| Hosted by CoCalc | Download
Kernel: SageMath 9.2

Volumes of Revolution - A Methodical Approach

var('x y') # usually we're rotating a portion of a function, so the piecewise feature is useful f = piecewise([([0, 1], x^2)]) g = piecewise([([0, 1], sqrt(x))])
p1 = plot(f, 0, 1, gridlines=True) t1 = text("I", (0.8, 0.2), fontsize='large', fontweight='bold') p2 = plot(g, 0, 1) t2 = text("II", (0.2, 0.8), fontsize='large', fontweight='bold') t3 = text("III", (0.4, 0.4), fontsize='large', fontweight='bold') (p1 + t1 + p2 + t2 + t3).show()
Image in a Jupyter notebook

This notebook will demonstrate many permutations of rotating the above regions about various axes using disk, washer and shell methods. Depending on the goal, we will integrate either with respect to x- or with respect to y.

Rotate regions I, II, and III about the x axis.

Region 1 is bounded by the curve y=x2y = x^2 above, the x axis below, and the line x=1x=1. Here is the surface of the volume created when rotated about the x-axis.

revolution_plot3d(f, (x,0,1), parallel_axis='x', show_curve=True, opacity=0.7).show(aspect_ratio=(1,1,1))

Using the disk method, with the disk perpendicular to the x-axis,

V=∫A  dxV = \int A \; dxA=πr2A = \pi r^2r=f(x)r = f(x)V=π∫01(f(x))2  dxV = \pi \int_{0}^{1} (f(x))^2 \; dx
pi * integrate(f^2, x, 0, 1)
1/5*pi

Region 2 is bounded by the curve y=xy = \sqrt{x} below, the y axis, and the line y=1y=1 above. Here is the surface of the volume created when rotated about the x-axis.

l = 1 sur1 = revolution_plot3d(l, (x,0,1), parallel_axis='x', opacity=0.5, rgbcolor=(1,0.5,0)) sur2 = revolution_plot3d(g, (x,0,1), parallel_axis='x', show_curve=True, mesh=True, opacity=0.5, rgbcolor=(0,1,0)) (sur1 + sur2).show()

Rotating region II about the x axis requires use of the washer method, i.e. one disk subtracted from another.

A=π(ro2−ri2)A = \pi (r_o^2 - r_i^2)ro=1, and ri=xr_o = 1 \text{, and } r_i = \sqrt{x}V=π∫0112−(x)2  dxV = \pi \int_{0}^{1} 1^2 - (\sqrt{x})^2 \; dx
# The volume of region II rotated about the x axis pi * integrate(1 - x, x, 0, 1)
1/2*pi

Region 3 is bounded above by the curve y=xy = \sqrt{x} and below by the curve y=x2y = x^2. Here the surface of the volume created, still about the -axis.

aor = line([(0,0,0), (1,0,0)], thickness=3, color='green') sur1 = revolution_plot3d(f, (x,0,1), parallel_axis='x', show_curve=True, opacity=0.5, color='red') sur2 = revolution_plot3d(g, (x,0,1), parallel_axis='x', show_curve=True, opacity=0.5, color='blue') (aor + sur1 + sur2).show(aspect_ratio=(1,1,1))

Link to Geogebra construction of the same shape: Region III rotated about x.

Using the washer method, with the washer / disk perpendicular to the x-axis,

V=∫A  dxV = \int A \; dxA=πro2−πri2A = \pi r_o^2 - \pi r_i^2ro=g(x) and ri=f(x)r_o = g(x) \text{ and } r_i = f(x)V=π∫01(g(x))2−(f(x))2  dxV = \pi \int_{0}^{1} (g(x))^2 - (f(x))^2 \; dx
# The volume of region III rotated about the x-axis. pi * integral(x - x^2, x, 0, 1)
1/6*pi

Rotate regions I, II, and III about the y axis.

Region I, bounded by y=x2y=x^2, the x axis, and the line x=1x=1, rotated about the y axis, looks like this:

# SageMath surface plots assume the form z=f(x, y), so the independent variable is z. Hence the parallel axis. aor = line([(0,0,0), (0,0,1)], thickness=2, color='red') surf1 = revolution_plot3d(f, (x,0,1), parallel_axis='z', show_curve=True, opacity=0.7) (aor + surf1).show(aspect_ratio=(1,1,1))

Using the shell method, the area of each shell is

A=2Ï€rhA = 2 \pi r h

where

r=x and h=f(x)=x2r = x \text{ and } h = f(x) = x^2

So

V=2π∫01xf(x)  dxV = 2 \pi \int_{0}^{1} x f(x) \; dx
2 * pi * integral(x * x^2, x, 0, 1)
1/2*pi

Region II, bounded by y=xy = \sqrt{x}, the y axis, and the line y=1y=1, rotated about the y axis.

aor = line([(0,0,0), (0,0,1)], thickness=2, color='red') surf1 = revolution_plot3d(g, (x,0,1), parallel_axis='z', show_curve=True, opacity=0.7) (aor + surf1).show(aspect_ratio=(1,1,1))

Using the shell method again,

r=x and h=1−xr = x \text{ and } h = 1 - \sqrt{x}V=2π∫01x(1−g(x))  dxV = 2 \pi \int_{0}^{1} x (1 - g(x)) \; dx
2 * pi * integral(x*(1-sqrt(x)), x, 0, 1)
1/5*pi

Region III, rotated about y.

aor = line([(0,0,0), (0,0,1)], thickness=2, color='black') sur1 = revolution_plot3d(f, (x,0,1), show_curve=True, opacity=0.4, color='blue', parallel_axis='z') sur2 = revolution_plot3d(g, (x,0,1), show_curve=True, opacity=0.8, color='red', parallel_axis='z') (aor + sur1 + sur2).show(aspect_ratio=(1,1,1))
r=x and h=x−x2r = x \text{ and } h = \sqrt{x} - x^2V=2π∫01x(x−x2)  dxV = 2 \pi \int_{0}^{1} x (\sqrt{x} - x^2) \; dx
2 * pi * integral(x*(sqrt(x) - x^2), x, 0, 1)
3/10*pi

The next group will be the same regions and rotations, but replacing the shell method with the disk or washer and integrating with respect to y. When integrating with respect to y we'll need to use the inverse functions.

f(x)=x2⇒f−1(y)=yf(x) = x^2 \Rightarrow f^{-1}(y) = \sqrt{y}g(x)=x⇒g−1(y)=y2g(x) = \sqrt{x} \Rightarrow g^{-1}(y) = y^2

Rotating region I around y, we get a washer with

ro=1 and ri=yr_o = 1 \text{ and } r_i = \sqrt{y}V=π∫0112−(y)2  dyV = \pi \int_{0}^{1} 1^2 - (\sqrt{y})^2 \; dy
pi * integral(1 - y, y, 0, 1)
1/2*pi

Rotating region II about y and integrating with respect to y yields a simple disk with r=y2r = y^2

V=π∫01(y2)2  dyV = \pi \int_{0}^{1} (y^2)^2 \; dy
pi * integral(y^4, y, 0, 1)
1/5*pi

And, rotating region III about y and integrating with respect to y gives a washer with

ro=y and ri=y2r_o = \sqrt{y} \text{ and } r_i = y^2

and

V=π∫01(y)2−(y2)2  dyV = \pi \int_{0}^{1} (\sqrt{y})^2 - (y^2)^2 \; dy
pi * integral(y - y^4, y, 0, 1)
3/10*pi

Rotate regions I, II, and III about y=1

Rotating region I about y=1 using the washer method, we have

ro=1 and ri=1−x2r_o = 1 \text{ and } r_i = 1-x^2V=π∫01ro2−ri2  dxV = \pi \int_{0}^{1} r_o^2 - r_i^2 \; dxV=π∫0112−(1−x2)2  dxV = \pi \int_{0}^{1} 1^2 - (1-x^2)^2 \; dx
pi * integral(1 - (1-x^2)^2, x, 0, 1)
7/15*pi

Rotating region II about y=1 is a simple disk method with r=1−xr = 1 - \sqrt{x}.

pi * integral((1 - sqrt(x))^2, x, 0, 1)
1/6*pi

For region III about y=1 we have a washer with

ro=1−x2 and ri=1−xr_o = 1-x^2 \text{ and } r_i = 1 - \sqrt{x}V=π∫01(1−x2)2−(1−x)2  dxV = \pi \int_{0}^{1} (1 - x^2)^2 - (1 - \sqrt{x})^2 \; dx
aor = line([(1,0,1), (0,0,1)], thickness=3, color='blue') sur1 = revolution_plot3d(f, (x,0,1), parallel_axis='x', axis=(0,1), show_curve=True, opacity=0.7, mesh=True, color='purple') sur2 = revolution_plot3d(g, (x,0,1), parallel_axis='x', axis=(0,1), show_curve=True, opacity=0.7, mesh=True, color='yellow') (aor + sur1 + sur2).show(aspect_ratio=(1,1,1))
pi * integral((1-x^2)^2 - (1-sqrt(x))^2, x, 0, 1)
11/30*pi

Rotate regions I, II, and III about x=1

Rotating region I about x=1 and using the shell method yields

A=2πrhA = 2 \pi r hr=1−x and h=x2r = 1 - x \text{ and } h = x^2V=2π∫01(1−x)x2  dxV = 2 \pi \int_{0}^{1} (1-x) x^2 \; dx
2 * pi * integral((1-x)*x^2, x, 0, 1)
1/6*pi

Rotating region II about x=1, also using shells.

A=2πrhA = 2 \pi r hr=1−x, and h=1−xr = 1 - x, \text{ and } h = 1 - \sqrt{x}V=2π∫01(1−x)(1−x)  dxV = 2 \pi \int_{0}^{1} (1 - x) (1 - \sqrt{x}) \; dx
2 * pi * integral((1 - x)*(1 - sqrt(x)), x, 0, 1)
7/15*pi

Region III rotated about x=1.

# this is the volume enclosed by y = x^2 and y = sqrt(x) rotated about x=1 aor = line([(1,0,0), (1,0,1)], thickness=3, color='green') surf1 = revolution_plot3d(f, (x,0,1), parallel_axis='z', axis=(1, 0), show_curve=True, opacity=0.7, color='red') surf2 = revolution_plot3d(g, (x,0,1), parallel_axis='z', axis=(1, 0), show_curve=True, opacity=0.7, color='blue') (aor + surf1 + surf2).show(aspect_ratio=(1,1,1))

Rotating region III about x=1 using shells.

r=1−x, and h=x−x2r = 1 - x, \text{ and } h = \sqrt{x} - x^2V=2π∫01(1−x)(x−x2)  dxV = 2 \pi \int_{0}^{1} (1 - x) (\sqrt{x} - x^2) \; dx
2 * pi * integral((1 - x)*(sqrt(x) - x^2), x, 0, 1)
11/30*pi