Calculus 2 Quiz: Washer Method X Or Y Axis
2 questions · exam conditions
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Washer Method X Or Y AxisQuestion 1 of 2

A region is bounded by y=x2+1y = x^2 + 1, y=5y = 5, and the y-axis in the first quadrant. When this region is revolved around the x-axis, what is the volume of the solid formed?

128π15\frac{128\pi}{15}
256π15\frac{256\pi}{15}
64π3\frac{64\pi}{3}
32π5\frac{32\pi}{5}
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Calculus 2 Quiz

Calculus 2 Quiz: Washer Method X Or Y Axis

Practice Washer Method X Or Y Axis 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 Washer Method X Or Y Axis, 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

A region is bounded by y=x2+1y = x^2 + 1, y=5y = 5, and the y-axis in the first quadrant. When this region is revolved around the x-axis, what is the volume of the solid formed?

  1. 128π15\frac{128\pi}{15}
  2. 256π15\frac{256\pi}{15} (correct answer)
  3. 64π3\frac{64\pi}{3}
  4. 32π5\frac{32\pi}{5}
Explanation: First, find the intersection points: x2+1=5x^2 + 1 = 5 gives x=2x = 2. The region extends from x=0x = 0 to x=2x = 2. Using the washer method, the outer radius is R(x)=5R(x) = 5 and inner radius is r(x)=x2+1r(x) = x^2 + 1. Volume = π∫02[52−(x2+1)2]dx=π∫02[25−x4−2x2−1]dx=π∫02[24−x4−2x2]dx=π[24x−x55−2x33]02=π[48−325−163]=π[720−192−8015]=256π15\pi \int_0^2 [5^2 - (x^2 + 1)^2] dx = \pi \int_0^2 [25 - x^4 - 2x^2 - 1] dx = \pi \int_0^2 [24 - x^4 - 2x^2] dx = \pi [24x - \frac{x^5}{5} - \frac{2x^3}{3}]_0^2 = \pi [48 - \frac{32}{5} - \frac{16}{3}] = \pi [\frac{720 - 192 - 80}{15}] = \frac{256\pi}{15}. Choice A incorrectly uses disk method instead of washer. Choice C forgets to subtract the inner radius. Choice D makes an error in the polynomial expansion.

Question 2

A solid is formed by revolving the region bounded by y=xy = \sqrt{x}, y=0y = 0, and x=4x = 4 around the line y=2y = 2. What is the volume of this solid?

  1. 56π3\frac{56\pi}{3} (correct answer)
  2. 64π3\frac{64\pi}{3}
  3. 48π5\frac{48\pi}{5}
  4. 32π3\frac{32\pi}{3}
Explanation: When revolving around y=2y = 2, the outer radius is the distance from y=0y = 0 to y=2y = 2, which is 2. The inner radius is the distance from y=xy = \sqrt{x} to y=2y = 2, which is 2−x2 - \sqrt{x}. Volume = π∫04[22−(2−x)2]dx=π∫04[4−(4−4x+x)]dx=π∫04[4x−x]dx=π[8x3/23−x22]04=π[643−8]=π[64−243]=56π3\pi \int_0^4 [2^2 - (2 - \sqrt{x})^2] dx = \pi \int_0^4 [4 - (4 - 4\sqrt{x} + x)] dx = \pi \int_0^4 [4\sqrt{x} - x] dx = \pi [\frac{8x^{3/2}}{3} - \frac{x^2}{2}]_0^4 = \pi [\frac{64}{3} - 8] = \pi [\frac{64 - 24}{3}] = \frac{56\pi}{3}. Choice B forgets to expand (2−x)2(2 - \sqrt{x})^2 correctly. Choice C makes an integration error. Choice D uses the disk method incorrectly.