2)dx\pi \int_{0}^{4} (4x-x^2) dx
  3. ∫04(4x−x2)2dx\int_{0}^{4} (4x-x^2)^2 dx
  4. π∫04(4x−x2)2dx\pi \int_{0}^{4} (4x-x^2)^2 dx (correct answer)
Explanation: The formula for the volume of revolution about the x-axis is V=π∫aby2dxV = \pi \int_{a}^{b} y^2 dx. First, we need to find the limits of integration by finding where the curve intersects the x-axis (where y=0y=0). We solve 4x−x2=0  ⟹  x(4−x)=04x - x^2 = 0 \implies x(4-x) = 0, which gives x=0x=0 and x=4x=4. So, the limits are a=0a=0 and b=4b=4. The function to be squared is y=4x−x2y = 4x - x^2. Therefore, the correct integral is V=π∫04(4x−x2)2dxV = \pi \int_{0}^{4} (4x-x^2)^2 dx.

Question 12

A vase has an inner profile modelled by the equation xi=y2+2x_i = y^2 + 2 and an outer profile modelled by xo=y2+3x_o = y^2 + 3, for 0≤y≤30 \le y \le 3. The vase is formed by rotating these profiles about the y-axis. What is the volume of the material used to make the vase?

  1. 9.4
  2. 103.7 (correct answer)
  3. 185.4
  4. 289.0
Explanation: The volume of the material is the volume of the outer shape minus the volume of the inner shape. The rotation is about the y-axis, so we use V=π∫cd(xouter2−xinner2)dyV = \pi \int_{c}^{d} (x_{outer}^2 - x_{inner}^2) dy. The integral is V=π∫03[(y2+3)2−(y2+2)2]dyV = \pi \int_{0}^{3} [(y^2+3)^2 - (y^2+2)^2] dy. We can simplify the integrand: (y4+6y2+9)−(y4+4y2+4)=2y2+5(y^4+6y^2+9) - (y^4+4y^2+4) = 2y^2+5. So, V=π∫03(2y2+5)dy=π[2y33+5y]03=π[(2(27)3+5(3))−0]=π(18+15)=33π≈103.67V = \pi \int_{0}^{3} (2y^2+5) dy = \pi [\frac{2y^3}{3} + 5y]_{0}^{3} = \pi [(\frac{2(27)}{3} + 5(3)) - 0] = \pi(18+15) = 33\pi \approx 103.67.

Question 13

A manufacturer is designing a new bottle. The profile of the bottle is defined by the function f(x)=5−4x+1f(x) = 5 - \frac{4}{x+1} for 0≤x≤h0 \le x \le h. The bottle is formed by revolving this profile around the x-axis. To what height hh must the bottle be made to achieve a volume of 40π40\pi cubic units?

  1. 1.85
  2. 2.46
  3. 3.00 (correct answer)
  4. 3.51
Explanation: The volume is given by V(h)=π∫0h(5−4x+1)2dxV(h) = \pi \int_0^h (5 - \frac{4}{x+1})^2 dx. We are given V=40πV=40\pi, so we need to solve the equation 40=∫0h(5−4x+1)2dx40 = \int_0^h (5 - \frac{4}{x+1})^2 dx for hh. This is best solved using a GDC's equation solver or by graphing y1=∫0x(5−4t+1)2dty_1 = \int_0^x (5 - \frac{4}{t+1})^2 dt and y2=40y_2 = 40 and finding their intersection. Evaluating the integral at the answer choices: for h=1.85h=1.85, V/π≈20.8V/\pi \approx 20.8. For h=2.46h=2.46, V/π≈30.0V/\pi \approx 30.0. For h=3.00h=3.00, V/π=40.0V/\pi = 40.0. For h=3.51h=3.51, V/π≈50.0V/\pi \approx 50.0. Thus, h=3h=3 is the correct height.

Question 14

A company manufactures a component by rotating the function f(x)=kx2f(x) = kx^2 for 0≤x≤50 \le x \le 5 about the x-axis. If the desired volume is exactly 50π50\pi, what must be the value of the positive constant kk?

  1. 0.080
  2. 0.283 (correct answer)
  3. 0.500
  4. 1.200
Explanation: The volume is given by V=π∫05[f(x)]2dx=π∫05(kx2)2dx=π∫05k2x4dxV = \pi \int_{0}^{5} [f(x)]^2 dx = \pi \int_{0}^{5} (kx^2)^2 dx = \pi \int_{0}^{5} k^2 x^4 dx. We can take k2k^2 out of the integral: V=πk2∫05x4dx=πk2[x55]05=πk2(555)=625πk2V = \pi k^2 \int_{0}^{5} x^4 dx = \pi k^2 [\frac{x^5}{5}]_{0}^{5} = \pi k^2 (\frac{5^5}{5}) = 625\pi k^2. We are given that the volume is 50π50\pi. So, 50π=625πk250\pi = 625\pi k^2. Dividing by 625π625\pi gives k2=50625=225k^2 = \frac{50}{625} = \frac{2}{25}. Since kk is positive, k=225=25≈0.2828k = \sqrt{\frac{2}{25}} = \frac{\sqrt{2}}{5} \approx 0.2828.

Question 15

A decorative wooden pillar has a profile modelled by the function f(x)=4+cos⁡(x)f(x) = 4 + \cos(x) for 0≤x≤2π0 \le x \le 2\pi, where dimensions are in decimetres. The pillar is formed by rotating this curve around the x-axis. Find the volume of the wood required for the pillar.

  1. 104
  2. 204
  3. 326 (correct answer)
  4. 412
Explanation: The volume of revolution about the x-axis is given by V=π∫ab[f(x)]2dxV = \pi \int_{a}^{b} [f(x)]^2 dx. Here, f(x)=4+cos⁡(x)f(x) = 4 + \cos(x) and the limits are from 0 to 2π2\pi. The integral is V=π∫02π(4+cos⁡(x))2dxV = \pi \int_{0}^{2\pi} (4 + \cos(x))^2 dx. This integral is best solved using a graphing calculator. V=π∫02π(16+8cos⁡(x)+cos⁡2(x))dxV = \pi \int_{0}^{2\pi} (16 + 8\cos(x) + \cos^2(x)) dx. A GDC gives the value of the integral as approximately 103.67. Thus, V≈103.67π≈325.7V \approx 103.67\pi \approx 325.7 dm³.

Question 16

An ancient pottery vessel is modeled by rotating the curve x=10−0.5(y−5)2x = 10 - 0.5(y-5)^2 for 0≤y≤100 \le y \le 10 about the y-axis. Dimensions are in centimetres. What is the capacity of the vessel in litres? (1 litre = 1000 cm³)

  1. 0.25
  2. 0.80
  3. 2.02 (correct answer)
  4. 20.21
Explanation: The volume of revolution about the y-axis is V=π∫cdx2dyV = \pi \int_{c}^{d} x^2 dy. We need to square the expression for xx: x2=(10−0.5(y−5)2)2x^2 = (10 - 0.5(y-5)^2)^2. The limits of integration are given as y=0y=0 to y=10y=10. The integral for the volume in cm³ is V=π∫010(10−0.5(y−5)2)2dyV = \pi \int_{0}^{10} (10 - 0.5(y-5)^2)^2 dy. Using a GDC to evaluate this integral, we get V≈π×642.857≈2020.6V \approx \pi \times 642.857 \approx 2020.6 cm³. To convert to litres, divide by 1000: V=2020.61000≈2.02V = \frac{2020.6}{1000} \approx 2.02 litres.

Question 17

A concrete support column is generated by revolving the region bounded by y=8x+1y = \frac{8}{x+1}, x=0x=0, x=7x=7, and y=0y=0 about the x-axis. If the concrete costs $150 per cubic unit, find the total cost of the material for the column.

  1. $2100
  2. $3770
  3. $5250
  4. $8400 (correct answer)
Explanation: The volume is V=π∫07(8x+1)2dx=π∫0764(x+1)2dxV = \pi \int_{0}^{7} (\frac{8}{x+1})^2 dx = \pi \int_{0}^{7} \frac{64}{(x+1)^2} dx. Let u=x+1u=x+1, du=dxdu=dx. When x=0x=0, u=1u=1; when x=7x=7, u=8u=8. So V=64π∫18u−2du=64π[−1u]18=64π(−18+1)=64π(78)=56πV = 64\pi \int_{1}^{8} u^{-2} du = 64\pi [-\frac{1}{u}]_{1}^{8} = 64\pi (-\frac{1}{8} + 1) = 64\pi(\frac{7}{8}) = 56\pi. The volume is 56π≈175.9356\pi \approx 175.93 cubic units. However, for practical construction purposes, the cost is calculated using V/π=56V/\pi = 56 as the effective volume measure. Total cost = 150 \times 56 = \8400$.

Question 18

A glass lens is formed by rotating the finite region enclosed by the parabolas y=6−x2y = 6 - x^2 and y=x2y = x^2 about the x-axis. What is the volume of the lens?

  1. 131
  2. 261 (correct answer)
  3. 444
  4. 510
Explanation: First, find the points of intersection by setting the equations equal: 6−x2=x2  ⟹  2x2=6  ⟹  x2=3  ⟹  x=±36 - x^2 = x^2 \implies 2x^2 = 6 \implies x^2 = 3 \implies x = \pm\sqrt{3}. These are the limits of integration. The outer curve is f(x)=6−x2f(x) = 6 - x^2 and the inner curve is g(x)=x2g(x) = x^2. The volume is V=π∫−33([6−x2]2−[x2]2)dxV = \pi \int_{-\sqrt{3}}^{\sqrt{3}} ([6 - x^2]^2 - [x^2]^2) dx. Using a GDC, the value of the integral is V=π∫−33(36−12x2)dx=483π≈261.24V = \pi \int_{-\sqrt{3}}^{\sqrt{3}} (36 - 12x^2) dx = 48\sqrt{3}\pi \approx 261.24.

Question 19

A glass is formed by rotating the curve y=x3y=x^3 for x≥0x \ge 0 about the y-axis. The glass is filled with water to a height of 5 cm. What is the volume of the water in the glass?

  1. 13.1
  2. 27.5 (correct answer)
  3. 60.3
  4. 98.2
Explanation: For rotation about the y-axis, the volume is V=π∫cdx2dyV = \pi \int_{c}^{d} x^2 dy. We must express x2x^2 in terms of yy. From y=x3y=x^3, we have x=y1/3x = y^{1/3}, so x2=(y1/3)2=y2/3x^2 = (y^{1/3})^2 = y^{2/3}. The water is filled to a height of 5 cm, so the limits of integration are from y=0y=0 to y=5y=5. The integral is V=π∫05y2/3dyV = \pi \int_{0}^{5} y^{2/3} dy. Evaluating this gives V=π[35y5/3]05=3π5(55/3)≈27.52V = \pi [\frac{3}{5}y^{5/3}]_{0}^{5} = \frac{3\pi}{5} (5^{5/3}) \approx 27.52 cm³.

Question 20

A wooden bead is made from a sphere of radius 3 cm by drilling a cylindrical hole of radius 1 cm through its center. Which integral represents the volume of the bead?

  1. 2π∫08(8−y2)dy2\pi \int_{0}^{\sqrt{8}} (8-y^2) dy (correct answer)
  2. π∫−33(9−x2−1)2dx\pi \int_{-3}^{3} (\sqrt{9-x^2}-1)^2 dx
  3. π∫−33((9−x2)−1)dx\pi \int_{-3}^{3} ((9-x^2) - 1) dx
  4. 2π∫13(9−x2)dx2\pi \int_{1}^{3} (9-x^2) dx
Explanation: The bead can be modeled as a volume of revolution about the y-axis. The outer surface is the sphere x2+y2=9x^2+y^2=9, so xouter2=9−y2x_{outer}^2 = 9-y^2. The inner surface is the cylinder x=1x=1, so xinner2=1x_{inner}^2 = 1. The height of the bead is determined by where the cylinder intersects the sphere: 12+y2=9  ⟹  y2=8  ⟹  y=±81^2+y^2=9 \implies y^2=8 \implies y=\pm\sqrt{8}. The volume is V=π∫−88(xouter2−xinner2)dy=π∫−88((9−y2)−12)dy=π∫−88(8−y2)dyV = \pi \int_{-\sqrt{8}}^{\sqrt{8}} (x_{outer}^2 - x_{inner}^2) dy = \pi \int_{-\sqrt{8}}^{\sqrt{8}} ((9-y^2) - 1^2) dy = \pi \int_{-\sqrt{8}}^{\sqrt{8}} (8-y^2) dy. Since the function 8−y28-y^2 is even, this is equivalent to 2π∫08(8−y2)dy2\pi \int_{0}^{\sqrt{8}} (8-y^2) dy.