Unit 5, lesson 65

Volumes by Cross-Sections, Disks, and Washers

How to find a volume by integrating the area of its slices, which explains the formulas for cylinders, cones, pyramids and balls and gives the disk and washer methods for solids of revolution.

  • Theory part 129
  • Practice part 130
  • Applications of Integration

Slice a loaf of bread and each slice is nearly a thin slab whose volume is its face area times its thickness. Add the slabs and you get the loaf. With thinner slices the estimate improves, and in the limit it becomes an integral of cross-sectional area. Every volume formula from school geometry, from the cone to the ball, falls out of this one idea.

Key terms

Cylinders and prisms

Definition Let BB be a plane region with area AA. A (right) cylinder with base BB and height h>0h > 0 is the solid swept out when BB moves a distance hh perpendicular to its own plane. Its volume is

V=Ah.V = A h.

A prism is a cylinder whose base is a polygon. A circular cylinder of radius rr has V=πr2hV = \pi r^2 h.

In plain words Any solid with the same flat shape all the way up, stacked straight, has volume base area times height. This is the building block for every slice.

Example A prism whose base is a right triangle with legs 33 cm and 44 cm, and whose height is 1010 cm, has volume 12⋅3⋅4⋅10=60\frac{1}{2} \cdot 3 \cdot 4 \cdot 10 = 60 cubic centimeters.

Cross-sectional area

Definition Let SS be a solid lying between the planes x=ax = a and x=bx = b. For each x∈[a,b]x \in [a, b], the cross section SxS_x is the intersection of SS with the plane perpendicular to the xx-axis at xx, and A(x)A(x) is its area. The same idea applies with any axis in place of the xx-axis.

In plain words Cut the solid with a flat knife at position xx, perpendicular to the axis, and measure the area of the cut face.

Example For a ball of radius 33 centered at the origin, the cross section at xx with −3≤x≤3-3 \le x \le 3 is a disk of radius 9−x2\sqrt{9 - x^2}, so A(x)=π(9−x2)A(x) = \pi(9 - x^2). At x=0x = 0 that is 9π9\pi, and at x=3x = 3 it is 00.

Volume as a limit of slices

Definition Let SS lie between x=ax = a and x=bx = b, and suppose its cross-sectional area A(x)A(x) is continuous on [a,b][a, b]. Partition [a,b][a, b] into subintervals of widths Δxi\Delta x_i and pick xi∗x_i^* in each. The volume of SS is

V=lim⁡max⁡Δxi→0∑i=1nA(xi∗) Δxi=∫abA(x) dx.V = \lim_{\max \Delta x_i \to 0} \sum_{i=1}^n A(x_i^*)\, \Delta x_i = \int_a^b A(x) \, dx.

In plain words Each thin slice is nearly a cylinder of volume A(xi∗) ΔxiA(x_i^*)\,\Delta x_i. Add them up and let the slices get infinitely thin.

Example A solid has the unit disk x2+y2≤1x^2 + y^2 \le 1 as its base, and every cross section perpendicular to the xx-axis is a square standing on the base. At xx the square’s side is the chord 21−x22\sqrt{1 - x^2}, so A(x)=4(1−x2)A(x) = 4(1 - x^2) and

V=∫−114(1−x2) dx=4(2−23)=163.V = \int_{-1}^1 4(1 - x^2) \, dx = 4\left(2 - \frac{2}{3}\right) = \frac{16}{3}.

Oblique solids

Definition If two solids lie between the same two parallel planes and every plane parallel to them cuts both solids in cross sections of equal area, then the solids have equal volume. When the common area function is continuous, this follows at once from the slice formula above. In particular an oblique cylinder, whose base slides along a slanted line instead of straight up, has volume V=AhV = A h, where hh is the perpendicular distance between the base planes.

In plain words Push a stack of coins so it leans. Every coin is the same as before, so the volume does not change. What counts is the vertical height, not the slanted length.

Example A stack of 2020 coins, each of radius 11 cm and thickness 0.20.2 cm, has volume 20⋅π⋅12⋅0.2=4π≈12.5720 \cdot \pi \cdot 1^2 \cdot 0.2 = 4\pi \approx 12.57 cubic centimeters, whether it stands straight or leans.

Pyramid

Definition A pyramid with base area AA and height hh (perpendicular distance from apex to base plane) has volume

V=13Ah.V = \frac{1}{3} A h.

The cross section at distance yy from the apex, parallel to the base, is similar to the base with ratio y/hy/h, so its area is Ay2/h2A y^2 / h^2, and

∫0hAy2h2 dy=13Ah.\int_0^h \frac{A y^2}{h^2} \, dy = \frac{1}{3}Ah.

In plain words A pyramid holds exactly one third of the prism with the same base and height, because its slices shrink like the square of the distance from the tip.

Example For a square base of side 66 and height 55, the slice at depth yy below the apex is a square of side 6y/56y/5, so

V=∫0536y225 dy=3625⋅1253=60.V = \int_0^5 \frac{36 y^2}{25} \, dy = \frac{36}{25} \cdot \frac{125}{3} = 60.

This agrees with 13⋅36⋅5=60\frac{1}{3} \cdot 36 \cdot 5 = 60.

Cone

Definition A circular cone with base radius rr and height hh has volume

V=13πr2h.V = \frac{1}{3}\pi r^2 h.

It is the solid obtained by rotating the triangle under the line y=rx/hy = rx/h for 0≤x≤h0 \le x \le h about the xx-axis. Its cross sections are disks of radius rx/hrx/h, so

V=π∫0h(rxh)2dx=13πr2h.V = \pi \int_0^h \left(\frac{r x}{h}\right)^2 dx = \frac{1}{3}\pi r^2 h.

In plain words A cone is a pyramid with a round base, so it also gets the factor one third.

Example With r=3r = 3 and h=4h = 4 the volume is

V=π∫04(3x4)2dx=π⋅916⋅643=12π.V = \pi\int_0^4 \left(\frac{3x}{4}\right)^2 dx = \pi \cdot \frac{9}{16} \cdot \frac{64}{3} = 12\pi.

Solids of revolution

Definition A solid of revolution is the solid swept out when a plane region rotates one full turn about a line in its plane, called the axis of rotation, where the region lies entirely on one side of the axis (it may touch the axis along its boundary). When the region lies between two graphs over an interval of the axis, each cross section perpendicular to the axis is a disk or a washer.

In plain words Spin a flat shape around a line, like clay on a potter’s wheel, and look at the solid it traces.

Example Rotating the rectangle 0≤x≤50 \le x \le 5, 0≤y≤20 \le y \le 2 about the xx-axis gives a circular cylinder of radius 22 and length 55, with volume π⋅22⋅5=20π\pi \cdot 2^2 \cdot 5 = 20\pi.

Disks and washers

Definition Let ff be continuous and nonnegative on [a,b][a, b]. Rotating the region under y=f(x)y = f(x) about the xx-axis gives a solid with disk cross sections, so

V=π∫abf(x)2 dx.V = \pi\int_a^b f(x)^2 \, dx.

If 0≤g(x)≤f(x)0 \le g(x) \le f(x) are continuous on [a,b][a, b], rotating the region between them about the xx-axis gives washers with outer radius f(x)f(x) and inner radius g(x)g(x), so

V=π∫ab(f(x)2−g(x)2)dx.V = \pi\int_a^b \left(f(x)^2 - g(x)^2\right) dx.

In plain words A disk is a solid coin of area πR2\pi R^2. A washer is a coin with a hole, of area πR2−πr2\pi R^2 - \pi r^2. Square the radii first, then subtract, never the other way around.

Example Rotating the region under y=xy = \sqrt{x} on [0,4][0, 4] about the xx-axis gives

V=π∫04(x)2 dx=π∫04x dx=8π.V = \pi\int_0^4 (\sqrt{x})^2 \, dx = \pi\int_0^4 x \, dx = 8\pi.

Rotating the region between y=xy = x and y=x2y = x^2 on [0,1][0, 1] gives washers with R=xR = x and r=x2r = x^2, so

V=π∫01(x2−x4) dx=π(13−15)=2π15.V = \pi\int_0^1 (x^2 - x^4) \, dx = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}.

Ball

Definition A ball of radius r>0r > 0 has volume

V=43πr3.V = \frac{4}{3}\pi r^3.

It is the solid of revolution formed by rotating the half disk under y=r2−x2y = \sqrt{r^2 - x^2}, −r≤x≤r-r \le x \le r, about the xx-axis, so

V=π∫−rr(r2−x2) dx=π(2r3−2r33)=43πr3.V = \pi\int_{-r}^r (r^2 - x^2) \, dx = \pi\left(2r^3 - \frac{2r^3}{3}\right) = \frac{4}{3}\pi r^3.

In plain words Spin a half disk about its straight edge and you get a ball. The disk method then gives the familiar formula.

Example For r=3r = 3 the disk method gives

V=π∫−33(9−x2) dx=π(54−18)=36π,V = \pi\int_{-3}^3 (9 - x^2) \, dx = \pi(54 - 18) = 36\pi,

matching 43π⋅27\frac{4}{3}\pi \cdot 27.

Bounds and radii

Definition For rotation about the horizontal line y=ky = k, the radius of a point on the curve y=f(x)y = f(x) is its distance ∣f(x)−k∣\lvert f(x) - k \rvert to the axis. For a region g(x)≤y≤f(x)g(x) \le y \le f(x) lying on one side of the axis, the outer radius R(x)R(x) is the larger of the distances from the axis to the two boundary curves and the inner radius r(x)r(x) is the smaller, so

V=π∫ab(R(x)2−r(x)2)dx,V = \pi\int_a^b \left(R(x)^2 - r(x)^2\right) dx,

where [a,b][a, b] is the interval of xx values covered by the region. For a vertical axis, use horizontal slices and integrate in yy.

In plain words Radii are distances to the axis, not yy values. Draw the axis, draw one slice, and measure from the axis to the far edge and to the near edge.

Example Rotate the region between y=xy = x and y=x2y = x^2 on [0,1][0, 1] about y=−1y = -1. The far edge is y=xy = x, so R=x+1R = x + 1, and the near edge is y=x2y = x^2, so r=x2+1r = x^2 + 1. The volume is

V=π∫01((x+1)2−(x2+1)2)dx=π∫01(2x−x2−x4) dx=π(1−13−15)=7π15.V = \pi\int_0^1 \left((x + 1)^2 - (x^2 + 1)^2\right) dx = \pi\int_0^1 (2x - x^2 - x^4) \, dx = \pi\left(1 - \frac{1}{3} - \frac{1}{5}\right) = \frac{7\pi}{15}.

Check yourself

  1. Rotate the region under y=x2y = x^2 on [0,1][0, 1] about the xx-axis. Find the volume.
Show the answer

The disks have radius x2x^2, so

V=π∫01x4 dx=π5.V = \pi\int_0^1 x^4 \, dx = \frac{\pi}{5}.
  1. Rotate the region between y=xy = \sqrt{x} and y=xy = x on [0,1][0, 1] about the xx-axis. Find the volume.
Show the answer

On [0,1][0, 1] you have x≥x\sqrt{x} \ge x, so R=xR = \sqrt{x} and r=xr = x. Then

V=π∫01(x−x2) dx=π6.V = \pi\int_0^1 (x - x^2) \, dx = \frac{\pi}{6}.
  1. A solid has as its base the triangle with vertices (0,0)(0, 0), (2,0)(2, 0) and (0,2)(0, 2). Cross sections perpendicular to the xx-axis are squares. Find its volume.
Show the answer

At xx the side is 2−x2 - x, so

V=∫02(2−x)2 dx=83.V = \int_0^2 (2 - x)^2 \, dx = \frac{8}{3}.

How this lesson connects

It also uses areas and volumes.