Disc Method About Other Lines — AP® Calculus AB Notes

Same discs, a different center

In the last topic every solid spun around the $x$ -axis or the $y$ -axis. Nothing in the disc method needs the axis of rotation to be one of the coordinate axes. A region can be revolved about the line $y = 1$ , or $y = -2$ , or $x = 3$ . The slices perpendicular to that line are still discs, as long as the region touches the line all the way across. What changes is the radius.

The radius of a disc is the distance from its center, on the axis, to its edge, on the curve. When the axis was the $x$ -axis, that distance happened to be $f(x)$ , the curve's height. When the axis is the line $y = k$ , the distance is the gap between the heights $k$ and $f(x)$ :

r = k - f(x) \text{ if the line is above the curve}, \qquad r = f(x) - k \text{ if it is below.}

Always subtract the smaller height from the larger, so the radius is positive. (Since the radius is squared, a sign slip does not change the volume, but it is a sign you have lost track of the picture.)

Left: the region between the parabola y = x² and the line y = 1, from x = −1 to 1, shaded, with y = 1 dashed as the axis of rotation and one thin vertical rectangle from the parabola up to the axis, of length 1 − x². Right: the solid made by revolving the region about y = 1, shaped like a lemon, with the axis through its center and one disc highlighted in blue whose radius is r = 1 − x². Below: V = the integral from −1 to 1 of π(1 − x²)² dx = 16π/15. — **Figure 1.** Revolving about $y = 1$ . Each rectangle runs from the parabola up to the axis, so the radius is $1 - x^2$ , not $x^2$ . LevlPrep original diagram.

Worked example. The region between $y = x^2$ and $y = 1$ is revolved about the line $y = 1$ . Find the volume.

Limits. $x^2 = 1$ at $x = \pm 1$ .

Radius. The axis $y = 1$ is above the parabola, so $r = 1 - x^2$ .

Volume.

V = \pi\int_{-1}^{1} \left(1 - x^2\right)^2\,dx = \pi\int_{-1}^{1} \left(1 - 2x^2 + x^4\right)\,dx = \pi\left(2 - \frac{4}{3} + \frac{2}{5}\right) = \frac{16\pi}{15}.

Two ways to get it wrong

Measuring from the wrong place. Using $r = x^2$ in the example measures from the $x$ -axis. That would spin the region under the parabola about the $x$ -axis, a different solid ( $\frac{2\pi}{5}$ ).

Squaring each term. $\pi\int (1^2 - (x^2)^2)\,dx$ is not the same as $\pi\int (1 - x^2)^2\,dx$ . The first gives $\frac{8\pi}{5}$ . It looks tempting because it resembles a formula you will meet next topic, but for a solid disc the radius is one quantity: square the whole difference.

Negative $k$ . About $y = -2$ , the radius to the curve $y = 2 - x^2$ is $(2 - x^2) - (-2) = 4 - x^2$ . Write the subtraction with parentheses before simplifying.

Vertical lines

When the axis is a vertical line $x = h$ , the slices perpendicular to it are horizontal, stacked in $y$ . Write the curve as $x$ in terms of $y$ ; the radius is the horizontal gap between $h$ and $x(y)$ .

Worked example. The region bounded by $y = x$ , $y = 0$ and $x = 3$ is revolved about the line $x = 3$ . Find the volume.

Slices. Horizontal, from $y = 0$ to $y = 3$ . At height $y$ the region runs from the line $x = y$ to the axis $x = 3$ .

Radius. $r = 3 - y$ .

Volume. $V$ $= \pi\int_0^3 (3 - y)^2\,dy$ $= \pi\left[-\frac{(3 - y)^3}{3}\right]_0^3$ $= \pi \cdot \frac{27}{3}$ $= 9\pi$ .

The solid is a cone with radius 3 and height 3, and $\frac{1}{3}\pi(3)^2(3) = 9\pi$ agrees.

The radius at a glance

Axis	Where it is	Radius	Integrate in
$y = k$	above the curve $y = f(x)$	$k - f(x)$	$dx$
$y = k$	below the curve	$f(x) - k$	$dx$
$x = h$	right of the curve $x = g(y)$	$h - g(y)$	$dy$
$x = h$	left of the curve	$g(y) - h$	$dy$

A quick check: the radius is 0 where the curve meets the axis, and largest where the curve is farthest from it. If your formula says otherwise, you are measuring from the wrong line.

With a calculator, find any crossing with the axis line numerically, store it, write the setup with the radius in parentheses, for example $\pi\int_0^c \left(4 - e^{x/2}\right)^2\,dx$ , and report three decimal places.