Volumes by Slicing; Disks and WashersDefinite integrals have been used to produce areas. We now show that they can be used to produce volumes as well.

Suppose that a solid extends along the x-axis and is bounded on the left and right by planes perpendicular to the x-axis at x = a and x = b.

x

A(x)

x - axisa b

Suppose that at each point x between a and b, we know the cross sectional area A(x) of the solid.

x

A(x)

x - axisa b

Theorem. If V is the volume of the solid, then

|

( )b

V A x dxa

= ∫

∆x

x - axisa b

The reason for this is that if we take the Riemann sum

( )1

nA x xi

i∆∑

=

( )A x xi

∆

we get an approximation that obviously tends to the volume of the solid as n tends to infinity.

The “infinitesimal” point of view

Whenever we want to represent some physical quantity as a definite integral, we can show that the Riemann sum which converges to that integral is an approximation to the physical quantity. This is what we did above. However it is often easier to set up the integral by using an imprecise but intuitive idea called “infinitesimals”.

An infinitesimal quantity is considered to be so small that it is essentially zero. Then the integral is used to “add” the infinite number of infinitesimals together. [There is actually a way to make infinitesimals mathematically real, but it is too complicated to go into here]

dx

dV = A(x)dx

x - axisa b

To return to the volume example for a minute, we can regard the red solid slice shown above as an infinitesimal volume dV.

Since the thickness, dx, is regarded as essentially 0, we think of dV as a rectangular solid whose volume is the cross sectional area times the thickness, that is A(x)dx - we don’t have to think of the thin sides of the solid dV as having slanted edges. We then “add” them all together by integration to get the entire volume

( )b

V A x dxa

= ∫

y - axis

We can proceed in a similar way if a volume is divided into vertical infinitesimal volumes.

dy

dV = A(y)dy

x - axisc

d

y - axis

( )d

V A y dyc

= ∫

Proceed as follows:

1. Set up the infinitesimal quantity to be integrated first. It will involve some independent infinitesimal dx, dy, etc depending on the physical problem we are attempting to solve, and various other quantities.

2. The independent infinitesimal dx, dy, etc. determines the integral, so all other quantities must be expressed in terms of the corresponding variable.

Thus if we have dx, the we are going to do an x integral, and every other quantity in the integral must be expressed in terms of x. On the other hand, if we have dy, then every other part of the quantity we are integrating must be expressed in terms of y, and so on.

Example. A solid has the unit circle x2 + y2 = 1 as its base. Its cross sections, taken perpendicular to the base and to the x - axis are squares. Find the volume.

Solution. First we look at the base. For each x between −1 and 1 we see that the base has width 2y. Thus the area of the square cross section at x is 4y2, the thickness is dx, and so the infinitesimal volume is dV = 4y2dx. Thus

(x, y)

x axis−1 1

y

131 1 162 24 4(1 ) 43 31 1 1

xV y dx x dx x = = − = − =∫ ∫ − − −

Compute the typical infinitesimal volume dV, then set up the integral that “sums” all these infinitesimals.

Express everything in the integrand in terms of x, since the expression involves dx.

Volumes of Revolution

One of the most common methods of generating solid bodies is by rotating an area about some axis of revolution. For example, suppose that we look at the area under a function f(x) from x = ato x = b, as shown below, and rotate this area 360 degrees about the x axis, generating the solid shown.

xa b

y

x

ab

yy=f(x)

x

xa b

y

x

ab

y

It is easy to see that the cross sections of this solid are disks with thickness dx and bases that are circles of radius y. Thus the infinitesimal volume dV of the cross section is

2 2( )dV y dx f x dxπ π= =

Therefore, the total volume is

2( )b

V f x dxaπ= ∫

y=f(x)

x

Example. Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis.

y = 0, x = π/4, x = π/2.cos( ),y x=

Solution. The area is shown tothe right. When revolved around the x axis, it makes a solid that resembles the nose of a bullet.

The volume is

( ) 22 2 2

44 4 4

2 12( ) cos( ) cos( ) sin( ) 12

V f x dx x dx x dx x

ππ π π

ππ π ππ π π π π = = = = = −∫ ∫ ∫

y = f(x)

If we now rotate the area between two curves about the x - axis, we will in general get cross sections that are washers, rather than disks.

xa b

y

x

ab

y

x

y=f(x)

y=g(x)

The area of the cross section is 2 2 2 2( ) ( ) ( ) ( )f x g x f x g xπ π π − = − so 2 2( ) ( ) .dV f x g x dxπ = −

Thus the volume is 2 2( ) ( ) .b

V f x g x dxa

π = −∫

Example. Find the volume of the solid that results when the region shown below is rotated about the x - axis.

Solution: Here the region is between y = x and y = 2 - x2, which meet at x = 1. Thus

( )22 2 4 22 5 4dV x x dx x x dxπ π = − − = − +

and1

5 31 5 384 25 4 45 3 150 0

x xV x x dx x ππ π = − + = − + =∫

It is just as likely that the solid is formed by rotating an area about the y-axis. Then the disks and washers are horizontal with thickness dy, and they must be added together by a y-integral.

Of course this means that all quantities associated with the area must be expressed in terms of y, and

( )d

V A y dyc

= ∫

Example. Find the volume of the solid that results when the region shown below is rotated about the y - axis.

Solution: Here the region is between x = 1/y and x = 2. Then yvaries from 1/2 to 2. Thus

21 122 42

dV dy dyy y

π π = − = − and

V

11

22

22 1 1 1 94 4 8 42 2 2V dy y

yyππ π π

= − = + = + − = ∫

Example. Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis. 2 , 0, 0, 1xy e y x x−= = = =

Solution. The area is shown tothe right.

The volume is

11 1 1

00 0 0

22 4 4 42( ) 14 4

x x xV f x dx e dx e dx e e

ππ ππ π π − − − − = = = =− = −∫ ∫ ∫

Example. Find the volume of the solid that results when the region enclosed by the given curves is revolved about the y-axis. 2 2, y x x y= =

Solution. The area is shown to the right. The cross section is a horizontal washer whose thickness is dy, so every thing must be expressed in terms of y.

y

In terms of y, the left and right curves are respectively 2 and . x y x y= = The infinitesimal volume is therefore

( ) ( )22 2 4dV y y dy y y dyπ π = − = −

To find the limits, we solve the equation

12 51 1 34 4

2 5 100 0 0

y yV y y dy y y dy ππ π π = − = − = − = ∫ ∫

2 4, or y y y y= =

The only real solutions of this are y = 0 and y = 1, which must therefore be the limits. So The volume is

the solid

Example. Find the volume of the solid that results when the same region is revolved about the x- axis.

Solution. The area is shown to the right. The cross section is a horizontal washer whose thickness is dx, so every thing must be expressed in terms of x. The curves are

x

The infinitesimal volume is therefore

( ) ( )22 2 4dV x x dx x x dxπ π = − = −

2, .y x y x= =

Thus the volume is the same for this solid, namely 3 .10π

Problem. Find the volume of the solid generated when the region enclosed by and y = 0 is revolved about the x - axis.

, 6y x y x= = −

y x=

6y x= −

To do this problem, we need to break the region into two parts as shown, then rotate both parts and add the results.

The curves meet when

2 26 or 0 12 36 13 36 ( 4)( 9)x x x x x x x x x= − = − + − = − + = − −

that is at x = 4 and x = 9.

y x=

6y x= −

( )2 44 42

81 2 00 0

xV x dx xdx ππ π π

= = = =∫ ∫

( )3 26 0 22 82 26 ( )

2 3 304 2 0

uV x dx u du u du

π ππ π π

= − = − = = =∫ ∫ ∫

Thus the total integral volume is 8 328 .3 3π π+ =