How to calculate the volume of a solid of revolution using disks and washers?

Here is a clear decision framework. **Disk vs. Washer:** - **Disk:** Use when the solid has no hole — the region being rotated touches the axis of rotation along its entire length. - **Washer:** Use when the solid has a hole — there is a gap between the region and the axis of rotation, creating an outer radius $R(x)$ and an inner radius $r(x)$. **$dx$ vs. $dy$:** - **$dx$ (perpendicular to $x$-axis):** Use when the region is bounded by functions of $x$ and you rotate around a horizontal axis. - **$dy$ (perpendicular to $y$-axis):** Use when the region is bounded by functions of $y$ and you rotate around a vertical axis. **Example: $y = \sqrt{x}$, $y = 0$, $x = 4$ rotated about the $x$-axis.** Since we rotate around the $x$-axis and the region is described by functions of $x$, we integrate with respect to $x$. The region is bounded above by $y = \sqrt{x}$ and below by $y = 0$ (the $x$-axis). The solid touches the axis, so we use the **disk method**. Radius: $R(x) = \sqrt{x} - 0 = \sqrt{x}$ Limits: $x = 0$ to $x = 4$ \begin{align*} V &= \pi \int_0^4 (\sqrt{x})^2 \, dx \\ &= \pi \int_0^4 x \, dx \\ &= \pi \left[\frac{x^2}{2}\ ight]_0^4 \\ &= \pi \cdot \frac{16}{2} = 8\pi \end{align*} **Second example: Same region rotated about the $y$-axis.** Now we integrate with respect to $y$. The region is bounded by $x = y^2$ (from $y = \sqrt{x}$), $x = 4$, and $y = 0$. Using the washer method with outer radius $R(y) = 4$ and inner radius $r(y) = y^2$: \begin{align*} V &= \pi \int_0^2 [(4)^2 - (y^2)^2] \, dy \\ &= \pi \int_0^2 (16 - y^4) \, dy \\ &= \pi \left[16y - \frac{y^5}{5}\ ight]_0^2 \\ &= \pi \left(32 - \frac{32}{5}\ ight) = \frac{128\pi}{5} \end{align*}