# Volumes of Revolution Practice Problems with Answers > 11 volume of revolution practice problems with answers and full solutions using the disk, washer and shell methods, including other axes. Source: https://practicecalculus.com/practice-problems/volumes-of-revolution Updated: 2026-11-06 $$ V = \pi\int_a^b \big(R(x)^2 - r(x)^2\big)\,dx \qquad V = 2\pi\int_a^b x\,h(x)\,dx $$ ## At a glance - **Problems:** 11, with answers and full solutions - **Level:** Calculus 1, Calculus 2 and AP Calculus, Unit 8 - **You need:** [Area between curves](/practice-problems/area-between-curves) and [definite integrals](/practice-problems/definite-integrals) - **Main idea:** Slice the solid into disks, washers or shells, find the volume of one slice, and add them up - **Practice with new numbers:** [Volumes with disks lesson](/learn/calculus-1/applications-of-integrals/volumes-disks) ## Key takeaways - Disk: $\pi\int R^2$. Washer: $\pi\int(R^2 - r^2)$. Shell: $2\pi\int(\text{radius})(\text{height})$. - A radius is a distance from the axis. For an axis like $y = -1$, add or subtract to get the distance. - Square each radius separately: $R^2 - r^2$, never $(R - r)^2$. These **volumes of revolution practice problems** cover the disk, washer and shell methods, rotation around the $y$-axis, and rotation around lines that are not axes. Try each one first. The answers come first so you can check quickly, then the full solutions. The three formulas, for rotation around the $x$-axis or the $y$-axis: $$\text{Disk: } V = \pi\int_a^b R^2\,dx \qquad \text{Washer: } V = \pi\int_a^b (R^2 - r^2)\,dx \qquad \text{Shell: } V = 2\pi\int_a^b x\,h(x)\,dx$$ For pictures of each slice, read the [volumes of revolution guide](/blog/volumes-of-revolution). ## Problems **Disks around the x-axis** 1. $y = \sqrt{x}$, $0 \le x \le 4$, around the $x$-axis 2. $y = x^2$, $0 \le x \le 1$, around the $x$-axis 3. $y = \sqrt{9 - x^2}$, $-3 \le x \le 3$, around the $x$-axis 4. $y = \sin x$, $0 \le x \le \pi$, around the $x$-axis **Washers** 5. The region between $y = x$ and $y = x^2$, around the $x$-axis 6. The region between $y = \sqrt{x}$ and $y = x$, around the $x$-axis **Around the y-axis** 7. The region between $y = x^2$, $y = 4$ and the $y$-axis ($x \ge 0$), around the $y$-axis, with disks 8. The region under $y = x - x^2$ on $[0, 1]$, around the $y$-axis, with shells 9. The region under $y = x^2$ on $[0, 2]$, around the $y$-axis, with shells **Other axes** 10. The region between $y = x$ and $y = x^2$, around the line $y = -1$ 11. The region between $y = x^2$ and $y = 1$, around the line $y = 1$ ## Answers | Problem | Answer | | --- | --- | | 1 | $8\pi$ | | 2 | $\frac{\pi}{5}$ | | 3 | $36\pi$ | | 4 | $\frac{\pi^2}{2}$ | | 5 | $\frac{2\pi}{15}$ | | 6 | $\frac{\pi}{6}$ | | 7 | $8\pi$ | | 8 | $\frac{\pi}{6}$ | | 9 | $8\pi$ | | 10 | $\frac{7\pi}{15}$ | | 11 | $\frac{16\pi}{15}$ | ## Step-by-step solutions **1.** Each slice is a disk of radius $\sqrt{x}$: $$\pi\int_0^4 x\,dx = \pi\cdot 8 = 8\pi$$ **2.** The radius is $x^2$, so $R^2 = x^4$: $$\pi\int_0^1 x^4\,dx = \frac{\pi}{5}$$ **3.** The radius squared is $9 - x^2$: $$\pi\int_{-3}^{3}(9 - x^2)\,dx = \pi(54 - 18) = 36\pi$$ This is a sphere of radius $3$, and $\frac{4}{3}\pi(3)^3 = 36\pi$. Good check. **4.** Use $\sin^2 x = \frac{1 - \cos 2x}{2}$: $$\pi\int_0^{\pi}\sin^2 x\,dx = \pi\cdot\frac{\pi}{2} = \frac{\pi^2}{2}$$ **5.** On $[0, 1]$, the line $y = x$ is farther from the axis, so $R = x$ and $r = x^2$: $$\pi\int_0^1(x^2 - x^4)\,dx = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}$$ **6.** On $[0, 1]$, $\sqrt{x} \ge x$, so $R = \sqrt{x}$ and $r = x$: $$\pi\int_0^1(x - x^2)\,dx = \pi\left(\frac{1}{2} - \frac{1}{3}\right) = \frac{\pi}{6}$$ **7.** Slice horizontally. At height $y$, the radius is $x = \sqrt{y}$, so $R^2 = y$: $$\pi\int_0^4 y\,dy = 8\pi$$ **8.** A shell at $x$ has radius $x$ and height $x - x^2$: $$2\pi\int_0^1 x(x - x^2)\,dx = 2\pi\left(\frac{1}{3} - \frac{1}{4}\right) = \frac{\pi}{6}$$ **9.** A shell at $x$ has radius $x$ and height $x^2$: $$2\pi\int_0^2 x^3\,dx = 2\pi\cdot 4 = 8\pi$$ Problems 7 and 9 describe different solids that happen to have the same volume. Problem 7 is the solid that fills the paraboloid bowl; problem 9 is the solid under the bowl, between the bowl and the cylinder $x = 2$. Together they fill a cylinder of radius $2$ and height $4$, with volume $16\pi$. **10.** Measure from $y = -1$. The outer radius is $x + 1$ and the inner radius is $x^2 + 1$: $$\pi\int_0^1\big[(x + 1)^2 - (x^2 + 1)^2\big]dx = \pi\int_0^1(2x - x^2 - x^4)\,dx = \pi\left(1 - \frac{1}{3} - \frac{1}{5}\right) = \frac{7\pi}{15}$$ **11.** The region touches the line $y = 1$, so use disks. The radius is $1 - x^2$, and the curves meet at $x = \pm 1$: $$\pi\int_{-1}^{1}(1 - x^2)^2\,dx = 2\pi\int_0^1(1 - 2x^2 + x^4)\,dx = 2\pi\cdot\frac{8}{15} = \frac{16\pi}{15}$$ ## Common mistakes on these problems - **Writing $(R - r)^2$.** In problem 5, $\pi\int(x - x^2)^2\,dx$ gives $\frac{\pi}{30}$, not $\frac{2\pi}{15}$. Square each radius separately. - **Measuring from the wrong axis.** In problem 10, the radii are $x + 1$ and $x^2 + 1$, not $x$ and $x^2$. - **Forgetting the $\pi$ or the $2\pi$.** The disk and washer formulas start with $\pi$. The shell formula starts with $2\pi$. - **Mixing variables.** Horizontal slices need everything in $y$, including the limits. ## Keep practicing For unlimited solids with instant feedback, try the [disks lesson](/learn/calculus-1/applications-of-integrals/volumes-disks), the [washers lesson](/learn/calculus-1/applications-of-integrals/volumes-washers) and the [cross sections lesson](/learn/calculus-1/applications-of-integrals/volumes-cross-sections). If a region runs off to infinity, the [improper integral problems](/practice-problems/improper-integrals) show how to handle it. ## Frequently asked questions ### How do you choose between the washer method and the shell method? Both give the same volume. Slices perpendicular to the axis give disks or washers; slices parallel to it give shells. Pick the one whose functions and limits are easier. For a region described in $x$ rotated about the $y$-axis, shells usually avoid solving for $x$. ### What is the difference between a disk and a washer? A disk is a solid slice with no hole, used when the region touches the axis. A washer has a hole, used when there is a gap between the region and the axis. The washer volume is $\pi(R^2 - r^2)$ times the thickness. ### How do you rotate around a line that is not an axis? Measure every radius from that line. For rotation around $y = -1$, a curve $y = f(x)$ above the line has radius $f(x) + 1$. Problem 10 shows this.