Volume of solid of revolution using integration — y=x² rotated about x-axis

Question

Find the volume of the solid generated by revolving the area bounded by $y = x^2$ , $x = 0$ , and $x = 2$ about the x-axis.

(JEE Advanced 2022, similar pattern)

Solution — Step by Step

When we rotate a curve $y = f(x)$ about the x-axis, each vertical slice at position $x$ creates a circular disc of radius $y = f(x)$ and thickness $dx$ .

Volume of one disc: $dV = \pi y^2 \, dx = \pi [f(x)]^2 \, dx$

Total volume:

V = \int_a^b \pi [f(x)]^2 \, dx

Here $f(x) = x^2$ , $a = 0$ , $b = 2$ :

V = \int_0^2 \pi (x^2)^2 \, dx = \pi \int_0^2 x^4 \, dx

V = \pi \left[\frac{x^5}{5}\right]_0^2 = \pi \left(\frac{32}{5} - 0\right)

\boxed{V = \frac{32\pi}{5} \text{ cubic units}}

Why This Works

The disc method is based on the idea that a solid of revolution can be sliced into infinitely thin circular discs perpendicular to the axis of rotation. Each disc has a known area ( $\pi r^2$ where $r = y$ for rotation about the x-axis), and we sum all such discs from $x = a$ to $x = b$ using integration.

The key insight: rotation about the x-axis means the radius of each disc is the y-coordinate of the curve at that point. Squaring the function and integrating gives the total volume.

Alternative Method — Shell method (rotate about y-axis for comparison)

If the same region were rotated about the y-axis instead, we’d use the shell method:

V = \int_0^2 2\pi x \cdot f(x) \, dx = 2\pi \int_0^2 x \cdot x^2 \, dx = 2\pi \int_0^2 x^3 \, dx

= 2\pi \left[\frac{x^4}{4}\right]_0^2 = 2\pi \cdot 4 = 8\pi

Note how the two volumes are different ( $32\pi/5$ vs $8\pi$ ) because the axis of rotation matters.

For JEE: use the disc method when the axis of rotation is the same axis you’re integrating along (rotate about x-axis, integrate w.r.t. $x$ ). Use the shell method when the axis of rotation is perpendicular to the integration variable. Choosing the right method can simplify the integral dramatically.

Common Mistake

The most common error: forgetting to square the function. Students write $V = \pi \int x^2 \, dx$ instead of $V = \pi \int (x^2)^2 \, dx = \pi \int x^4 \, dx$ . The disc formula has $[f(x)]^2$ — you must square the function, not just write it as-is. The area of a circle is $\pi r^2$ , and the radius is $y = f(x)$ , so the square is essential.

Question

Solution — Step by Step

Why This Works

Alternative Method — Shell method (rotate about y-axis for comparison)

Common Mistake

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