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Volumes by disks and washers practice problems

Solids of revolution sliced perpendicular to the axis. 18 problems with worked solutions; in 18 of them every equation is proved by a computer algebra system.

8 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 5.17

easy
The region between \( \displaystyle y = x \) and \( \displaystyle y = x^{2} \) is revolved about the x-axis. Find the volume.
  1. \[ - x \left(x - 1\right) = - x^{2} + x \]
    The curves meet at x = 0 and x = 1.✓ Proved
  2. Cross sections are washers: outer radius x, inner radius x**2.
    Reviewed
  3. \[ \int\limits_{0}^{1} \pi \left(- x^{4} + x^{2}\right)\, dx = \frac{2 \pi}{15} \]
    V = ∫ A(x) dx.✓ Proved
Answer \( \frac{2 \pi}{15} \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 5.18

easy
The region between \( \displaystyle y = 2 x \) and \( \displaystyle y = x^{2} \) is revolved about the x-axis. Find the volume.
  1. \[ - x \left(x - 2\right) = - x^{2} + 2 x \]
    The curves meet at x = 0 and x = 2.✓ Proved
  2. Cross sections are washers: outer radius 2*x, inner radius x**2.
    Reviewed
  3. \[ \int\limits_{0}^{2} \pi \left(- x^{4} + 4 x^{2}\right)\, dx = \frac{64 \pi}{15} \]
    V = ∫ A(x) dx.✓ Proved
Answer \( \frac{64 \pi}{15} \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 5.16

easy
The region under \( \displaystyle y = \frac{1}{x} \) from \( \displaystyle x = 1 \) to \( \displaystyle x = 3 \) is revolved about the x-axis. Find the volume.
  1. Cross sections perpendicular to the x-axis are disks of radius f(x), with area πf(x)².
    Reviewed
  2. \[ \int\limits_{1}^{3} \frac{\pi}{x^{2}}\, dx = \frac{2 \pi}{3} \]
    V = ∫ A(x) dx.✓ Proved
Answer \( \frac{2 \pi}{3} \)

✓ Nihil obstat Lines: 1 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 5.19

easy
The region under \( \displaystyle y = \sqrt{x} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 4 \) is revolved about the x-axis. Find the volume.
  1. Cross sections perpendicular to the x-axis are disks of radius f(x), with area πf(x)².
    Reviewed
  2. \[ \int\limits_{0}^{4} \pi x\, dx = 8 \pi \]
    V = ∫ A(x) dx.✓ Proved
Answer \( 8 \pi \)

✓ Nihil obstat Lines: 1 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 5.21

easy
The region under \( \displaystyle y = x \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 4 \) is revolved about the x-axis. Find the volume.
  1. Cross sections perpendicular to the x-axis are disks of radius f(x), with area πf(x)².
    Reviewed
  2. \[ \int\limits_{0}^{4} \pi x^{2}\, dx = \frac{64 \pi}{3} \]
    V = ∫ A(x) dx.✓ Proved
Answer \( \frac{64 \pi}{3} \)

✓ Nihil obstat Lines: 1 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 5.22

easy
The region under \( \displaystyle y = x^{2} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 1 \) is revolved about the x-axis. Find the volume.
  1. Cross sections perpendicular to the x-axis are disks of radius f(x), with area πf(x)².
    Reviewed
  2. \[ \int\limits_{0}^{1} \pi x^{4}\, dx = \frac{\pi}{5} \]
    V = ∫ A(x) dx.✓ Proved
Answer \( \frac{\pi}{5} \)

✓ Nihil obstat Lines: 1 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 5.23

easy
The region under \( \displaystyle y = \sin{\left(x \right)} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = \pi \) is revolved about the x-axis. Find the volume.
  1. Cross sections perpendicular to the x-axis are disks of radius f(x), with area πf(x)².
    Reviewed
  2. \[ \int\limits_{0}^{\pi} \pi \sin^{2}{\left(x \right)}\, dx = \frac{\pi^{2}}{2} \]
    V = ∫ A(x) dx.✓ Proved
Answer \( \frac{\pi^{2}}{2} \)

✓ Nihil obstat Lines: 1 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 5.25

easy
The region under \( \displaystyle y = e^{x} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 4 \) is revolved about the x-axis. Find the volume.
  1. Cross sections perpendicular to the x-axis are disks of radius f(x), with area πf(x)².
    Reviewed
  2. \[ \int\limits_{0}^{4} \pi e^{2 x}\, dx = - \frac{\pi \left(1 - e^{8}\right)}{2} \]
    V = ∫ A(x) dx.✓ Proved
Answer \( - \frac{\pi \left(1 - e^{8}\right)}{2} \)

✓ Nihil obstat Lines: 1 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More volumes by disks and washers practice

The region under \( \displaystyle y = \frac{1}{x} \) from \( \displaystyle x = 1 \) to \( \displaystyle x = 4 \) is revolved about the x-axis. Find the volume.
Problem 5.20easy✓ Nihil obstat
The region under \( \displaystyle y = \sqrt{x} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 2 \) is revolved about the x-axis. Find the volume.
Problem 5.24easy✓ Nihil obstat
The region under \( \displaystyle y = x^{2} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 4 \) is revolved about the x-axis. Find the volume.
Problem 5.26easy✓ Nihil obstat
The region under \( \displaystyle y = x^{2} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 2 \) is revolved about the x-axis. Find the volume.
Problem 5.27easy✓ Nihil obstat
The region under \( \displaystyle y = e^{x} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 1 \) is revolved about the x-axis. Find the volume.
Problem 5.28easy✓ Nihil obstat
The region under \( \displaystyle y = x \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 2 \) is revolved about the x-axis. Find the volume.
Problem 5.29easy✓ Nihil obstat
The region under \( \displaystyle y = \frac{1}{x} \) from \( \displaystyle x = 1 \) to \( \displaystyle x = 2 \) is revolved about the x-axis. Find the volume.
Problem 5.30easy✓ Nihil obstat
The region under \( \displaystyle y = x \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 1 \) is revolved about the x-axis. Find the volume.
Problem 5.76easy✓ Nihil obstat
The region under \( \displaystyle y = \sqrt{x} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 1 \) is revolved about the x-axis. Find the volume.
Problem 5.77easy✓ Nihil obstat
The region under \( \displaystyle y = e^{x} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 2 \) is revolved about the x-axis. Find the volume.
Problem 5.93easy✓ Nihil obstat