Volume of Revolution — Washer Method

When the region you're rotating has a hole in it (because two curves bound it), you can't use plain disks. Instead, each cross-section is a washer — an annular ring with an outer radius and an inner radius. Subtract the inner area from the outer to get the washer's area, then integrate.

The Washer Method Formula

For an outer curve $y = R(x)$ and inner curve $y = r(x)$, both rotated around the $x$-axis:

$V = \int_{a}^{b} \pi\bigl(R(x)^{2} - r(x)^{2}\bigr)\,dx$

The geometry: at each $x$, the disk of radius $R$ has the disk of radius $r$ "punched out" of its center.

Step-by-Step Solution

Given: Region between $y = x$ (outer) and $y = x^{2}$ (inner) on $[0, 1]$, rotated around the $x$-axis.

Find: The volume of the resulting solid (a kind of "lens shell").

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Step 1 — Confirm which curve is outer.

On $(0, 1)$, pick a test point $x = 0.5$:

$f(0.5) = 0.5 \quad\text{vs.}\quad g(0.5) = 0.25$

So $y = x$ is above $y = x^{2}$ on $(0, 1)$. When rotated, $y = x$ becomes the outer radius and $y = x^{2}$ becomes the inner radius.

Step 2 — Confirm intersection points.

$x = x^{2} \;\Longrightarrow\; x(1 - x) = 0 \;\Longrightarrow\; x = 0 \text{ or } x = 1$

Both endpoints — perfect.

Step 3 — Set up the washer integral.

$V = \int_{0}^{1} \pi\bigl((x)^{2} - (x^{2})^{2}\bigr)\,dx = \int_{0}^{1} \pi\bigl(x^{2} - x^{4}\bigr)\,dx$

Step 4 — Pull $\pi$ out and find the antiderivative.

$V = \pi\int_{0}^{1}\bigl(x^{2} - x^{4}\bigr)\,dx = \pi\left[\dfrac{x^{3}}{3} - \dfrac{x^{5}}{5}\right]_{0}^{1}$

Step 5 — Evaluate at the bounds.

At $x = 1$: $\dfrac{1}{3} - \dfrac{1}{5}$. At $x = 0$: $0$.

$V = \pi\left(\dfrac{1}{3} - \dfrac{1}{5}\right) - 0$

Step 6 — Combine the fractions over a common denominator.

$\dfrac{1}{3} - \dfrac{1}{5} = \dfrac{5}{15} - \dfrac{3}{15} = \dfrac{2}{15}$

So:

$V = \pi \cdot \dfrac{2}{15} = \dfrac{2\pi}{15}$

Step 7 — Decimal value.

$V = \dfrac{2\pi}{15} \approx \dfrac{6.2832}{15} \approx 0.4189\;\text{cubic units}$

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Answer: The volume of the washer-shaped solid is

$\boxed{\;V = \dfrac{2\pi}{15} \approx 0.4189\;\text{cubic units}\;}$

The solid is a thin shell — a "shell of a paraboloid" wrapped inside a cone. At any $x$, the cross-section is an annulus with outer radius $x$ and inner radius $x^{2}$.

Try It

• The visualization shows the outer curve ($y = x$) in pink and the inner curve ($y = x^{2}$) in cyan.
• The shaded region between them is what gets rotated to form the solid.
• Toggle show washers to see the cross-sections rendered as rings.
• Notice that at $x = 0$ and $x = 1$, the two curves meet — the washer collapses to a point at each end.