Calculus 1

The disk method

Spin a flat region about a line and it sweeps out a solid whose every slice across the line is a circle. How do those circles add up to a volume?

You need

Area between curvesCalculus 1 Volume of prisms and cylindersMath 7

Unlocks

Surface area of a solid of revolutionCalculus 2 The washer methodCalculus 1

Where you are 10 in Applications of integrals

Calculus8

Applications of integrals10

▴ 5 above Arc length Average value of a function

The disk method

Surface area of a solid of revolution The washer method

When you reach for it

A solid of revolution is what a flat region sweeps out when it makes one full turn about a line, the axis of revolution. Turned wood is the everyday case: a spindle or a chair leg is cut by spinning a block about its center line while a chisel shapes its outline. Geometry gives the volume when the region is a rectangle, a right triangle or a half-disk. For any other outline, the curve itself has to be integrated.

You reach for the disk method when the region lies against the axis all along its length, so that every slice perpendicular to the axis is a filled circle, a disk. It is the special case of Volumes with known cross sections in which every cross section is a disk, with area A = \pi R^2.

The procedure

Day 1: disks along a coordinate axis

The slice is a disk. Let f be continuous on [a, b], and let the region lie between the graph y = f(x) and the x-axis for a \le x \le b. Turn it about the x-axis and cut the solid with the plane perpendicular to the axis at x. The region meets that plane in the segment from (x, 0) to (x, f(x)), and a segment turned about one of its own endpoints sweeps a disk centered on the axis. Its radius is the distance from the axis to the curve,

R(x) = |f(x)|,

and its area is A(x) = \pi [R(x)]^2 (Figure 1). Since |f(x)|^2 = [f(x)]^2, a region below the axis sweeps the same disks as its mirror image above it.

Figure 1The region of Example 1 and its reflection across the x-axis. Turning the region about the axis sweeps the solid; the slice at x is a disk of radius R(x).

The integral. A solid whose slice at x has area A(x) has volume \int_a^b A(x)\,dx (Volumes with known cross sections). With disk slices this is the disk method:

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

The picture behind it is a stack of thin coins. Cut [a, b] into pieces of width \Delta x. The slab over one piece is nearly a cylinder of radius R(x_i^*) and height \Delta x, with volume \pi [R(x_i^*)]^2\,\Delta x, base area times height as in Volume of prisms and cylinders. Adding the coins gives a Riemann sum, and the integral is its limit as the coins get thinner (The definite integral as a limit of Riemann sums). The constant \pi comes out in front, and what is left is an ordinary definite integral (The second fundamental theorem of calculus), usually after squaring R and expanding.

It agrees with Geometry. A constant radius R(x) = r on [0, h] gives \int_0^h \pi r^2\,dx = \pi r^2 h, the cylinder. Example 2 recovers the cone and the ball.

About the y-axis. If the region lies between a curve x = g(y) and the y-axis for c \le y \le d, turning it about the y-axis gives slices perpendicular to the y-axis, which are horizontal. The slice at height y is a disk of radius R(y) = |g(y)|, and

V = \int_c^d \pi [R(y)]^2\,dy.

The variable of integration is the one that runs along the axis. A boundary given as y = f(x) must first be solved for x, just as for the horizontal strips of Area between curves.

Day 2: other axes, and objects with a profile

A line that is not an axis. The radius is always the distance from the axis of revolution to the curve, measured along the slice. For a horizontal axis y = k and a region between the curve y = f(x) and the line y = k,

R(x) = |f(x) - k|,

which is f(x) - k when the curve is above the line and k - f(x) when it is below. For a vertical axis x = h and a region between a curve x = g(y) and the line x = h,

R(y) = |h - g(y)|.

With k = 0 or h = 0 these are the Day 1 formulas. Moving the axis moves the solid, so one region gives different volumes about different lines.

Choosing the variable. Disks are slices perpendicular to the axis of revolution, so the axis decides the variable: a horizontal axis (y = k, including the x-axis) means integrating in x, and a vertical axis (x = h, including the y-axis) means integrating in y. The limits are where the region begins and ends along the axis. A slice parallel to the axis is not a disk.

A disk needs the region to reach the axis. The slice is a full disk only when the region’s segment in that plane runs all the way to the axis. If a gap separates them, the slice is a disk with a hole, which is the next lesson, the The washer method.

A profile curve. A turned object is a solid of revolution about its center line. If its radius at distance x along the center line is R(x) for 0 \le x \le L, its volume is \pi\int_0^L [R(x)]^2\,dx. When the radius is known only from measured diameters, a Riemann sum of disk volumes estimates it.

Worked examples

Example 1

Day 1: a parabola about the x-axis

Find the volume of the solid formed when the region under y = x^2 + 1 and above the x-axis for 0 \le x \le 2 is revolved about the x-axis.

Solution

The region rests on the x-axis, so the slice at x is a disk of radius R(x) = x^2 + 1 (Figure 1). Squaring, [R(x)]^2 = x^4 + 2x^2 + 1, and

\begin{aligned} V &= \pi\int_0^2 \left(x^4 + 2x^2 + 1\right)dx \\ &= \pi\left(\frac{x^5}{5} + \frac{2x^3}{3} + x\right)\Big|_0^2 \\ &= \pi\left(\frac{32}{5} + \frac{16}{3} + 2\right) \\ &= \frac{206\pi}{15} \approx 43.1. \end{aligned}

A check: R grows from 1 to 5, so the solid fits between a cylinder of radius 1 and one of radius 5, both of length 2. Their volumes are 2\pi \approx 6.3 and 50\pi \approx 157, and the answer lies between them.

Example 2

Day 1: Lujain recovers the cone and the ball

Lujain checks two formulas from Geometry with the disk method: the cone of radius r and height h, and the ball of radius r.

Solution

The cone first. The line \displaystyle y = \frac{r}{h}x runs from the origin to (h, r). The region under it for 0 \le x \le h is a right triangle with a leg of length h on the x-axis and a leg of length r at x = h, and turning it about the x-axis sweeps a cone with its apex at the origin. With \displaystyle R(x) = \frac{r}{h}x,

\begin{aligned} V &= \pi\int_0^h \frac{r^2}{h^2}x^2\,dx \\ &= \pi\frac{r^2}{h^2}\cdot\frac{h^3}{3} = \frac{1}{3}\pi r^2 h. \end{aligned}

Now the ball. The upper half of the circle x^2 + y^2 = r^2 is \displaystyle y = \sqrt{r^2 - x^2} for -r \le x \le r, and the half-disk under it turns into a ball. Squaring removes the root, [R(x)]^2 = r^2 - x^2, and the integrand is even, so

\begin{aligned} V &= 2\pi\int_0^r \left(r^2 - x^2\right)dx \\ &= 2\pi\left(r^3 - \frac{r^3}{3}\right) = \frac{4}{3}\pi r^3. \end{aligned}

Both are the formulas from Geometry, for the cone and the sphere.

Example 3

Day 1: a radius whose square is easier

Find the volume when the region under y = \sec x and above the x-axis for \displaystyle 0 \le x \le \frac{\pi}{3} is revolved about the x-axis.

Solution

On \displaystyle \left[0, \frac{\pi}{3}\right], \cos x > 0, so \sec x is continuous and positive there and R(x) = \sec x. Its square, \sec^2 x, is the derivative of \tan x:

\begin{aligned} V &= \pi\int_0^{\pi/3} \sec^2 x\,dx = \pi\tan x\,\Big|_0^{\pi/3} \\ &= \sqrt{3}\,\pi \approx 5.44. \end{aligned}

The area of the region would need an antiderivative of \sec x itself, which is harder to find; the volume asks only for \sec^2 x.

Example 4

Day 1: Lujain’s paraboloid, about the y-axis

The region in the first quadrant bounded by y = 2x^2, the line y = 8 and the y-axis is revolved about the y-axis. Find the volume, and compare it with the cylinder that just holds the solid.

Solution

The axis is vertical, so the slices are horizontal and the variable is y. The curve meets y = 8 where 2x^2 = 8, at x = 2, so the region runs from y = 0 to y = 8. At height y it stretches from the y-axis out to the curve. Solving y = 2x^2 with x \ge 0 gives \displaystyle x = \sqrt{\frac{y}{2}}, so \displaystyle R(y) = \sqrt{\frac{y}{2}} and \displaystyle [R(y)]^2 = \frac{y}{2}:

V = \pi\int_0^8 \frac{y}{2}\,dy = \pi\cdot\frac{y^2}{4}\,\Big|_0^8 = 16\pi.

The solid sits inside the cylinder of radius 2 and height 8, whose volume is \pi(2)^2(8) = 32\pi, and fills exactly half of it. Slicing in x here would give vertical slices, parallel to the axis, and those are not disks.

Example 5

Day 2: about the line y = -1

The region bounded by \displaystyle y = \frac{1}{x}, the line y = -1, and the lines x = 1 and x = 3 is revolved about the line y = -1. Find the volume.

Solution

The axis is horizontal, so the variable is x, from 1 to 3. The bottom of the region lies on the axis, so each slice is a disk reaching from the axis up to the curve (Figure 2). The curve is above the line, so the radius is the curve’s height minus the line’s height:

R(x) = \frac{1}{x} - (-1) = \frac{1}{x} + 1.

Squaring, \displaystyle [R(x)]^2 = \frac{1}{x^2} + \frac{2}{x} + 1, and

\begin{aligned} V &= \pi\int_1^3 \left(\frac{1}{x^2} + \frac{2}{x} + 1\right)dx \\ &= \pi\left(-\frac{1}{x} + 2\ln x + x\right)\Big|_1^3 \\ &= \pi\left(\frac{2}{3} + 2\ln 3 + 2\right) \\ &= \pi\left(\frac{8}{3} + 2\ln 3\right) \approx 15.3. \end{aligned}

The height \displaystyle \frac{1}{x} above the x-axis is not the radius here; the radius is measured from the axis of revolution, wherever that line is.

Figure 2The region of Example 5 turned about the line y = -1. The radius runs from the axis up to the curve, so it is \displaystyle \frac{1}{x} + 1, not \displaystyle \frac{1}{x}.

Example 6

Day 2: Ammar turns a region about x = 4

Ammar revolves the region bounded by \displaystyle y = \frac{x^2}{4}, the x-axis and the line x = 4 about the line x = 4. Find the volume.

Solution

The axis is vertical, so the slices are horizontal and the variable is y. The curve meets x = 4 at height \displaystyle \frac{16}{4} = 4, so 0 \le y \le 4. At height y the region runs from the curve across to the line x = 4, which is the axis, so the slice is a disk (Figure 3). Solving \displaystyle y = \frac{x^2}{4} with x \ge 0 gives \displaystyle x = 2\sqrt{y}, and the radius is the distance from there to the axis:

R(y) = 4 - 2\sqrt{y}.

Then \displaystyle [R(y)]^2 = 16 - 16\sqrt{y} + 4y, and

\begin{aligned} V &= \pi\int_0^4 \left(16 - 16y^{1/2} + 4y\right)dy \\ &= \pi\left(16y - \frac{32}{3}y^{3/2} + 2y^2\right)\Big|_0^4 \\ &= \pi\left(64 - \frac{256}{3} + 32\right) \\ &= \frac{32\pi}{3} \approx 33.5. \end{aligned}

Slices in x would be vertical, parallel to the axis x = 4, and so not disks.

Figure 3The region of Example 6 turned about the line x = 4. The axis is vertical, so the disks are horizontal and the variable is y.

Example 7

Day 2: Sanad’s spindle

Sanad turns a wooden spindle 20 cm long. At x cm from one end, measured along its center line, its radius is

R(x) = 2 + \frac{x(20 - x)}{50}\ \text{cm}, \qquad 0 \le x \le 20.

Find its volume, its mass if the wood has a density of 0.6 grams per cubic centimeter, and how much wood Sanad cuts away from a cylindrical blank of radius 4 cm and the same length.

Solution

The spindle is the solid swept by the region under the profile y = R(x) turned about its center line. Its radius is 2 at both ends and largest, 4, at x = 10. Centering at the bulge shortens the algebra: substitute u = x - 10, so du = dx, the limits 0 and 20 become -10 and 10 (u-substitution), and x(20 - x) = 100 - u^2. Then \displaystyle R = 4 - \frac{u^2}{50} and

[R]^2 = 16 - \frac{4u^2}{25} + \frac{u^4}{2500},

an even function of u, so

\begin{aligned} V &= 2\pi\int_0^{10} \left(16 - \frac{4u^2}{25} + \frac{u^4}{2500}\right)du \\ &= 2\pi\left(160 - \frac{160}{3} + 8\right) \\ &= \frac{688\pi}{3} \approx 720.5\ \text{cm}^3. \end{aligned}

Its mass is \displaystyle 0.6 \times \frac{688\pi}{3} \approx 432 grams. The blank has volume \pi(4)^2(20) = 320\pi, so Sanad cuts away \displaystyle 320\pi - \frac{688\pi}{3} = \frac{272\pi}{3} \approx 284.8 cubic centimeters, a little over a quarter of the blank.

Boundaries

The radius must be continuous on the closed interval (or continuous on pieces, with the integral split at the breaks). A profile with a vertical asymptote inside, such as \displaystyle y = \frac{1}{x} near x = 0, gives an improper integral that needs a separate limit argument.
Every slice must be a full disk: the region must reach the axis of revolution at every point of the interval. Where a gap separates the region from the axis, the slices are disks with holes, and the The washer method is needed.
The region must lie on one side of the axis. If it straddles the axis, the two sides sweep overlapping solids, and the slice is one disk whose radius is the larger of the two distances, not the sum of two disks.
The radius is measured from the axis of revolution along the slice. It equals f(x) only when the axis is the x-axis and f(x) \ge 0.
The slices are perpendicular to the axis, which fixes the variable: x for a horizontal axis, y for a vertical one.

Where this goes next

A region that does not reach the axis sweeps a solid with a hole, and subtracting an inner disk from an outer one gives the The washer method. Building the same solids from thin cylindrical shells parallel to the axis is the The shell method. The band that each slice’s rim sweeps out leads to the Surface area of a solid of revolution. The pattern underneath, a total built from thin pieces whose sizes are known, returns in Work as an integral.

Exercises

Day 1: about the x-axis

Find the volume when the region under y = x + 1 and above the x-axis for 0 \le x \le 3 is revolved about the x-axis, and check it as a large cone minus a small one.
The region under y = x^3 and above the x-axis for 0 \le x \le 2 is revolved about the x-axis. Find the volume.
The region under \displaystyle y = \sqrt{x + 2} and above the x-axis for -2 \le x \le 2 is revolved about the x-axis. Find the volume.
The region under y = e^x and above the x-axis for 0 \le x \le 1 is revolved about the x-axis. Find the volume.
The region under \displaystyle y = \frac{1}{x + 1} and above the x-axis for 0 \le x \le 2 is revolved about the x-axis. Find the volume.
The region under y = \sin x and above the x-axis for 0 \le x \le \pi is revolved about the x-axis. Find the volume. (Use \displaystyle \sin^2 x = \frac{1 - \cos 2x}{2}.)
The region under \displaystyle y = \sqrt{x}\,e^{x^2/2} and above the x-axis for 0 \le x \le 1 is revolved about the x-axis. Find the volume.
The region inside the upper half of the ellipse \displaystyle \frac{x^2}{9} + \frac{y^2}{4} = 1 and above the x-axis is revolved about the x-axis. Find the volume of the resulting solid.

Day 1: about the y-axis

The region in the first quadrant bounded by y = 3x^2, the line y = 3 and the y-axis is revolved about the y-axis. Find the volume.
The region bounded by the curve x = 2y - y^2 and the y-axis is revolved about the y-axis. Find the volume.

Day 2: other axes

The region bounded by y = x^3, the line y = -1 and the line x = 1 is revolved about the line y = -1. Find the volume.
The region bounded by y = x^2 and the line y = 2 is revolved about the line y = 2. Find the volume.
The region bounded by y = \sin x, the line y = 1 and the y-axis, for \displaystyle 0 \le x \le \frac{\pi}{2}, is revolved about the line y = 1. Find the volume.
The region bounded by \displaystyle y = \sqrt{x}, the x-axis and the line x = 1 is revolved about the line x = 1. Find the volume.
The triangle with vertices (0, 0), (2, 0) and (2, 4) is revolved (a) about the x-axis, (b) about the line x = 2. Find both volumes with the disk method and check each as a cone.

Day 2: turned objects

Ammar measures a turned spindle 12 cm long. At x = 2, 6 and 10 cm from one end its diameters are 3.0, 4.0 and 2.0 cm. Estimate its volume with a midpoint sum of disks.
Lujain turns a wooden knob 6 cm long. Measured from its center line, its profile is \displaystyle y = 1 + \sqrt{9 - x^2} for -3 \le x \le 3, in centimeters. Find its volume. (The integral of \displaystyle \sqrt{9 - x^2} over [-3, 3] is the area of a half-disk.)

Reasoning

The region under y = x^2 and above the x-axis for 0 \le x \le 2 is revolved about the line y = -1. Explain why the slices are not disks, and find the volume as the solid of Example 1 with a cylinder removed.
The region between the x-axis and y = -e^x for 0 \le x \le 1 lies below the axis. Explain why its volume about the x-axis equals the volume in Exercise 4, and why V = \pi\int_a^b [f(x)]^2\,dx holds whether f is positive or negative.
A cap of a ball of radius r is the part on one side of a plane, of height h measured from the plane to the ball’s surface, with 0 \le h \le 2r. By revolving part of \displaystyle y = \sqrt{r^2 - x^2}, show that its volume is \displaystyle \frac{\pi h^2(3r - h)}{3}, and check the cases h = r and h = 2r.
Every radius of a solid of revolution is doubled while its length stays the same. What happens to its volume? What if instead the solid is stretched to twice its length, with the radius at each new point equal to the old radius at the matching point? Explain with the integral.

Answers

21\pi: \displaystyle \pi\int_0^3 (x + 1)^2\,dx = \frac{\pi}{3}(x + 1)^3\Big|_0^3 = \frac{\pi}{3}(64 - 1). The line meets the axis at x = -1, so the solid is the cone from x = -1 to x = 3 (radius 4, height 4) minus the cone from x = -1 to x = 0 (radius 1, height 1): \displaystyle \frac{64\pi}{3} - \frac{\pi}{3} = 21\pi.
\displaystyle \frac{128\pi}{7}: \displaystyle \pi\int_0^2 x^6\,dx = \pi\cdot\frac{2^7}{7}.
8\pi: [R(x)]^2 = x + 2, and \displaystyle \pi\int_{-2}^{2} (x + 2)\,dx = \pi\cdot\frac{(x + 2)^2}{2}\Big|_{-2}^{2} = 8\pi.
\displaystyle \frac{\pi(e^2 - 1)}{2} \approx 10.0: \displaystyle \pi\int_0^1 e^{2x}\,dx = \frac{\pi}{2}e^{2x}\Big|_0^1.
\displaystyle \frac{2\pi}{3}: \pi\int_0^2 (x + 1)^{-2}\,dx is \displaystyle \pi\left(-\frac{1}{x + 1}\right)\Big|_0^2 = \pi\left(1 - \frac{1}{3}\right).
\displaystyle \frac{\pi^2}{2}: \displaystyle \pi\int_0^\pi \frac{1 - \cos 2x}{2}\,dx is \displaystyle \pi\left(\frac{x}{2} - \frac{\sin 2x}{4}\right)\Big|_0^\pi = \pi\cdot\frac{\pi}{2}.
\displaystyle \frac{\pi(e - 1)}{2} \approx 2.70: [R(x)]^2 = xe^{x^2}. With u = x^2, du = 2x\,dx, \displaystyle \pi\int_0^1 xe^{x^2}\,dx = \frac{\pi}{2}\int_0^1 e^u\,du = \frac{\pi}{2}(e - 1).
16\pi: \displaystyle y^2 = 4 - \frac{4x^2}{9}, and \displaystyle \pi\int_{-3}^{3} \left(4 - \frac{4x^2}{9}\right)dx = \pi(24 - 8).
\displaystyle \frac{3\pi}{2}: the curve meets y = 3 at x = 1, \displaystyle R(y) = \sqrt{\frac{y}{3}}, and \displaystyle \pi\int_0^3 \frac{y}{3}\,dy = \pi\cdot\frac{9}{6}.
\displaystyle \frac{16\pi}{15}: the curve meets the y-axis at y = 0 and y = 2, with R(y) = 2y - y^2 \ge 0 between, and \pi\int_0^2 \left(4y^2 - 4y^3 + y^4\right)dy is \displaystyle \pi\left(\frac{32}{3} - 16 + \frac{32}{5}\right).
\displaystyle \frac{16\pi}{7}: the curve meets y = -1 at x = -1, and R(x) = x^3 + 1, so \pi\int_{-1}^{1} \left(x^6 + 2x^3 + 1\right)dx is \displaystyle \pi\left(\frac{2}{7} + 0 + 2\right).
\displaystyle \frac{64\sqrt{2}\,\pi}{15} \approx 19.0: they meet at \displaystyle x = \pm\sqrt{2}, R(x) = 2 - x^2, and by symmetry \displaystyle 2\pi\int_0^{\sqrt{2}} \left(4 - 4x^2 + x^4\right)dx is \displaystyle 2\pi\sqrt{2}\left(4 - \frac{8}{3} + \frac{4}{5}\right).
\displaystyle \pi\left(\frac{3\pi}{4} - 2\right) \approx 1.12: R(x) = 1 - \sin x, and \pi\int_0^{\pi/2} \left(1 - 2\sin x + \sin^2 x\right)dx is \displaystyle \pi\left(\frac{\pi}{2} - 2 + \frac{\pi}{4}\right).
\displaystyle \frac{8\pi}{15}: slice in y for 0 \le y \le 1. The region runs from x = y^2 to the axis x = 1, so R(y) = 1 - y^2 and \pi\int_0^1 \left(1 - 2y^2 + y^4\right)dy is \displaystyle \pi\left(1 - \frac{2}{3} + \frac{1}{5}\right).
(a) \displaystyle \frac{32\pi}{3}, (b) \displaystyle \frac{16\pi}{3}. About the x-axis R(x) = 2x and \displaystyle \pi\int_0^2 4x^2\,dx = \frac{32\pi}{3}, a cone of radius 4 and height 2. About x = 2, slice in y: the hypotenuse is \displaystyle x = \frac{y}{2}, so \displaystyle R(y) = 2 - \frac{y}{2} and \displaystyle \pi\int_0^4 \left(2 - \frac{y}{2}\right)^2 dy = \frac{16\pi}{3}, a cone of radius 2 and height 4.
29\pi \approx 91.1 cubic centimeters: the radii 1.5, 2 and 1 sit at the midpoints of pieces 4 cm wide, so 4\pi\left(1.5^2 + 2^2 + 1^2\right) = 4\pi(7.25).
42\pi + 9\pi^2 \approx 220.8 cubic centimeters: \displaystyle [R(x)]^2 = 10 - x^2 + 2\sqrt{9 - x^2}. Here \int_{-3}^{3} (10 - x^2)\,dx = 42 and \displaystyle \int_{-3}^{3} \sqrt{9 - x^2}\,dx = \frac{9\pi}{2}, the area of a half-disk of radius 3, so V = \pi(42 + 9\pi).
\displaystyle \frac{176\pi}{15}. At each x the region’s points lie between 1 and x^2 + 1 from the axis, so the slice is a disk of radius x^2 + 1 with a disk of radius 1 removed, not a full disk. The full disks are those of Example 1 moved down by 1, and the removed disks make a cylinder of radius 1 and length 2, so \displaystyle V = \frac{206\pi}{15} - 2\pi.
The slice at x is the segment from (x, -e^x) to (x, 0), and turning it about the axis sweeps a disk of radius e^x, the same disk as the segment from (x, 0) to (x, e^x) in Exercise 4. In general the radius is |f(x)|, and |f(x)|^2 = [f(x)]^2, so squaring makes the sign of f irrelevant.
The cap at the right end of the ball is the part with r - h \le x \le r, so V = \pi\int_{r - h}^{r} \left(r^2 - x^2\right)dx, which is \displaystyle \pi\left(r^2h - \frac{r^3 - (r - h)^3}{3}\right). Since r^3 - (r - h)^3 = 3r^2h - 3rh^2 + h^3, this is \displaystyle \pi\left(rh^2 - \frac{h^3}{3}\right) = \frac{\pi h^2(3r - h)}{3}. With h = r it is \displaystyle \frac{2}{3}\pi r^3, half the ball; with h = 2r it is \displaystyle \frac{4}{3}\pi r^3, the whole ball.
Doubling every radius multiplies each [R(x)]^2 by 4, so the volume is multiplied by 4. Stretching to twice the length gives the new radius \displaystyle R\left(\frac{x}{2}\right) on [2a, 2b], and \displaystyle \pi\int_{2a}^{2b} \left[R\left(\frac{x}{2}\right)\right]^2 dx becomes 2\pi\int_a^b [R(u)]^2\,du with \displaystyle u = \frac{x}{2}, dx = 2\,du: the volume doubles.