Topic 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis: find the volume of a solid of revolution about a coordinate axis using the disc method.

What this topic is asking

The College Board (Topic 8.9) finds the volume of a solid of revolution about the $x$- or $y$-axis using the disc method. When a region touching the axis is revolved, the cross sections are discs, and you integrate $\pi$ times the radius squared.

The disc method

A worked disc volume

Choosing the variable of integration

The axis of revolution sets the variable. Revolving about the $x$-axis makes discs stacked along $x$, so integrate in $x$ with the radius as the function's height $y = R(x)$. Revolving about the $y$-axis makes discs stacked along $y$, so integrate in $y$, which requires writing the curve as $x = R(y)$ and using $y$-limits. A frequent error is revolving about the $y$-axis but integrating in $x$, or failing to solve the curve for $x$ in terms of $y$. Match the integration variable to the axis: discs are perpendicular to the axis of revolution.

Why the radius is squared with a pi

Each disc is a thin cylinder whose circular face has area $\pi R^2$; its volume is that area times the thickness. Summing the discs gives $\pi\int R^2$. The squaring means any error in the radius is amplified, so identifying $R$ correctly, as the distance from the axis of revolution to the curve, is essential. When the region borders the axis directly there is no inner radius to subtract, which is what distinguishes the disc method from the washer method (where a gap between the region and the axis creates a hole). Confirm the region touches the axis before using a single disc; otherwise you need a washer.

The disc method as a cross-section method

The disc method is a special case of the general volume-by-cross-section formula $V = \int A\,dx$: revolving a region about an axis it touches produces circular cross sections, whose area is $\pi R^2$. Seeing it this way unifies the volume topics: cross-section problems with given shapes (squares, triangles, semicircles) and solids of revolution both come from integrating a cross-sectional area along an axis. The only difference is where the area formula comes from, a given shape versus a circle generated by revolution. This perspective also makes the washer case natural: when revolution leaves a hole, the circular cross section becomes an annulus of area $\pi R^2 - \pi r^2$.

Keeping the radius positive and well-defined

The radius in a disc integral is a distance, so it is non-negative, and squaring it removes any sign concern from the integrand. What matters is that $R$ is the correct distance from the axis to the boundary curve at each value of the integration variable. When revolving about the $x$-axis, $R(x)$ is the function's height $y = f(x)$; when revolving about the $y$-axis, $R(y)$ is the curve solved as $x = g(y)$. Writing the radius explicitly as this distance, then squaring inside the integral, avoids the slip of integrating the function without squaring or of using the wrong variable's expression for the radius.