AP Calculus AB and BC

Solid of Revolution Builder: Disk, Washer, and Shell

A solid of revolution is the shape made by spinning a flat region around an axis; its volume is a definite integral over thin slices: disks (pi r squared) for one curve on the axis, washers for a gap between two curves, shells for strips parallel to the axis. Drag the bounds and watch each integral assemble.

Method

Disk, washer, and shell are not interchangeable here; this region calls for the disk.

The region and the axis of revolution

The solid as stacked disks

Disk method

V = \pi\int_a^b \big[R(x)\big]^2\,dx

V = \pi\int_{0}^{4}\left(\sqrt{x}\,\right)^{2}dx = \pi\int_{0}^{4} x\,dx

Exact volume of the full region

V = 8\pi \approx 25.1327

lower bound a

0.00

upper bound b

4.00

volume (exact)

25.1327

At the setup's own bounds the numeric integral 25.1327 matches the exact closed form shown above.

Rotating about the x-axis, a slice taken perpendicular to the axis meets the single curve y = sqrt(x) and sweeps a full disk of radius sqrt(x). One curve and no gap, so the disk method fits: the area of each face is pi times the radius squared.

A solid of revolution is what you get when you take a flat region in the plane and spin it a full turn around a straight line, the axis of revolution. The region sweeps out a 3D shape, and the question is almost always the same: what is its volume? The move that answers it is the one behind every integral. Slice the solid into thin pieces, write the volume of one piece, and add the pieces up with a definite integral (Topics 8.9 through 8.12).

Where you take the slice decides the method. Cut perpendicular to the axis and each slice is a flat coin whose face is a full circle or a ring. Cut parallel to the axis and each slice is a thin cylindrical tube. Those two choices give the three formulas the builder assembles for you.

Method	Use it when	Slice area	Volume
Disk	one curve, the slice reaches the axis	$\pi\,[R(x)]^2$	$\pi\int_a^b [R(x)]^2\,dx$
Washer	two curves leave a hole at the axis	$\pi\left([R]^2-[r]^2\right)$	$\pi\int_a^b \left([R(x)]^2-[r(x)]^2\right)\,dx$
Shell	strip runs parallel to the axis	$2\pi\,x\,h(x)$	$2\pi\int_a^b x\,h(x)\,dx$

What the AP exam actually requires

The CED builds volumes of revolution from the disk and washer methods only (Topics 8.9 through 8.12). The shell method is a standard technique many courses add for rotations about an axis parallel to the slices, but the CED does not test it. The builder includes shell so you can see the cleaner setup, and it names, on every region, the reason the other two methods do not fit. Choosing the method is the whole skill; the arithmetic is the easy part.

How to read it

Drag the two bound handles that sit on the x-axis (or press the arrow keys, holding Shift for fine steps) to change the interval the region sweeps over.

Only one method button lights up for each region. A single curve above the axis enables Disk, a second curve that carves a gap enables Washer, and a strip parallel to the axis enables Shell. Hover a greyed-out button to read exactly why that method does not fit this region.
Drag a bound off the region's own limits and the volume readout relabels itself from exact to numeric. At the full interval it prints the closed form, for example $8\pi \approx 25.13$ for $y=\sqrt{x}$ ; anywhere else it prints a composite-Simpson estimate for the sub-interval you dragged to.
On the washer region ( $y=x$ over $y=x^2$ ) the middle panel carves a real hole down the axis, and the integrand is $x^2 - x^4$ , the difference of the squared radii, not $(x-x^2)^2$ . Squaring the gap is the single most common washer error.
Switch to the parabola-about- $y$ setup and the axis turns vertical: the solid redraws as nested cylindrical shells of radius $x$ and height $x^2$ , and the entrance sweep replays each time you change regions.
The lower handle $a$ can never pass the upper handle $b$ ; a minimum gap is held so the interval stays valid no matter how far you drag.
Hold Shift while pressing an arrow key to nudge a bound by $0.01$ instead of $0.1$ , so you can land the handle back onto the region's own limit and watch the numeric volume snap onto the exact value. The sine region's upper limit $\pi$ is the exception: it does not fall on the grid, so use the 'Reset bounds to the full region' button to return to the exact label there.

When the region is bounded by a single curve and the axis itself, every slice taken perpendicular to the axis is a full disk. Its radius $R(x)$ is the distance from the axis out to the curve, and its face has area $\pi[R(x)]^2$ . The first setup, $y=\sqrt{x}$ rotated about the x-axis, is the clean case: the radius is just $\sqrt{x}$ , and squaring it collapses the integrand to $x$ .

V = \pi\int_0^4 \left(\sqrt{x}\right)^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi

The builder prints this as $V = 8\pi \approx 25.1327$ and stacks eleven of these disks along the axis so you can see the coin shape before the integral fuses them into one number. The fourth setup, $y=\sin x$ on $[0,\pi]$ , is still a disk problem, but the slice area $\pi\sin^2 x$ needs a trig identity before it will integrate.

V = \pi\int_0^{\pi} \sin^2 x\,dx = \pi\int_0^{\pi}\frac{1-\cos 2x}{2}\,dx = \frac{\pi}{2}\left[x-\frac{\sin 2x}{2}\right]_0^{\pi} = \frac{\pi^2}{2}

That lands on $\frac{\pi^2}{2} \approx 4.9348$ , exactly the value the readout shows at the full bounds. Rewriting $\sin^2 x$ with the power-reduction identity is the same step you would reach for on a free-response volume question.

When a second curve runs between the region and the axis, the slice is no longer solid; it is a ring, or washer. The outer radius $R(x)$ reaches the far curve and the inner radius $r(x)$ reaches the near one, so the ring's area is the big disk minus the hole. On the second setup, $y=x$ stays above $y=x^2$ across $[0,1]$ , so the line gives the outer radius and the parabola gives the inner radius.

V = \pi\int_0^1 \left(x^2 - \left(x^2\right)^2\right)\,dx = \pi\int_0^1 \left(x^2 - x^4\right)\,dx = \pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1 = \frac{2\pi}{15}

Subtract the squares, not the gap

The integrand is $[R]^2-[r]^2 = x^2 - x^4$ , never $(R-r)^2 = (x-x^2)^2$ . Squaring the difference of the two radii is the most common washer mistake, and it gives a different, wrong volume. Square each radius on its own first, then subtract. The builder bakes $x^2 - x^4$ into the shown integral so you can check your own setup against it and confirm $\frac{2\pi}{15} \approx 0.4189$ .

Rotating the region under $y=x^2$ about the y-axis is the awkward case for disks: a slice perpendicular to the y-axis would force you to rewrite the curve as $x=\sqrt{y}$ and integrate in $y$ . Slicing parallel to the axis avoids that. A thin vertical strip at position $x$ sweeps a cylindrical shell of radius $x$ and height $h(x)=x^2$ ; unrolled, the shell is a flat sheet of area $2\pi x \cdot h(x)$ .

V = 2\pi\int_0^2 x\cdot x^2\,dx = 2\pi\int_0^2 x^3\,dx = 2\pi\left[\frac{x^4}{4}\right]_0^2 = 8\pi

The shell method keeps $x$ as the variable and skips the inversion, which is why the builder marks it as the fitting method here. Remember, though, that the AP CED requires only disk and washer (Topics 8.9 through 8.12). On the exam this same region is handled as a washer integrated in $y$ , with outer radius $2$ and inner radius $\sqrt{y}$ :

V = \pi\int_0^4 \left(2^2 - \left(\sqrt{y}\right)^2\right)\,dy = \pi\int_0^4 (4-y)\,dy = \pi\left[4y-\frac{y^2}{2}\right]_0^4 = 8\pi

Both routes land on $8\pi$ , which is the sanity check worth internalizing: the volume of a solid does not depend on how you choose to slice it. Shell or washer, the number is the same; the methods only differ in how much algebra they cost.

The bottom panel assembles the integral in stages. First the generic skeleton for the method, then the same integral with this region's curves and bounds filled in, then the exact closed form beside its decimal. Below that, three readouts track the current lower bound $a$ , upper bound $b$ , and the volume.

At the region's own limits the volume is labeled exact and equals the closed form. Drag a bound anywhere else and the label flips to numeric: the builder switches to a composite-Simpson estimate of the volume over the sub-interval you chose, because no closed form is stored for an arbitrary interval. The closed form above keeps reporting the full-region value in its own limits, so the two are never confused.

Why the two numbers agree at the full bounds

Composite Simpson matches the exact value to the displayed precision on these polynomial and sine integrands, so when you drag the handles back onto the region's own limits the numeric estimate snaps onto the closed form. That agreement is the tool being honest about what it computes versus what it looks up: the decimal is always a real integral of the shown integrand, not a cached answer.