Volume of Revolution — Disk Method

When you rotate a 2D region around an axis, you get a 3D solid. The disk method slices the solid into thin disks perpendicular to the axis of rotation. Each disk has radius equal to the function value at that $x$, so its volume is $\pi r^{2}\,dx = \pi [f(x)]^{2}\,dx$. Integrating gives the total volume.

The Disk Method Formula

For a curve $y = f(x)$ on $[a, b]$ rotated around the $x$-axis:

$V = \int_{a}^{b} \pi\,[f(x)]^{2}\,dx$

The integrand is the area of a single disk at position $x$, and $dx$ is its infinitesimal thickness.

Step-by-Step Solution

Given: $f(x) = \sqrt{x}$, rotate around the $x$-axis from $x = 0$ to $x = 4$.

Find: The volume of the solid of revolution.

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Step 1 — Identify the disk radius.

At a given $x$, the curve $y = \sqrt{x}$ is at height $\sqrt{x}$ above the axis. When you rotate this point around the $x$-axis, it traces a circle of radius $r = \sqrt{x}$.

Step 2 — Write the disk volume.

$dV = \pi r^{2}\,dx = \pi (\sqrt{x})^{2}\,dx = \pi x\,dx$

Step 3 — Set up the integral.

$V = \int_{0}^{4} \pi x\,dx$

Step 4 — Pull the constant $\pi$ outside.

$V = \pi\int_{0}^{4} x\,dx$

Step 5 — Find the antiderivative of $x$.

$\int x\,dx = \dfrac{x^{2}}{2} + C$

Step 6 — Apply the Fundamental Theorem.

$V = \pi \cdot \dfrac{x^{2}}{2}\bigg|_{0}^{4} = \pi\left(\dfrac{16}{2} - 0\right) = \pi \cdot 8 = 8\pi$

Step 7 — Decimal value.

$V = 8\pi \approx 8 \times 3.14159 \approx 25.133\;\text{cubic units}$

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Answer: The volume of the solid of revolution is

$\boxed{\;V = 8\pi \approx 25.13\;\text{cubic units}\;}$

The solid looks like a "horn" or paraboloid that grows from a point at the origin to a circular face of radius 2 at $x = 4$. Its cross-section at any $x$ is a disk of radius $\sqrt{x}$.

Try It

• Adjust the right bound widget to see the volume grow as you include more of the curve.
• Toggle show 3D outline to visualize the rotated solid as a series of stacked disks rendered in perspective.
• The integrand $\pi x$ grows linearly, so the volume grows as $\pi x^{2}/2$ — quadratically with the bound. Doubling the bound gives 4× the volume.