Shared05 - Volume part 2 Assignment / Volume part 2 Notes.sagewsOpen in CoCalc
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Prerequisites:

  • Intro to Sage
  • Graphing and Solving Equations
  • Symbolic Integration
  • Volume, Part 1

Volume, part 2: Solids of Revolution

Recall from last week: If a solid is placed along the x-axis from x=ax=a to x=bx=b and the area of the cross section perpendicular to the x-axis at xx is A(x)A(x), then the volume of the solid is V=abA(x)dx\displaystyle V=\int_a^b A(x)\, dx.

If the solid may be obtained by rotating a plane region around an axis in that plane, we get a "solid of revolution."

If a solid of revolution is obtained by rotating around the x-axis the region between the graph of a function ff and the x-axis over the interval [a,b][a,b], then the cross section perpendicular to the x-axis at xx is a disk with area πr2\pi r^2, where the radius is r=f(x)r=f(x). Therefore, the volume of this solid is V=abπf(x)2dxV=\int_a^b \pi f(x)^2\, dx

Example 1

Find the volume of the solid obtained by rotating around the x-axis the region between f(x)=xf(x)=\sqrt{x} and the x-axis from x=0x=0 to x=4x=4.

First, let's look at a graph of the region we're rotating.

f(x)=sqrt(x)
plot(f,xmin=0,xmax=4,fill='axis')

Sage can plot the surface we get by rotating the curve only. To get the solid, imagine filling in this surface.

The volume of this solid is V=04π(x)2dx=8π\displaystyle V=\int_0^4 \pi(\sqrt{x})^2\, dx=8\pi.

integral(pi*x,x,0,4)
8*pi

Example 2

Find the volume of the solid obtained by rotating around the horizontal line y=1y=1 the region between f(x)=xf(x)=\sqrt{x} and the line y=1y=1 over the interval [1,4][1,4].

Here is a graph of the region we're rotating.

f(x)=sqrt(x)
g(x)=1
plot(f,xmin=1,xmax=4,fill=g(x))+plot(g,xmin=0,xmax=5,color='black',ymin=0,linestyle='--')

Once again, the cross sections perpendicular to the axis of rotation are disks. However, the radius is not given by the value of the function; instead, the radius is the distance from the axis of rotation to the function, i.e., f(x)1f(x)-1.

Thus, the volume is V=14π(x1)2dx=7π6\displaystyle V=\int_1^4 \pi(\sqrt{x}-1)^2\, dx=\frac{7\pi}{6}.

integral(pi*(sqrt(x)-1)^2,x,1,4)
7/6*pi

Here is the 3D picture. Imagine the cone-shaped thing is filled in.

3D rendering not yet implemented

Rotating around the y-axis

If we rotate the region between a curve and the y-axis and our axis of rotation is the y-axis, then we need xx given as a function of yy, x=g(y)x=g(y). When we take cross sections perpendicular to the y-axis, we get disks again, and the radius of the disk is given by g(y)g(y). Thus, the volume of the solid of revolution is V=cdπg(y)2dyV=\int_c^d \pi g(y)^2\, dy

Example 3

Find the volume of the solid obtained by rotating around the y-axis the region between f(x)=xf(x)=\sqrt{x} and the y-axis from y=0y=0 to y=2y=2.

Here is a graph of the region.

plot(sqrt(x),xmin=0,xmax=4,fill=2)

We are given yy as a function of xx, but we need xx as a function of yy. So we solve the equation y=xy=\sqrt{x} for xx: x=y2x=y^2.

Then the volume of our solid is V=02π(y2)2dy=32π5\displaystyle V=\int_0^2 \pi(y^2)^2\, dy=\frac{32\pi}{5}.

%var y
integral(pi*y^4,y,0,2)
32/5*pi

Here is a picture of the surface.

3D rendering not yet implemented

The Washer Method

In the previous example, cross sections perpendicular to the axis of rotation were disks, because the region we were rotating touched the axis of rotation.

If there is a gap between our region and our axis of rotation, then the resulting cross sections are not disks, instead they are annuli (singular, annulus). An annulus is the region between two circles in the same plane, like a washer.

If the outer circle has radius RR and the inner circle has radius rr, then the area of the annulus is A=πR2πr2A=\pi R^2-\pi r^2.


Here is a picture of an annulus with outer radius 2 and inner radius 1. Its area is A=π22π12=3πA=\pi\cdot 2^2-\pi\cdot 1^2=3\pi.

If we form a solid of revolution by rotating around the x-axis the region between two functions ff and gg from x=ax=a to x=bx=b, then the cross sections perpendicular to the x-axis are annuli, where the outer radius is given by whichever function is greater and the inner radius is given by the other function.

Let's assume f(x)g(x)f(x)\ge g(x) for axba\le x\le b. Then the area of an annular cross section is A(x)=πf(x)2πg(x)2A(x)=\pi f(x)^2-\pi g(x)^2, and the volume of the solid is V=abπf(x)2πg(x)2dxV=\int_a^b\pi f(x)^2-\pi g(x)^2\, dx

Example 4

Find the volume of the solid obtained by rotating around the x-axis the region between f(x)=xf(x)=\sqrt{x} and g(x)=x2g(x)=x^2 over the interval [0,1][0,1].

First, notice that ff is the function on top on this interval (that is, f(x)g(x)f(x)\ge g(x) for all xx in [0,1][0,1]).

f(x)=sqrt(x)
g(x)=x^2
plot(f,xmin=0,xmax=1,fill=g(x))+plot(g,xmin=0,xmax=1,color='red')

Now the value of the integral is V=01π(x)2π(x2)2dx=3π10\displaystyle V=\int_0^1 \pi(\sqrt{x})^2-\pi(x^2)^2\, dx=\frac{3\pi}{10}.

integral(pi*x-pi*x^4,x,0,1)
3/10*pi

Here is a picture (the solid is between the green and blue surfaces).

3D rendering not yet implemented

Example 5

Let's rotate the same region around the horizontal line y=2y=2 and find the new volume.

plot(f,xmin=0,xmax=1,fill=g(x))+plot(g,xmin=0,xmax=1,color='red')+plot(2,xmin=-.2,xmax=1.2,color='black',linestyle='--')

Cross section perpendicular to the line y=2y=2 are annuli again.

The outer radius is the distance from the center (y=2y=2) to the function that's furthest away from the center (in this case, y=x2y=x^2). This distance is 2x22-x^2.

The inner radius is the distance from the center to the closest function (in this case, y=xy=\sqrt{x}). This distance is 2x2-\sqrt{x}.

Therefore, the volume is V=01π(2x2)2π(2x)2dx=31π30\displaystyle V=\int_0^1 \pi(2-x^2)^2-\pi(2-\sqrt{x})^2\, dx=\frac{31\pi}{30}.

integral(pi*(2-x^2)^2-pi*(2-sqrt(x))^2,x,0,1)
31/30*pi
3D rendering not yet implemented

Example 6

Find the volume of the solid obtained by rotating around the vertical line x=4x=4 the region between y=2x+1y=2x+1 and y=x2+1y=x^2+1 from x=0x=0 to x=2x=2.

Here's the region:

f(x)=2*x+1
g(x)=x^2+1
plot(f,xmin=0,xmax=2,fill=g(x))+plot(g,xmin=0,xmax=2,color='red')+line([(4,0),(4,5.1)],color='black',linestyle='--')

Now we need to take cross sections perpendicular to the line x=4x=4. When we do so, we get annuli. That is, we cut the solid at a given value of yy to get a annulus. So we need to integrate in the y-direction, which means we need everything in terms of y.

If y=2x+1y=2x+1, then x=y12x=\frac{y-1}{2}. If y=x2+1y=x^2+1, then x=y1x=\sqrt{y-1}. When x=0x=0, then y=1y=1, and when x=2x=2, then y=5y=5.

The outer radius of a cross section is the horizontal distance from x=4x=4 to the function furthest away, which is x=y12x=\frac{y-1}{2}. This distance is 4y124-\frac{y-1}{2}.

The inner radius is the horizontal distance from x=4x=4 to the closest function, which is x=y1x=\sqrt{y-1}. This distance is 4y14-\sqrt{y-1}.

Therefore, the volume is V=15π(4y12)2π(4y1)2dy=8π\displaystyle V=\int_1^5 \pi\left(4-\frac{y-1}{2}\right)^2-\pi\left(4-\sqrt{y-1}\right)^2\, dy=8\pi.

%var y
integral (pi*(4-(y-1)/2)^2-pi*(4-sqrt(y-1))^2,y,1,5)
8*pi

Here is a picture.

3D rendering not yet implemented

Example 7

Find the volume of the solid obtained by rotating around the x-axis the region between y=sin(x)+1y=\sin(x)+1 and y=cos(x)+1y=\cos(x)+1 from x=3π4x=-\frac{3\pi}{4} and x=5π4x=\frac{5\pi}{4}.

Here is the region:

f(x)=sin(x)+1
g(x)=cos(x)+1
plot(f,xmin=-3*pi/4,xmax=5*pi/4,fill=g(x),ticks=pi/4,tick_formatter=pi)+plot(g,xmin=-3*pi/4,xmax=5*pi/4,color='red')

Cross sections perpendicular to the x-axis are annuli.

The outer radius is given by the function on top. In this example, the function on top actually changes. This happens when the two functions cross at π4\frac{\pi}{4} [How do we know this is where they cross?]. Of course, the function on the bottom changes as well.

We need to break our problem into two parts:

  1. From x=3π4x=-\frac{3\pi}{4} to x=π4x=\frac{\pi}{4}, y=cos(x)+1y=\cos(x)+1 is on top and y=sin(x)+1y=\sin(x)+1 is on the bottom.

  2. From x=π4x=\frac{\pi}{4} to x=5π4x=\frac{5\pi}{4}, these are switched.

So the volume is V=3π/4π/4π(cos(x)+1)2π(sin(x)+1)2dx+π/45π/4π(sin(x)+1)2π(cos(x)+1)2dx=8π2\displaystyle V=\int_{-3\pi/4}^{\pi/4}\pi(\cos(x)+1)^2-\pi(\sin(x)+1)^2\, dx+\int_{\pi/4}^{5\pi/4}\pi(\sin(x)+1)^2-\pi(\cos(x)+1)^2\, dx=8\pi\sqrt{2}.

integral(pi*(cos(x)+1)^2-pi*(sin(x)+1)^2,x,-3*pi/4,pi/4)+integral(pi*(sin(x)+1)^2-pi*(cos(x)+1)^2,x,pi/4,5*pi/4)
8*sqrt(2)*pi
3D rendering not yet implemented

Example 8 - Two Functions with a Gap

Find the volume of the solid obtained by rotating around the vertical line x=3x=3 the region between the curves y=x2y=x^2 and y=2x+4y=2x+4 from x=0x=0 to x=2x=2.

Here is a picture of the region. Notice the gap between the two functions.

f(x)=x^2
g(x)=2*x+4
plot(f,xmin=0,xmax=2,color='red')+plot(g,xmin=0,xmax=2,fill=f)+line([(3,0),(3,8)],linestyle='dashed',color='black')

Since we are rotating around a vertical line, we need to integrate in the y-direction, and since there is a gap between the two functions, the outer and inner radius are not given by the same functions the entire time.

From y=0y=0 to y=4y=4, the outer radius is a constant 33, while the inner radius is 3y3-\sqrt{y}.

From y=4y=4 to y=8y=8, the outer radius is 3(12y2)3-(\frac{1}{2}y-2), and the inner radius is a constant 32=13-2=1.

Thus, we need to do two integrals and add them together.

V=04π32π(3y)2dy+48π(3(12y2))2π12dy=112π3 V=\int_0^4 \pi\cdot 3^2-\pi \left(3-\sqrt{y}\right)^2\, dy+\int_4^8 \pi \left(3-\left(\frac{1}{2}y-2\right)\right)^2-\pi\cdot 1^2\, dy=\frac{112\pi}{3}

%var y
integral(pi*3^2-pi*(3-sqrt(y))^2,y,0,4)+integral(pi*(3-(1/2*y-2))^2-pi*1^2,y,4,8)
112/3*pi

Summary

To find volumes for solids of revolution:

  • Draw a picture of the region to be rotated and indicate the axis of rotation.

  • If you're rotating around the x-axis or another horizontal line, then you will integrate with respect to x. You need functions in terms of x and x-values for the limits of integration.

  • If you're rotating around the y-axis or another vertical line, then you will integrate with respect to y. You need functions in terms of y and y-values for the limits of integration.

  • Decide if your cross sections are disks or annuli. Is there a gap between the region to be rotated and the axis of rotation? Then you have annuli.

  • Determine the radius for a disk or the outer and inner radii for an annulus.

  • Integrate!

Final Hint: If you get a negative volume for a washer question, you probably reversed the inner and outer radii.