Find the volume of a solid generated by rotating a region about the x-axis or y-axis using the disc method

What this dot point is asking

SCSA wants you to set up and evaluate volumes of revolution by the disc method, choosing the axis, the variable of integration and the limits correctly.

What a solid of revolution is

A solid of revolution is generated by rotating a plane region a full turn about a fixed straight line, the axis of revolution. Because of the rotation, every cross-section perpendicular to the axis is a circle (or, between two curves, an annulus), so the solid is rotationally symmetric. Familiar shapes arise this way: rotating a rectangle gives a cylinder, rotating a right triangle gives a cone, and rotating a semicircle gives a sphere. The disc method computes the volume of any such solid by summing the areas of these circular cross-sections along the axis, which is why the technique reduces to a single definite integral of $\pi r^2$.

The disc method

When a region under $y = f(x)$ is rotated about the $x$-axis, each thin slice becomes a disc of radius $f(x)$ and thickness $dx$, with area $\pi[f(x)]^2$. Summing gives

The radius is the distance from the axis to the curve, which here is just $f(x)$. The squared radius is essential: forgetting to square is the classic error.

The reasoning behind the formula is a slicing argument. Cut the solid into thin slabs perpendicular to the axis of rotation. Each slab is approximately a cylinder (a disc) of radius $f(x)$ and thickness $\Delta x$, so its volume is approximately $\pi[f(x)]^2\,\Delta x$. Adding the slabs and letting $\Delta x \to 0$ turns the sum into the definite integral $\pi\int_a^b [f(x)]^2\,dx$. Seeing the volume as an accumulation of disc areas explains why the integrand is the cross-sectional area $\pi r^2$ and why the variable of integration must run along the axis of rotation.

Rotating about the y-axis

Rotating about the $y$-axis makes the radius the horizontal distance $x$, so you express $x$ as a function of $y$ and integrate over $y$:

The limits $c$ and $d$ are now $y$-values. Match the variable of integration to the axis of rotation.

Rotating the region between two curves

Setting limits

The limits are where the region begins and ends along the axis of rotation, often the intersection points of the bounding curves with each other or with the axis. A sketch confirms them.

Matching the integral to the axis

The single most important set-up decision is matching the variable of integration to the axis of rotation. For rotation about the $x$-axis, the discs are stacked along $x$, so integrate $\pi\int [f(x)]^2\,dx$ with $x$-limits. For rotation about the $y$-axis, the discs are stacked along $y$, so express the radius $x$ as a function of $y$ and integrate $\pi\int [x(y)]^2\,dy$ with $y$-limits. Mixing the two, for instance integrating in $x$ for a $y$-axis rotation, gives a meaningless answer. A clear sketch with a single representative disc drawn perpendicular to the axis makes the correct choice obvious.

Rotation about a line other than an axis

A harder variant rotates the region about a horizontal or vertical line that is not a coordinate axis, such as $y = 1$ or $x = 3$. The method is the same, but the radius becomes the distance from that line to the curve rather than from the axis. Rotating $y = f(x)$ about the line $y = c$ uses a radius of $|f(x) - c|$, so the volume is $\pi\int_a^b [f(x) - c]^2\,dx$. Reading the radius as a distance, rather than just as the function value, is the key adjustment, and it also covers the washer case where an inner and outer distance are each measured from the same line.