Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)
Published: October 2025
Category: Calculus | Integration Applications

🔹 What Is Volume of Revolution?

When a 2D curve is rotated about an axis, it sweeps out a 3D solid.

For example, if the region under the curve y=f(x)y = f(x)y=f(x) between x=ax = ax=a and x=bx = bx=b is rotated about the x-axis, it forms a solid of revolution — a concept central to applications of integration.

This technique allows us to calculate the exact volume of such solids by summing infinitesimal cylindrical slices using calculus.

🔹 The Core Formula

🔸 When Rotated About the x-Axis:

V=π∫ab[f(x)]2 dxV = \pi \int_a^b [f(x)]^2 \, dxV=π∫ab​[f(x)]2dx

🔸 When Rotated About the y-Axis:

If x=g(y)x = g(y)x=g(y), then V=π∫cd[g(y)]2 dyV = \pi \int_c^d [g(y)]^2 \, dyV=π∫cd​[g(y)]2dy

This works because each slice forms a thin disk (or washer) with radius = function value, and thickness = infinitesimal change in the axis variable.

🔹 Step-by-Step Example 1 — Simple Parabola

Find the volume generated when the region under y=x2y = x^2y=x2 from x=0x = 0x=0 to x=2x = 2x=2 is revolved about the x-axis. V=π∫02(x2)2 dx=π∫02×4 dxV = \pi \int_0^2 (x^2)^2 \, dx = \pi \int_0^2 x^4 \, dxV=π∫02​(x2)2dx=π∫02​x4dx V=π[x55]02=32π5V = \pi \left[\frac{x^5}{5}\right]_0^2 = \frac{32\pi}{5}V=π[5×5​]02​=532π​

✅ Volume = 32π5 units3\frac{32\pi}{5} \, \text{units}^3532π​units3

🔹 Step-by-Step Example 2 — About the y-Axis

Find the volume formed by rotating y=xy = \sqrt{x}y=x​ from x=0x = 0x=0 to x=4x = 4x=4 about the y-axis.

First, express xxx in terms of yyy: x=y2x = y^2x=y2

When x=0⇒y=0x = 0 \Rightarrow y = 0x=0⇒y=0; when x=4⇒y=2x = 4 \Rightarrow y = 2x=4⇒y=2. V=π∫02(y2)2 dy=π∫02y4 dy=π[y55]02=32π5V = \pi \int_0^2 (y^2)^2 \, dy = \pi \int_0^2 y^4 \, dy = \pi \left[\frac{y^5}{5}\right]_0^2 = \frac{32\pi}{5}V=π∫02​(y2)2dy=π∫02​y4dy=π[5y5​]02​=532π​

✅ The same numerical result, but about a different axis — a beautiful symmetry in calculus!

🔹 Example 3 — Washer Method (Hollow Solids)

If the region between two curves y=f(x)y = f(x)y=f(x) and y=g(x)y = g(x)y=g(x) is revolved about the x-axis, the volume is: V=π∫ab([f(x)]2−[g(x)]2)dxV = \pi \int_a^b \left( [f(x)]^2 – [g(x)]^2 \right) dxV=π∫ab​([f(x)]2−[g(x)]2)dx

This accounts for the outer and inner radii of the solid — creating a hollow shape, much like a tube or shell.

🔹 Common Mistakes to Avoid

1. ❌ Forgetting to square the radius function ([f(x)]2[f(x)]^2[f(x)]2).
2. ❌ Mixing up axes — always check if it’s x-axis or y-axis revolution.
3. ❌ Using wrong limits when changing variables (e.g., x↔yx \leftrightarrow yx↔y).
4. ❌ Ignoring negative function values — use [f(x)]2[f(x)]^2[f(x)]2 to ensure positive radii.

🔹 Real-World Applications

• Engineering: volume of tanks, pipes, and domes
• Physics: center of mass and rotational motion
• Architecture: designing curved surfaces (e.g., domes and towers)
• Exams: IB Math HL Paper 3, AP Calculus BC, A Level P3, STEP, MAT

🔹 Advanced Challenge (for STEP / A Level P3)

Find the volume of revolution of the region bounded by y=e−xy = e^{-x}y=e−x, the x-axis, and x=0x = 0x=0 to x=1x = 1x=1 about the x-axis. V=π∫01(e−x)2dx=π∫01e−2xdx=π2(1−e−2)V = \pi \int_0^1 (e^{-x})^2 dx = \pi \int_0^1 e^{-2x} dx = \frac{\pi}{2}(1 – e^{-2})V=π∫01​(e−x)2dx=π∫01​e−2xdx=2π​(1−e−2)

✅ Volume = π2(1−e−2)\frac{\pi}{2}(1 – e^{-2})2π​(1−e−2)

🌟 Why Volume of Revolution Matters

This topic beautifully unites geometry, algebra, and calculus — converting 2D ideas into three-dimensional understanding.

Students who master this develop strong intuition for:

• Integration as “accumulation”
• Modeling real-world physical shapes
• Visualizing calculus beyond the page

📘 Go Beyond Formula Application

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In the Mathematics Elevate Mentorship Program, you’ll:
✅ Visualize 3D solids through real applications,
✅ Solve advanced integration problems intuitively,
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