Calculus 1 Quiz: Washer Method X Or Y Axis
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Washer Method X Or Y AxisQuestion 1 of 20

The region R is bounded by the graphs of y=2xy=2\sqrt{x} and y=xy=x. Which integral gives the volume of the solid generated by revolving R about the x-axis?

π∫02(y2−y2/4)dy\pi \int_0^2 (y^2 - y^2/4) dy
π∫04(2x−x)2dx\pi \int_0^4 (2\sqrt{x} - x)^2 dx
π∫04(x2−4x)dx\pi \int_0^4 (x^2 - 4x) dx
π∫04(4x−x2)dx\pi \int_0^4 (4x - x^2) dx
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Calculus 1 Quiz

Calculus 1 Quiz: Washer Method X Or Y Axis

Practice Washer Method X Or Y Axis in Calculus 1 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 Washer Method X Or Y Axis, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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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

The region R is bounded by the graphs of y=2xy=2\sqrt{x} and y=xy=x. Which integral gives the volume of the solid generated by revolving R about the x-axis?

  1. π∫02(y2−y2/4)dy\pi \int_0^2 (y^2 - y^2/4) dy
  2. π∫04(2x−x)2dx\pi \int_0^4 (2\sqrt{x} - x)^2 dx
  3. π∫04(x2−4x)dx\pi \int_0^4 (x^2 - 4x) dx
  4. π∫04(4x−x2)dx\pi \int_0^4 (4x - x^2) dx (correct answer)
Explanation: When finding the volume of a solid of revolution about the x-axis, you use the disk method: V=π∫ab[R(x)]2−[r(x)]2 dxV = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx, where R(x) is the outer radius and r(x) is the inner radius. First, find where the curves intersect by solving 2x=x2\sqrt{x} = x. Squaring both sides: 4x=x24x = x^2, so x2−4x=0x^2 - 4x = 0, giving x=0x = 0 and x=4x = 4. Between these points, 2x>x2\sqrt{x} > x (you can verify by testing x=1x = 1), so y=2xy = 2\sqrt{x} is the upper curve. When revolving about the x-axis, the outer radius is the distance from the x-axis to the upper curve: R(x)=2xR(x) = 2\sqrt{x}. The inner radius is the distance to the lower curve: r(x)=xr(x) = x. The volume integral becomes: V=π∫04[(2x)2−x2] dx=π∫04(4x−x2) dxV = \pi \int_0^4 [(2\sqrt{x})^2 - x^2] \, dx = \pi \int_0^4 (4x - x^2) \, dx This matches answer choice D. Choice A uses dy integration, which would require the washer method with horizontal slices—but the setup is incorrect. Choice B uses (2x−x)2(2\sqrt{x} - x)^2, which represents the area between curves squared, not the difference of areas of circles. Choice C has the terms reversed: it subtracts the larger radius squared from the smaller one, giving a negative volume. Strategy tip: Always identify which curve is on top, set up your radii correctly, and remember the disk/washer method uses the difference of the squares of the radii, not the square of their difference.

Question 2

The region in the first quadrant bounded by y=4−x2y=4-x^2 and y=3xy=3x is revolved around the x-axis. Which integral represents the volume?

  1. π∫04((4−x2)2−(3x)2)dx\pi \int_0^4 ((4-x^2)^2 - (3x)^2) dx
  2. π∫01(4−x2−3x)2dx\pi \int_0^1 (4-x^2-3x)^2 dx
  3. π∫01((4−x2)2−(3x)2)dx\pi \int_0^1 ((4-x^2)^2 - (3x)^2) dx (correct answer)
  4. π∫01((3x)2−(4−x2)2)dx\pi \int_0^1 ((3x)^2 - (4-x^2)^2) dx
Explanation: When finding volumes of revolution around the x-axis, you need to visualize the solid created when a region rotates around that axis. The key is using the washer method when you have two bounding curves. First, find where the curves intersect by setting 4−x2=3x4-x^2 = 3x. Rearranging gives x2+3x−4=0x^2 + 3x - 4 = 0, which factors as (x+4)(x−1)=0(x+4)(x-1) = 0. Since we're in the first quadrant, we use x=1x = 1 (rejecting x=−4x = -4). At x=0x = 0, we have y=4y = 4 for the parabola and y=0y = 0 for the line, so the parabola y=4−x2y = 4-x^2 is above y=3xy = 3x from x=0x = 0 to x=1x = 1. When this region revolves around the x-axis, you get washers with outer radius R=4−x2R = 4-x^2 and inner radius r=3xr = 3x. The washer method formula is V=π∫ab(R2−r2)dxV = \pi \int_a^b (R^2 - r^2) dx, giving us π∫01((4−x2)2−(3x)2)dx\pi \int_0^1 ((4-x^2)^2 - (3x)^2) dx. Looking at the wrong answers: A) uses incorrect limits (0 to 4 instead of 0 to 1), B) attempts to subtract the functions before squaring, which doesn't match the washer geometry, and D) reverses the radii, putting the smaller function as the outer radius. Study tip: Always sketch the region first, identify which function is on top, find intersection points for your limits, then apply the washer formula π∫(R2−r2)dx\pi \int (R^2 - r^2) dx where R is the outer radius.

Question 3

Let R be the region in the first quadrant bounded by y=xy=x, y=2xy=2x, and y=2y=2. Find the volume of the solid generated by revolving R about the y-axis.

  1. 4π3\frac{4\pi}{3}
  2. 2π2\pi (correct answer)
  3. 8π3\frac{8\pi}{3}
  4. 2π3\frac{2\pi}{3}
Explanation: To revolve about the y-axis, we integrate with respect to y. The region is bounded by y=0y=0 (x-axis) and y=2y=2. The functions in terms of y are x=yx=y and x=y/2x=y/2. On the interval y∈[0,2]y \in [0,2], y≥y/2y \ge y/2. The outer radius is R(y)=yR(y)=y and the inner radius is r(y)=y/2r(y)=y/2. The volume is V=π∫02(y2−(y/2)2)dy=π∫02(y2−y2/4)dy=π∫02(34y2)dy=π[14y3]02=π(84)=2πV = \pi \int_0^2 (y^2 - (y/2)^2) dy = \pi \int_0^2 (y^2 - y^2/4) dy = \pi \int_0^2 (\frac{3}{4}y^2) dy = \pi [\frac{1}{4}y^3]_0^2 = \pi(\frac{8}{4}) = 2\pi.

Question 4

The region bounded by y=x2+2y=x^2+2 and y=6y=6 is revolved about the x-axis. What is the volume of the resulting solid?

  1. 256π5\frac{256\pi}{5}
  2. 704π15\frac{704\pi}{15}
  3. 1408π15\frac{1408\pi}{15} (correct answer)
  4. 128π3\frac{128\pi}{3}
Explanation: When you encounter a solid of revolution problem, you're using the disk or washer method to find volume. Since this region is bounded by two curves and revolved about the x-axis, you'll need the washer method because there's a hollow center. First, find where the curves intersect by setting x2+2=6x^2 + 2 = 6, which gives x2=4x^2 = 4, so x=±2x = \pm 2. The region extends from x=−2x = -2 to x=2x = 2, with y=6y = 6 as the outer curve and y=x2+2y = x^2 + 2 as the inner curve. For the washer method, the volume is V=π∫−22[R(x)2−r(x)2] dxV = \pi \int_{-2}^{2} [R(x)^2 - r(x)^2] \, dx, where R(x)=6R(x) = 6 (outer radius) and r(x)=x2+2r(x) = x^2 + 2 (inner radius). V=π∫−22[62−(x2+2)2] dx=π∫−22[36−(x4+4x2+4)] dxV = \pi \int_{-2}^{2} [6^2 - (x^2 + 2)^2] \, dx = \pi \int_{-2}^{2} [36 - (x^4 + 4x^2 + 4)] \, dx V=π∫−22(32−4x2−x4) dxV = \pi \int_{-2}^{2} (32 - 4x^2 - x^4) \, dx Evaluating: V=π[32x−4x33−x55]−22V = \pi \left[32x - \frac{4x^3}{3} - \frac{x^5}{5}\right]_{-2}^{2} Since the integrand is even, you can compute from 0 to 2 and double it: V=2π[64−323−325]=2π[320−160−9615]=1408π15V = 2\pi \left[64 - \frac{32}{3} - \frac{32}{5}\right] = 2\pi \left[\frac{320 - 160 - 96}{15}\right] = \frac{1408\pi}{15} Answer C is correct. Answer A likely comes from using only the outer radius, B from calculation errors in the integration, and D from forgetting to expand (x2+2)2(x^2 + 2)^2 properly. Remember: always identify which curve is outer/inner, and be careful expanding binomial expressions in washer problems.

Question 5

The region bounded by x=y2x=y^2 and x=y+2x=y+2 is revolved about the y-axis. Which integral represents the volume of the solid?

  1. π∫−12(y+2−y2)2dy\pi \int_{-1}^2 (y+2-y^2)^2 dy
  2. π∫−12(y4−(y+2)2)dy\pi \int_{-1}^2 (y^4 - (y+2)^2) dy
  3. 2π∫−12y(y+2−y2)dy2\pi \int_{-1}^2 y(y+2 - y^2) dy
  4. π∫−12((y+2)2−y4)dy\pi \int_{-1}^2 ((y+2)^2 - y^4) dy (correct answer)
Explanation: When finding volumes of solids of revolution about the y-axis, you need to determine whether to use the disk/washer method or cylindrical shells. Since this region is bounded by two curves and we're revolving about the y-axis, the washer method is most natural here. First, find where the curves intersect by solving y2=y+2y^2 = y + 2, which gives y2−y−2=0y^2 - y - 2 = 0, so (y−2)(y+1)=0(y-2)(y+1) = 0. The curves intersect at y=−1y = -1 and y=2y = 2. Between these y-values, the line x=y+2x = y + 2 is always to the right of the parabola x=y2x = y^2 (the outer radius), while x=y2x = y^2 forms the inner radius. Using the washer method, the volume is V=π∫−12[R(y)2−r(y)2]dyV = \pi \int_{-1}^2 [R(y)^2 - r(y)^2] dy, where R(y)=y+2R(y) = y + 2 (outer radius) and r(y)=y2r(y) = y^2 (inner radius). This gives V=π∫−12[(y+2)2−(y2)2]dy=π∫−12[(y+2)2−y4]dyV = \pi \int_{-1}^2 [(y+2)^2 - (y^2)^2] dy = \pi \int_{-1}^2 [(y+2)^2 - y^4] dy, which is answer D. Option A squares the entire difference (y+2−y2)2(y+2-y^2)^2, which would be incorrect for the washer method. Option B has the radii reversed, making the inner radius larger than the outer radius. Option C uses the shell method formula 2π∫y⋅h(y)dy2\pi \int y \cdot h(y) dy, but this approach is more complex here since you'd need to split the integral due to the curves' orientations. Strategy tip: For revolution about the y-axis with vertically simple regions, always check which radius is larger at each y-value, then apply the washer formula: outer radius squared minus inner radius squared.

Question 6

The region bounded by y=exy=e^x, y=1y=1, and x=2x=2 is revolved about the x-axis. What is the volume of the resulting solid?

  1. π(e4−12−1)\pi (\frac{e^4-1}{2} - 1)
  2. π(e4−12−2)\pi(\frac{e^4-1}{2} - 2) (correct answer)
  3. π(e4−2)\pi (e^4 - 2)
  4. π∫02(ex−1)2dx\pi \int_0^2 (e^x-1)^2 dx
Explanation: The region is bounded by x=0x=0 (where y=exy=e^x intersects y=1y=1) and x=2x=2. For revolution about the x-axis, the outer radius is R(x)=exR(x) = e^x and the inner radius is r(x)=1r(x)=1. The volume is V=π∫02((ex)2−12)dx=π∫02(e2x−1)dxV = \pi \int_0^2 ((e^x)^2 - 1^2) dx = \pi \int_0^2 (e^{2x} - 1) dx. Evaluating the integral gives π[12e2x−x]02=π((12e4−2)−(12e0−0))=π(e42−2−12)=π(e4−12−2)\pi [\frac{1}{2}e^{2x} - x]_0^2 = \pi ((\frac{1}{2}e^4 - 2) - (\frac{1}{2}e^0 - 0)) = \pi (\frac{e^4}{2} - 2 - \frac{1}{2}) = \pi(\frac{e^4-1}{2} - 2).

Question 7

A solid is generated by revolving the region bounded by y=x2y=x^2, y=4y=4, and x=0x=0 about the x-axis. What is the volume of the solid?

  1. 128π5\frac{128\pi}{5} (correct answer)
  2. 256π15\frac{256\pi}{15}
  3. 32π5\frac{32\pi}{5}
  4. 32π32\pi
Explanation: The region is bounded by x=0x=0 and the intersection of y=x2y=x^2 and y=4y=4, which is x=2x=2. For revolution about the x-axis, the outer radius is the distance from the axis (y=0) to the outer curve (y=4), so R(x)=4R(x)=4. The inner radius is the distance from the axis to the inner curve (y=x2y=x^2), so r(x)=x2r(x)=x^2. The volume is V=π∫02(42−(x2)2)dx=π∫02(16−x4)dxV = \pi \int_0^2 (4^2 - (x^2)^2) dx = \pi \int_0^2 (16 - x^4) dx. Evaluating gives π[16x−15x5]02=π(32−325)=128π5\pi [16x - \frac{1}{5}x^5]_0^2 = \pi (32 - \frac{32}{5}) = \frac{128\pi}{5}.

Question 8

Let RR be the region enclosed by the graphs of y=x+2y = x+2 and y=x2y = x^2. What is the volume of the solid generated by revolving RR about the x-axis?

  1. 36π5\frac{36\pi}{5}
  2. 72π5\frac{72\pi}{5} (correct answer)
  3. 108π5\frac{108\pi}{5}
  4. 216π15\frac{216\pi}{15}
Explanation: First, find the points of intersection by setting the two equations equal: x2=x+2⇒x2−x−2=0⇒(x−2)(x+1)=0x^2 = x+2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0. The points of intersection are x=−1x=-1 and x=2x=2. In the interval [−1,2][-1, 2], the line y=x+2y=x+2 is above the parabola y=x2y=x^2. When revolving around the x-axis, the outer radius is R(x)=x+2R(x) = x+2 and the inner radius is r(x)=x2r(x) = x^2. The volume VV is given by the washer method formula: V=π∫−12[R(x)2−r(x)2]dx=π∫−12[(x+2)2−(x2)2]dxV = \pi \int_{-1}^{2} [R(x)^2 - r(x)^2] dx = \pi \int_{-1}^{2} [(x+2)^2 - (x^2)^2] dx V=π∫−12(x2+4x+4−x4)dxV = \pi \int_{-1}^{2} (x^2 + 4x + 4 - x^4) dx V=π[−x55+x33+2x2+4x]−12V = \pi \left[ -\frac{x^5}{5} + \frac{x^3}{3} + 2x^2 + 4x \right]_{-1}^{2} V=π((−325+83+8+8)−(15−13+2−4))V = \pi \left( \left(-\frac{32}{5} + \frac{8}{3} + 8 + 8\right) - \left(\frac{1}{5} - \frac{1}{3} + 2 - 4\right) \right) V=π((−325+83+16)−(15−13−2))V = \pi \left( \left(-\frac{32}{5} + \frac{8}{3} + 16\right) - \left(\frac{1}{5} - \frac{1}{3} - 2\right) \right) V=π(−96+40+24015−3−5−3015)=π(18415−−3215)=216π15=72π5V = \pi \left( \frac{-96+40+240}{15} - \frac{3-5-30}{15} \right) = \pi \left( \frac{184}{15} - \frac{-32}{15} \right) = \frac{216\pi}{15} = \frac{72\pi}{5} Distractor A results from an arithmetic error in the final step. Distractor C may result from calculation errors, for example, incorrectly evaluating the definite integral. Distractor D is the unsimplified fraction.

Question 9

Let kk be a positive constant. The region bounded by the parabola y=kx2y=kx^2 and the line y=ky=k is revolved about the x-axis. If the volume of the resulting solid is 16π5\frac{16\pi}{5}, what is the value of kk?

  1. 222\sqrt{2}
  2. 44
  3. 22
  4. 2\sqrt{2} (correct answer)
Explanation: First, find the bounds of integration by setting the functions equal: kx2=k  ⟹  x2=1  ⟹  x=±1kx^2=k \implies x^2=1 \implies x=\pm 1. The line y=ky=k is the upper boundary, so the outer radius is R(x)=kR(x)=k. The parabola y=kx2y=kx^2 is the lower boundary, so the inner radius is r(x)=kx2r(x)=kx^2. The volume VV is given by: V=π∫−11[k2−(kx2)2]dx=π∫−11(k2−k2x4)dxV = \pi \int_{-1}^{1} [k^2 - (kx^2)^2] dx = \pi \int_{-1}^{1} (k^2 - k^2x^4) dx V=πk2∫−11(1−x4)dx=πk2[x−x55]−11V = \pi k^2 \int_{-1}^{1} (1 - x^4) dx = \pi k^2 \left[ x - \frac{x^5}{5} \right]_{-1}^{1} V=πk2((1−15)−(−1−−15))=πk2(45−(−45))=8πk25V = \pi k^2 \left( (1 - \frac{1}{5}) - (-1 - \frac{-1}{5}) \right) = \pi k^2 \left( \frac{4}{5} - (-\frac{4}{5}) \right) = \frac{8\pi k^2}{5} We are given that the volume is 16π5\frac{16\pi}{5}. So, we set them equal: 8πk25=16π5  ⟹  8k2=16  ⟹  k2=2\frac{8\pi k^2}{5} = \frac{16\pi}{5} \implies 8k^2 = 16 \implies k^2 = 2 Since kk is positive, k=2k=\sqrt{2}. Distractor C, k=2k=2, might arise from an algebraic error such as 4k=84k=8. Distractor B arises from other miscalculations. Distractor A results from a different algebraic mistake.

Question 10

Let R1R_1 be the region bounded by y=f(x)y=f(x) and y=g(x)y=g(x) on the interval [a,b][a,b], where f(x)≥g(x)≥0f(x) \ge g(x) \ge 0. Let V1V_1 be the volume of the solid generated by revolving R1R_1 about the x-axis. Let R2R_2 be the region bounded by y=f(x)+cy=f(x)+c and y=g(x)+cy=g(x)+c on [a,b][a,b], where cc is a positive constant. Let V2V_2 be the volume of the solid generated by revolving R2R_2 about the x-axis. Which statement correctly relates V2V_2 to V1V_1?

  1. V2>V1V_2 > V_1 (correct answer)
  2. V2=V1V_2 = V_1
  3. V2<V1V_2 < V_1
  4. The relationship cannot be determined without knowing the specific functions.
Explanation: When you encounter problems involving solids of revolution with vertical shifts, think about how the washer method changes when both curves move up by the same amount. For region R1R_1, the volume is V1=π∫ab[f(x)]2−[g(x)]2 dxV_1 = \pi \int_a^b [f(x)]^2 - [g(x)]^2 \, dx When both functions shift up by constant cc, region R2R_2 has volume: V2=π∫ab[f(x)+c]2−[g(x)+c]2 dxV_2 = \pi \int_a^b [f(x)+c]^2 - [g(x)+c]^2 \, dx Expanding the squared terms: V2=π∫ab[f(x)2+2cf(x)+c2]−[g(x)2+2cg(x)+c2] dxV_2 = \pi \int_a^b [f(x)^2 + 2cf(x) + c^2] - [g(x)^2 + 2cg(x) + c^2] \, dx Simplifying: V2=π∫ab[f(x)2−g(x)2]+2c[f(x)−g(x)] dxV_2 = \pi \int_a^b [f(x)^2 - g(x)^2] + 2c[f(x) - g(x)] \, dx V2=V1+2πc∫ab[f(x)−g(x)] dxV_2 = V_1 + 2\pi c \int_a^b [f(x) - g(x)] \, dx Since f(x)≥g(x)f(x) \geq g(x) on [a,b][a,b], we have f(x)−g(x)≥0f(x) - g(x) \geq 0, making the integral positive. Therefore V2>V1V_2 > V_1. A is correct because the additional term 2πc∫ab[f(x)−g(x)] dx2\pi c \int_a^b [f(x) - g(x)] \, dx is positive. B is wrong because the volumes would only be equal if f(x)=g(x)f(x) = g(x) everywhere, which contradicts having a region between them. C is wrong because it ignores that shifting up increases the outer radius more than the inner radius in the washer method. D is wrong because the relationship always holds when f(x)≥g(x)f(x) \geq g(x) and c>0c > 0. Study tip: Remember that vertical shifts in revolution problems don't just translate volume—they amplify it because you're squaring the shifted functions. The washer method makes this effect multiplicative, not additive.

Question 11

The region in the first quadrant bounded by y=2xy=2\sqrt{x}, the line x=ax=a (for some a>0a>0), and the x-axis is revolved about the y-axis. The volume of this solid is found to be equal to the volume of the solid generated by revolving the same region about the x-axis. What is the value of aa?

  1. 54\frac{5}{4}
  2. 45\frac{4}{5}
  3. 1625\frac{16}{25}
  4. 2516\frac{25}{16} (correct answer)
Explanation: First, find the volume VxV_x from revolving about the x-axis. This is a disk method problem. The radius is R(x)=2xR(x)=2\sqrt{x}. Vx=π∫0a(2x)2dx=π∫0a4xdx=π[2x2]0a=2πa2V_x = \pi \int_0^a (2\sqrt{x})^2 dx = \pi \int_0^a 4x dx = \pi [2x^2]_0^a = 2\pi a^2 Next, find the volume VyV_y from revolving about the y-axis. We must use the washer method. The curve is x=y2/4x = y^2/4. The region is bounded on the right by x=ax=a and on the left by x=y2/4x=y^2/4. The y-values range from 00 to 2a2\sqrt{a}. The outer radius is R(y)=aR(y)=a and the inner radius is r(y)=y2/4r(y)=y^2/4. Vy=π∫02a[a2−(y2/4)2]dy=π∫02a(a2−y416)dyV_y = \pi \int_0^{2\sqrt{a}} [a^2 - (y^2/4)^2] dy = \pi \int_0^{2\sqrt{a}} (a^2 - \frac{y^4}{16}) dy Vy=π[a2y−y580]02a=π(a2(2a)−(2a)580)=π(2a5/2−32a5/280)=π(2a5/2−25a5/2)=8π5a5/2V_y = \pi \left[ a^2y - \frac{y^5}{80} \right]_0^{2\sqrt{a}} = \pi \left( a^2(2\sqrt{a}) - \frac{(2\sqrt{a})^5}{80} \right) = \pi \left( 2a^{5/2} - \frac{32a^{5/2}}{80} \right) = \pi \left( 2a^{5/2} - \frac{2}{5}a^{5/2} \right) = \frac{8\pi}{5}a^{5/2} Now, set Vx=VyV_x = V_y: 2πa2=8π5a5/22\pi a^2 = \frac{8\pi}{5}a^{5/2} Since a>0a>0, we can divide by 2πa22\pi a^2: 1=45a1/2  ⟹  a=54  ⟹  a=25161 = \frac{4}{5}a^{1/2} \implies \sqrt{a} = \frac{5}{4} \implies a = \frac{25}{16} Distractor A is the value of a\sqrt{a}, not aa. Distractor C is the reciprocal of the correct answer. Distractor B is from an algebraic error in solving for a\sqrt{a}.

Question 12

Let RR be the region in the first quadrant bounded by the graphs of y=x2+1y=x^2+1, y=5y=5, and x=1x=1. What is the volume of the solid generated by revolving RR about the y-axis?

  1. 4π4\pi
  2. 15π2\frac{15\pi}{2}
  3. 9π2\frac{9\pi}{2} (correct answer)
  4. 8π8\pi
Explanation: To revolve around the y-axis, we must integrate with respect to yy. The region is bounded by x=1x=1 on the left, y=5y=5 on the top, and y=x2+1y=x^2+1 on the bottom-right. We need to express xx in terms of yy: y=x2+1  ⟹  x=y−1y=x^2+1 \implies x=\sqrt{y-1} (since we are in the first quadrant). The bounds for yy are from the intersection of x=1x=1 and y=x2+1y=x^2+1 (which is y=12+1=2y=1^2+1=2) up to y=5y=5. So, we integrate from y=2y=2 to y=5y=5. For a given yy in this interval, the outer radius is the distance from the y-axis to the curve x=y−1x=\sqrt{y-1}, so R(y)=y−1R(y)=\sqrt{y-1}. The inner radius is the distance from the y-axis to the line x=1x=1, so r(y)=1r(y)=1. The volume is: V=π∫25[(y−1)2−12]dy=π∫25(y−1−1)dy=π∫25(y−2)dyV = \pi \int_{2}^{5} [(\sqrt{y-1})^2 - 1^2] dy = \pi \int_{2}^{5} (y-1-1) dy = \pi \int_{2}^{5} (y-2) dy V=π[y22−2y]25=π[(252−10)−(42−4)]=π[52−(−2)]=9π2V = \pi \left[ \frac{y^2}{2} - 2y \right]_2^5 = \pi \left[ (\frac{25}{2}-10) - (\frac{4}{2}-4) \right] = \pi \left[ \frac{5}{2} - (-2) \right] = \frac{9\pi}{2} Distractor A results from using incorrect bounds, such as integrating from 1 to 5. Distractor B is the volume of the solid using the disk method with radius R(y)=y−1R(y)=\sqrt{y-1}, ignoring the hole created by revolving around x=1x=1. Distractor D is the volume of the solid of revolution for the region bounded by y=x2+1y=x^2+1 and y=5y=5 (without the x=1x=1 boundary).

Question 13

Let RR be the region enclosed by y=x2y=x^2 and y=xy=\sqrt{x}. Let VxV_x be the volume of the solid generated by revolving RR about the x-axis, and let VyV_y be the volume of the solid generated by revolving RR about the y-axis. What is the ratio VxVy\frac{V_x}{V_y}?

  1. 11 (correct answer)
  2. 12\frac{1}{2}
  3. 22
  4. 32\frac{3}{2}
Explanation: When you encounter a problem involving volumes of revolution about different axes, look for symmetry relationships that might simplify your work rather than computing both volumes separately. First, let's find where the curves intersect. Setting x2=xx^2 = \sqrt{x}, we get x4=xx^4 = x, so x4−x=0x^4 - x = 0, giving us x(x3−1)=0x(x^3 - 1) = 0. The curves intersect at x=0x = 0 and x=1x = 1. For the region between these curves, we have x≥x2\sqrt{x} \geq x^2 on the interval [0,1][0,1]. Here's the key insight: there's a beautiful symmetry at play. If you make the substitution u=xu = \sqrt{x} (so x=u2x = u^2), the curve y=xy = \sqrt{x} becomes u=yu = y, and the curve y=x2y = x^2 becomes y=u4y = u^4. This transformation swaps the roles of the two curves while preserving the enclosed area. This symmetry means that revolving about the x-axis gives the same volume as revolving about the y-axis, so Vx=VyV_x = V_y and VxVy=1\frac{V_x}{V_y} = 1. Looking at the wrong answers: B (12\frac{1}{2}) and C (22) might tempt you if you incorrectly set up one of the integrals or confused which curve is on top. Answer D (32\frac{3}{2}) doesn't correspond to any natural computational error but serves as a distractor. Study tip: When comparing volumes of revolution about different axes, always check for symmetry first. The substitution u=xu = \sqrt{x} often reveals hidden relationships between x-axis and y-axis rotations, potentially saving you from lengthy calculations.

Question 14

The region in the first quadrant bounded by y=1/xy=1/x, x=1x=1, and y=4y=4 is revolved about the y-axis. Which integral represents the volume of the solid?

  1. π∫14(1/y2−1)dy\pi \int_1^4 (1/y^2 - 1) dy
  2. π∫14(1−1/y2)dy\pi \int_1^4 (1 - 1/y^2) dy (correct answer)
  3. π∫1/41(1/x−1)2dx\pi \int_{1/4}^1 (1/x - 1)^2 dx
  4. π∫14(1−1/y)dy\pi \int_1^4 (1 - 1/y) dy
Explanation: When you encounter a volume of revolution problem, the key is identifying which axis of rotation to use and setting up the correct method. Since we're revolving around the y-axis, you'll want to express everything in terms of y and use the washer method. First, sketch the region. The curve y=1/xy = 1/x intersects the line x=1x = 1 at point (1,1)(1,1), and the horizontal line y=4y = 4 intersects the curve at x=1/4x = 1/4. So our region extends from y=1y = 1 to y=4y = 4. For revolution about the y-axis, each horizontal slice creates a washer with outer radius and inner radius. At height yy, the outer radius extends from the y-axis to the line x=1x = 1, giving radius = 1. The inner radius extends from the y-axis to the curve x=1/yx = 1/y (solving y=1/xy = 1/x for xx), giving radius = 1/y1/y. Using the washer formula V=π∫[R2−r2]dyV = \pi \int [R^2 - r^2] dy, we get π∫14(12−(1/y)2)dy=π∫14(1−1/y2)dy\pi \int_1^4 (1^2 - (1/y)^2) dy = \pi \int_1^4 (1 - 1/y^2) dy. Choice A has the radii reversed, giving a negative volume. Choice C attempts to use x as the variable of integration, which doesn't match y-axis rotation. Choice D uses (1−1/y)(1 - 1/y) instead of the correct (1−1/y2)(1 - 1/y^2), forgetting to square the radii. Remember: for y-axis rotation, integrate with respect to y, and always square both the outer and inner radii in the washer method.

Question 15

Let R be the region bounded by y=xy=x, the x-axis, the y-axis, and the line x=kx=k for some k>0k>0. Let S be the region bounded by y=xy=x, the y-axis, and the line y=ky=k. If S is revolved around the y-axis and R is revolved around the x-axis, the resulting solids have the same volume. This statement is true for:

  1. no value of kk
  2. only k=1k=1
  3. only k=33k=\sqrt[3]{3}
  4. all k>0k>0 (correct answer)
Explanation: When you encounter volumes of revolution problems, you're applying the disk/washer method or cylindrical shell method to find volumes of 3D solids formed by rotating 2D regions around axes. Let's find the volumes systematically. For region R (bounded by y=xy=x, x-axis, y-axis, and x=kx=k), when revolved around the x-axis, we use the disk method: VR=π∫0kx2 dx=π[x33]0k=πk33V_R = \pi \int_0^k x^2 \, dx = \pi \left[\frac{x^3}{3}\right]_0^k = \frac{\pi k^3}{3} For region S (bounded by y=xy=x, y-axis, and y=ky=k), when revolved around the y-axis, we again use the disk method. Since y=xy=x means x=yx=y, the radius at height yy is yy: VS=π∫0ky2 dy=π[y33]0k=πk33V_S = \pi \int_0^k y^2 \, dy = \pi \left[\frac{y^3}{3}\right]_0^k = \frac{\pi k^3}{3} Both volumes equal πk33\frac{\pi k^3}{3}, so they're equal for all k>0k>0. Why the wrong answers fail: Choice A) suggests no value works, but we just showed they're always equal. Choice B) claims only k=1k=1 works, missing that the relationship holds universally. Choice C) proposes k=33k=\sqrt[3]{3} as the unique solution, but this stems from incorrectly setting up or solving the volume equations. The key insight here is recognizing the symmetry between these two setups. Both regions are essentially the same triangle oriented differently, and the revolution methods preserve this geometric relationship. Always set up your integrals carefully and look for underlying symmetries that might make problems simpler than they initially appear.

Question 16

The volume of a solid generated by revolving the region between two functions f(x)f(x) and g(x)g(x) (where f(x)≥g(x)≥0f(x) \ge g(x) \ge 0) on [a,b][a, b] about the x-axis is given by V=π∫ab(f(x)2−g(x)2)dxV = \pi \int_a^b (f(x)^2 - g(x)^2) dx. This formula is equivalent to:

  1. π∫ab(f(x)−g(x))2dx\pi \int_a^b (f(x) - g(x))^2 dx
  2. π(∫abf(x)dx−∫abg(x)dx)2\pi (\int_a^b f(x) dx - \int_a^b g(x) dx)^2
  3. The volume of the solid from f(x)f(x) minus the volume of the solid from g(x)g(x). (correct answer)
  4. The area of the region squared, multiplied by the length of the interval.
Explanation: When you encounter volume problems involving solids of revolution between two curves, think about what's physically happening: you're creating a solid with a hollow center, like a washer or ring. The formula V=π∫ab(f(x)2−g(x)2)dxV = \pi \int_a^b (f(x)^2 - g(x)^2) dx represents the washer method. At each cross-section, you have a disk of radius f(x)f(x) with a hole of radius g(x)g(x). The area of this washer is πf(x)2−πg(x)2=π(f(x)2−g(x)2)\pi f(x)^2 - \pi g(x)^2 = \pi(f(x)^2 - g(x)^2). Integrating gives the total volume. This is equivalent to finding the volume of the entire solid generated by f(x)f(x) and subtracting the volume of the hollow interior generated by g(x)g(x). That's exactly what option C describes - making it the correct interpretation. Option A, π∫ab(f(x)−g(x))2dx\pi \int_a^b (f(x) - g(x))^2 dx, would give you the volume using the shell method or a different setup, but it's not equivalent to the washer method formula given. Option B, π(∫abf(x)dx−∫abg(x)dx)2\pi (\int_a^b f(x) dx - \int_a^b g(x) dx)^2, incorrectly squares the entire difference of integrals rather than integrating the difference of squares. This conflates area calculations with volume. Option D confuses the geometric relationship entirely - volume isn't simply area squared times length in revolution problems. Remember: the washer method always equals "big volume minus little volume." When you see f(x)2−g(x)2f(x)^2 - g(x)^2 in a revolution integral, think of it as subtracting the inner solid from the outer solid.

Question 17

Let the region R be bounded by the curves y=x2y=x^2 and y=cy=c for a constant c>0c>0. The volume of the solid generated by revolving R about the y-axis is VyV_y. The volume of the solid generated by revolving R about the x-axis is VxV_x. For what value of cc is Vx=VyV_x = V_y?

  1. 25625\frac{256}{25}
  2. 516\frac{5}{16}
  3. 165\frac{16}{5}
  4. 25256\frac{25}{256} (correct answer)
Explanation: When you encounter problems involving volumes of revolution with different axes, you need to set up and compare two distinct integrals using the disk/washer method. First, let's find the region R. The curves y=x2y = x^2 and y=cy = c intersect when x2=cx^2 = c, so x=±cx = \pm\sqrt{c}. The region extends from x=−cx = -\sqrt{c} to x=cx = \sqrt{c}. For revolution about the x-axis, we use Vx=π∫−cc(c2−(x2)2)dx=π∫−cc(c2−x4)dxV_x = \pi \int_{-\sqrt{c}}^{\sqrt{c}} (c^2 - (x^2)^2) dx = \pi \int_{-\sqrt{c}}^{\sqrt{c}} (c^2 - x^4) dx. By symmetry, this becomes Vx=2π∫0c(c2−x4)dx=2π[c2x−x55]0c=2π(c2c−c5/25)=8πc5/25V_x = 2\pi \int_0^{\sqrt{c}} (c^2 - x^4) dx = 2\pi[c^2x - \frac{x^5}{5}]_0^{\sqrt{c}} = 2\pi(c^2\sqrt{c} - \frac{c^{5/2}}{5}) = \frac{8\pi c^{5/2}}{5}. For revolution about the y-axis, we use the shell method: Vy=2π∫0cx(c−x2)dx=2π∫0c(cx−x3)dx=2π[cx22−x44]0c=2π(c22−c24)=πc22V_y = 2\pi \int_0^{\sqrt{c}} x(c - x^2) dx = 2\pi \int_0^{\sqrt{c}} (cx - x^3) dx = 2\pi[\frac{cx^2}{2} - \frac{x^4}{4}]_0^{\sqrt{c}} = 2\pi(\frac{c^2}{2} - \frac{c^2}{4}) = \frac{\pi c^2}{2}. Setting Vx=VyV_x = V_y: 8πc5/25=πc22\frac{8\pi c^{5/2}}{5} = \frac{\pi c^2}{2}. Dividing by πc2\pi c^2: 8c5=12\frac{8\sqrt{c}}{5} = \frac{1}{2}. Solving: c=516\sqrt{c} = \frac{5}{16}, so c=25256c = \frac{25}{256}. Choice A (25625\frac{256}{25}) would result from incorrectly inverting the final answer. Choice B (516\frac{5}{16}) represents c\sqrt{c} rather than cc. Choice C (165\frac{16}{5}) comes from algebraic errors in the equation setup. Always check your algebra carefully when equating volumes, and remember that shell and disk methods often yield different-looking integrals for the same region.

Question 18

The region enclosed by the parabolas y=2−x2y = 2-x^2 and y=x2y=x^2 is revolved about the x-axis. What is the volume of the solid generated?

  1. 8π3\frac{8\pi}{3}
  2. 16π3\frac{16\pi}{3} (correct answer)
  3. 8π8\pi
  4. 32π3\frac{32\pi}{3}
Explanation: First, find the points of intersection: x2=2−x2  ⟹  2x2=2  ⟹  x2=1  ⟹  x=±1x^2 = 2-x^2 \implies 2x^2=2 \implies x^2=1 \implies x = \pm 1. The region is symmetric about the y-axis. For xx in [−1,1][-1, 1], the parabola y=2−x2y=2-x^2 is above y=x2y=x^2. Thus, the outer radius is R(x)=2−x2R(x)=2-x^2 and the inner radius is r(x)=x2r(x)=x^2. The volume is: V=π∫−11[(2−x2)2−(x2)2]dxV = \pi \int_{-1}^{1} [(2-x^2)^2 - (x^2)^2] dx V=π∫−11[(4−4x2+x4)−x4]dx=π∫−11(4−4x2)dxV = \pi \int_{-1}^{1} [(4 - 4x^2 + x^4) - x^4] dx = \pi \int_{-1}^{1} (4 - 4x^2) dx Using symmetry, we can write: V=2π∫01(4−4x2)dx=2π[4x−43x3]01V = 2\pi \int_{0}^{1} (4 - 4x^2) dx = 2\pi \left[ 4x - \frac{4}{3}x^3 \right]_{0}^{1} V=2π((4−43)−0)=2π(83)=16π3V = 2\pi \left( (4 - \frac{4}{3}) - 0 \right) = 2\pi \left( \frac{8}{3} \right) = \frac{16\pi}{3} Distractor A might result from forgetting to use symmetry and only integrating from 0 to 1, or another calculation error. Distractor C or D would result from incorrect algebraic expansion or integration.

Question 19

Let R be the region in the first quadrant bounded by the curves y=x3y=x^3 and y=xy=x. Which of the following integrals represents the volume of the solid formed by revolving R about the y-axis?

  1. π∫01(y2/3−y2)dy\pi \int_0^1 (y^{2/3} - y^2) dy (correct answer)
  2. π∫01(x2−x6)dx\pi \int_0^1 (x^2 - x^6) dx
  3. π∫01(y3−y)2dy\pi \int_0^1 (\sqrt[3]{y} - y)^2 dy
  4. π∫01(y2−y2/3)dy\pi \int_0^1 (y^2 - y^{2/3}) dy
Explanation: To revolve around the y-axis, we must integrate with respect to y. The functions must be expressed in terms of y: x=y3x = \sqrt[3]{y} and x=yx=y. The intersection points are found by setting y=y3y=y^3, which for the first quadrant gives y=0y=0 and y=1y=1. On the interval y∈[0,1]y \in [0,1], y3≥y\sqrt[3]{y} \ge y, so the outer radius is R(y)=y3R(y) = \sqrt[3]{y} and the inner radius is r(y)=yr(y)=y. The volume is given by the washer method formula V=π∫01(R(y)2−r(y)2)dy=π∫01((y3)2−y2)dy=π∫01(y2/3−y2)dyV = \pi \int_0^1 (R(y)^2 - r(y)^2) dy = \pi \int_0^1 ((\sqrt[3]{y})^2 - y^2) dy = \pi \int_0^1 (y^{2/3} - y^2) dy.

Question 20

Let RR be the region in the first quadrant bounded by y=ln⁡xy = \ln x, the line x=ex=e, and the x-axis. What is the volume of the solid formed when RR is revolved about the y-axis?

  1. π∫1e(ln⁡x)2dx\pi \int_{1}^{e} (\ln x)^2 dx
  2. π∫01(e2−e2y)dy\pi \int_{0}^{1} (e^2 - e^{2y}) dy (correct answer)
  3. π∫01(e−ey)2dy\pi \int_{0}^{1} (e - e^y)^2 dy
  4. π∫01(e2y−e2)dy\pi \int_{0}^{1} (e^{2y} - e^2) dy
Explanation: For revolution about the y-axis, we need to express xx in terms of yy and integrate with respect to yy. The curve y=ln⁡xy = \ln x becomes x=eyx = e^y. The line x=ex=e is a vertical boundary. The x-axis (y=0y=0) is the bottom boundary. The region is bounded on the right by x=ex=e and on the left by x=eyx=e^y. The bounds for integration in yy are from y=0y=0 up to the intersection of x=ex=e and x=eyx=e^y, which is where e=ey  ⟹  y=1e=e^y \implies y=1. The outer radius is R(y)=eR(y) = e and the inner radius is r(y)=eyr(y) = e^y. The volume is V=π∫01[R(y)2−r(y)2]dy=π∫01[e2−(ey)2]dy=π∫01(e2−e2y)dyV = \pi \int_{0}^{1} [R(y)^2 - r(y)^2] dy = \pi \int_{0}^{1} [e^2 - (e^y)^2] dy = \pi \int_{0}^{1} (e^2 - e^{2y}) dy. Distractor A represents the volume using the disk method for revolving the region under y=ln⁡xy=\ln x about the x-axis. Distractor C uses the incorrect formula (R−r)2(R-r)^2. Distractor D swaps the outer and inner radii, which would result in a negative volume.