Solid of Revolution: Volumes by Disk/Washer Method

A solid formed by rotating a region (area) around a fixed axis (called the axis of revolution).
The resulting shape has circular cross-sections perpendicular to the axis.

Used when a region is rotated about the x-axis (or a horizontal line) and the region is bounded above and below by functions of $x$.
Formula: For region bounded by $y = f(x)$ (top) and $y = g(x)$ (bottom) from $x=a$ to $x=b$, rotated about x-axis: $$ V = \pi \int_a^b \left( [f(x)]^2 - [g(x)]^2 \right) , dx $$
If only one curve $y=f(x)$ and $y=0$, then $V = \pi \int_a^b [f(x)]^2 , dx$.

Example (Tutorial 1(a)): Region bounded by $y = x$, $y=0$, $x=1$, $x=2$.

$V = \pi \int_1^2 (x)^2 , dx = \pi \left[ \frac{x^3}{3} \right]_1^2 = \frac{7\pi}{3} \text{ unit}^3$.

For a region bounded above by $f(x)$ and below by $g(x)$: $$ A = \int_a^b (f(x) - g(x)) , dx $$

Use functions of $y$: region bounded by $x = f(y)$ and $x = g(y)$ from $y=c$ to $y=d$.
Formula: $$ V = \pi \int_c^d \left( [f(y)]^2 - [g(y)]^2 \right) , dy $$

Example (Tutorial 2): Region bounded by $y = x^2$, $y = 9x^2$ ($x \ge 0$), $y = 1$, rotated about y-axis.

Express curves in terms of $y$: $x = \sqrt{y}$, $x = \sqrt{y/9} = \frac{\sqrt{y}}{3}$.
$V = \pi \int_0^1 \left( (\sqrt{y})^2 - \left(\frac{\sqrt{y}}{3}\right)^2 \right) dy = \pi \int_0^1 \left( y - \frac{y}{9} \right) dy = \pi \left[ \frac{8}{9} \cdot \frac{y^2}{2} \right]_0^1 = \frac{4\pi}{9} \text{ unit}^3$.

Use the Washer Method: When rotating about a horizontal line $y = c$ or vertical line $x = d$, adjust radii as $R(x) = |f(x) - c|$ (or $R(y) = |f(y) - d|$).
For rotation about $x = k$ (vertical line), use functions of $y$: $$ V = \pi \int_c^d \left( (f_{\text{outer}}(y) - k)^2 - (g_{\text{inner}}(y) - k)^2 \right) , dy $$
For rotation about $y = k$ (horizontal line), use functions of $x: $$ V = \pi \int_a^b \left( (k - g(x))^2 - (k - f(x))^2 \right) , dx $$

Example (Tutorial 4(b)(ii)): Region bounded by $y = x^2$ and $y = x$, rotated about $x = 2$.

Intersection points: $x^2 = x$ → $x=0,1$. Functions of $y$: $x = \sqrt{y}$ (right), $x = y$ (left).
Distance from $x=2$: outer radius $= 2 - y$, inner radius $= 2 - \sqrt{y}$.
$V = \pi \int_0^1 \left[ (2 - y)^2 - (2 - \sqrt{y})^2 \right] dy$ Simplify: $(4 - 4y + y^2) - (4 - 4\sqrt{y} + y) = -4y + y^2 + 4\sqrt{y} - y = 4\sqrt{y} - 5y + y^2$.
$V = \pi \left[ \frac{8}{3} y^{3/2} - \frac{5}{2} y^2 + \frac{y^3}{3} \right]_0^1 = \pi \left( \frac{8}{3} - \frac{5}{2} + \frac{1}{3} \right) = \frac{7\pi}{6} \text{ unit}^3$.

Disk Method: One boundary (no hole); Washer Method: Two boundaries (hole in the middle).
Always check the axis: horizontal → integrate with respect to $x$; vertical → integrate with respect to $y$.
For rotation about a line not at origin, shift the curves by subtracting the line’s coordinate.