Calculus 2 Quiz: Disc Method X Or Y Axis
5 questions · exam conditions
0:00
Disc Method X Or Y AxisQuestion 1 of 5

The region bounded by y=xy = \sqrt{x}, y=0y = 0, and x=4x = 4 is revolved around the x-axis. If the resulting solid is sliced by planes perpendicular to the x-axis, and the cross-sectional area at position x=ax = a is 9π9\pi square units, what is the value of aa?

a=3a = 3
a=9a = 9
a=9πa = \frac{9}{\pi}
a=9πa = \sqrt{9\pi}
← Back to quizzes

Calculus 2 Quiz

Calculus 2 Quiz: Disc Method X Or Y Axis

Practice Disc 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 Disc Method X Or Y Axis, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

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.

All questions

Question 1

The region bounded by y=xy = \sqrt{x}, y=0y = 0, and x=4x = 4 is revolved around the x-axis. If the resulting solid is sliced by planes perpendicular to the x-axis, and the cross-sectional area at position x=ax = a is 9π9\pi square units, what is the value of aa?

  1. a=3a = 3
  2. a=9a = 9 (correct answer)
  3. a=9πa = \frac{9}{\pi}
  4. a=9πa = \sqrt{9\pi}
Explanation: When the region is revolved around the x-axis, cross-sections perpendicular to the x-axis are circular discs with radius r=xr = \sqrt{x}. The area of such a disc is A(x)=πr2=π(x)2=πxA(x) = \pi r^2 = \pi(\sqrt{x})^2 = \pi x. Setting this equal to 9π9\pi: πx=9π\pi x = 9\pi, so x=9x = 9. Therefore a=9a = 9. Choice A results from incorrectly taking the square root of 9. Choice C results from forgetting to divide out the π\pi factor. Choice D results from incorrectly solving x=9π\sqrt{x} = 9\pi instead of πx=9π\pi x = 9\pi.

Question 2

The curve x=y2−4x = y^2 - 4 intersects the y-axis at two points. The region between this curve and the y-axis is revolved around the y-axis. What is the volume of the resulting solid?

  1. 64π3\frac{64\pi}{3}
  2. 256π15\frac{256\pi}{15}
  3. 128π5\frac{128\pi}{5} (correct answer)
  4. 32π3\frac{32\pi}{3}
Explanation: The curve x=y2−4x = y^2 - 4 intersects the y-axis when x=0x = 0, so 0=y2−40 = y^2 - 4, giving y=±2y = \pm 2. For −2≤y≤2-2 \leq y \leq 2, the curve is to the left of the y-axis. When revolved around the y-axis, we get discs with radius r=∣y2−4∣=4−y2r = |y^2 - 4| = 4 - y^2 (since y2≤4y^2 \leq 4 in this interval). The volume is V=π∫−22(4−y2)2 dy=π∫−22(16−8y2+y4) dy=π[16y−8y33+y55]−22=π[32−643+325]⋅2=128π5V = \pi \int_{-2}^{2} (4 - y^2)^2 \, dy = \pi \int_{-2}^{2} (16 - 8y^2 + y^4) \, dy = \pi \left[16y - \frac{8y^3}{3} + \frac{y^5}{5}\right]_{-2}^{2} = \pi \left[32 - \frac{64}{3} + \frac{32}{5}\right] \cdot 2 = \frac{128\pi}{5}. Other choices result from sign errors or incorrect radius calculations.

Question 3

The area between y=4−x2y = \sqrt{4-x^2} and the x-axis for −2≤x≤2-2 \leq x \leq 2 is revolved around the x-axis. If the volume is kπk\pi, what is the value of kk?

  1. 163\frac{16}{3}
  2. 1616
  3. 88
  4. 323\frac{32}{3} (correct answer)
Explanation: When you encounter a problem involving revolving a region around an axis, you're dealing with the disk/washer method for finding volumes of revolution. The key is identifying the radius function and setting up the correct integral. First, recognize that y=4−x2y = \sqrt{4-x^2} is the upper half of a circle with radius 2 centered at the origin. When you revolve this semicircle around the x-axis, you create a solid sphere with radius 2. Using the disk method, the volume is V=π∫−22[R(x)]2 dxV = \pi \int_{-2}^{2} [R(x)]^2 \, dx, where R(x)=4−x2R(x) = \sqrt{4-x^2} is the radius at each x-value. This gives us: V=π∫−22(4−x2)2 dx=π∫−22(4−x2) dxV = \pi \int_{-2}^{2} (\sqrt{4-x^2})^2 \, dx = \pi \int_{-2}^{2} (4-x^2) \, dx Evaluating: V=π[4x−x33]−22=π[(8−83)−(−8+83)]=π[16−163]=32π3V = \pi \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \pi \left[\left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right)\right] = \pi \left[16 - \frac{16}{3}\right] = \frac{32\pi}{3} Therefore, k=323k = \frac{32}{3}, which is answer D. Answer A (163\frac{16}{3}) likely comes from forgetting to account for the negative portion of the interval. Answer B (1616) might result from incorrectly computing 4x4x over the interval without the x3x^3 term. Answer C (88) could come from various computational errors in the integration or evaluation. Remember: when revolving around the x-axis, always square the y-function inside your integral, and be careful with your interval arithmetic when evaluating definite integrals.

Question 4

Consider the region in the first quadrant bounded by y=x2y = x^2, y=2xy = 2x, and the y-axis. When this region is revolved around the y-axis, what integral represents the volume using the disc method?

  1. π∫04(y2)2 dy\pi \int_0^4 \left(\frac{y}{2}\right)^2 \, dy
  2. π∫04[(y2)2−(y)2] dy\pi \int_0^4 \left[\left(\frac{y}{2}\right)^2 - (\sqrt{y})^2\right] \, dy (correct answer)
  3. π∫04[(y)2−(y2)2] dy\pi \int_0^4 \left[(\sqrt{y})^2 - \left(\frac{y}{2}\right)^2\right] \, dy
  4. π∫02[(2x)2−(x2)2] dx\pi \int_0^2 \left[(2x)^2 - (x^2)^2\right] \, dx
Explanation: The curves intersect when x2=2xx^2 = 2x, giving x=0x = 0 and x=2x = 2, so yy ranges from 0 to 4. For revolution around the y-axis, we need the washer method (a form of disc method) since there's a hole. The outer radius is from y=2xy = 2x (so x=y2x = \frac{y}{2}) and the inner radius is from y=x2y = x^2 (so x=yx = \sqrt{y}). The volume integral is π∫04[(y2)2−(y)2] dy\pi \int_0^4 \left[\left(\frac{y}{2}\right)^2 - (\sqrt{y})^2\right] \, dy. Choice A forgets the inner radius. Choice C reverses the outer and inner radii. Choice D uses the wrong variable of integration and incorrect setup.

Question 5

The region bounded by y=exy = e^x, y=1y = 1, and x=ln⁡3x = \ln 3 is revolved around the line y=1y = 1. Using the disc method, what integral represents the volume?

  1. π∫0ln⁡3(ex−1)2 dx\pi \int_0^{\ln 3} (e^x - 1)^2 \, dx (correct answer)
  2. π∫13(ln⁡y−1)2 dy\pi \int_1^3 (\ln y - 1)^2 \, dy
  3. π∫0ln⁡3(ex)2 dx\pi \int_0^{\ln 3} (e^x)^2 \, dx
  4. π∫13(ln⁡y)2 dy\pi \int_1^3 (\ln y)^2 \, dy
Explanation: When revolving around y=1y = 1, the radius of each disc is the distance from the curve y=exy = e^x to the line y=1y = 1, which is r=ex−1r = e^x - 1. The region extends from x=0x = 0 (where e0=1e^0 = 1) to x=ln⁡3x = \ln 3. Therefore, the volume integral is π∫0ln⁡3(ex−1)2 dx\pi \int_0^{\ln 3} (e^x - 1)^2 \, dx. Choice B attempts to use yy as the variable but has an incorrect radius formula. Choice C forgets to account for the shift to y=1y = 1. Choice D uses the wrong variable and wrong radius.