6 volumes of solids of revolution ii x

Volume and Solids of Revolution II

Following are more examples of solids whose
cross–sectional areas are tractable so that we may
use the Cavalieri’s principle to find their volumes.

One group of such solids has cross–sectional areas
that are formed from the linear cross–sections of the
base regions of the solids.

A solid has a semi–circular
disc of radius r as base.
Its cross-sections,
perpendicular to the
diameter of the base, are
isosceles right triangles as
shown. Find its volume.
Example A.

Example A.
A solid has a semi–circular
disc of radius r as base.
Its cross-sections,
perpendicular to the
diameter of the base, are
isosceles right triangles as
shown. Find its volume.
r

We calculate the volume of half of the solid.
r

r
isosceles right triangles

rr

r
Set the x measurement as shown from x = 0 to x = r.
rr
x=0
x=r
x

√r2 – x2
r
Given an x, the cross-sectional length is √r2 – x2.
rr
x=0
x=r
x

So the cross-section of the solid is a right isosceles
triangle with area ½ (√r2 – x2)2 = ½ (r2 – x2).
√r2 – x2
r
rr
x=0
x=r
x

So the volume of the entire solid (two halves) is
∫x=0
r
dxr2 – x2
√r2 – x2
r
rr
x=0
x=r
x

∫x=0
r
dxr2 – x2
= r2x – x3
3 x=0
r
√r2 – x2
r
rr
x=0
x=r
x

∫x=0
r
dxr2 – x2
= r2x – x3
3 x=0
r
= 2r3
3
x=0
x=r
x√r2 – x2
r
rr

Note that if the vertical leg of the triangle is above
the circular arc instead of above the diameter,

the circular arc instead of above the diameter,
(These two solids have the same cross-sections hence the same volume.)
r

the circular arc instead of above the diameter, the
resulting solid looks different but has the same
volume as before because they have the same
cross-sections.
(These two solids have the same cross-sections hence the same volume.)
r
‘

There are other ways to find volumes by the
Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus. Basically, we
subdivide the solid into small manageable pieces
and sum the volumes of all these pieces.

and sum the volumes of all these pieces. Then we
cut the solid into smaller and smaller such pieces
and pass the sums into a definite integral.

For example, a solid of revolution is generated by a
template,

For example, a solid of revolution is generated by a
template, then a strip of cross–section that is parallel
to the axis of rotation will generate a shell that is
approximately a cylindrical shell.

The volume of a cylindrical shell depends on the
radius r, the thickness Δx, and the height h.
Δx
r
h

To approximate the volume of a cylindrical shell,
slice it open and unroll it into a flat slab.
Δx
r
h
Δx

Δx
r
h
Δx
It’s almost a rectangular box.

Δx
r
h
Δx
It’s almost a rectangular box.
2πr
Δx
h≈
Δx

It’s almost a rectangular box. We use the volume of
the rectangular box which is 2πr*Δx*h to approximate
the volume of this slab.
Δx
r
h
Δx
2πr
Δx
h≈
Δx

Δx
r
h
2πr
Δx
h≈
Therefore the volume of the cylindrical shell is
roughly 2πrhΔx.

Δx
r
h
2πr
Δx
h≈
roughly 2πrhΔx.
x=a x=b
y=f(x)
Suppose a solid of revolution is formed by rotating a
template that is bounded by y = f(x), from x = a to
x = b, around a vertical axis.

Δx
r
h
2πr
Δx
h≈
roughly 2πrhΔx.
x=a x=b
y=f(x)
Suppose a solid of revolution is formed by rotating a
template that is bounded by y = f(x), from x = a to
x = b, around a vertical axis.
x=a
x=b
y=f(x)

x=a x=b
Partition [a, b] into n equal size subintervals
{x0=a, x1, x2, ..xi, .. xn=b}, xi be an arbitrary point in
[xi–1, xi].
*
xi
*

x=a x=b
[xi–1, xi]. Let Δx be the width of the subinterval.
*
xi
*

x=a x=b
Then the shell generated by the strip over [xi–1, xi]
is approximately the cylindrical shell with height f(xi),
width Δx, and radius D(xi).
*
xi
*
*
*

x=a x=b
y=f(xi)
*
xi
D(x )
*
*
*
*
*

x=a x=b
y=f(xi)
*
xi
D(x )
*
*
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

x=a x=b
y=f(xi)
*
xi
D(x )
2πD(xi)f(xi)Δx.
*
*
x=a x=bxi
D(xi )
* *
Hence the volume of this cylindrical shell is
*
*
**
y=f(xi)*
*

≈ 2πD(xi)f(xi)Δx.* *
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

The sum of the volumes of all such cylindrical shells
approximates the volume of the solid of revolution V.
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

Volume of V ≈
i=1
n
∑
2πD(xi)f(xi)Δx* *
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

Volume of V ≈
i=1
n
∑
In fact the volume is its limit as Δx0 or as n ∞.
2πD(xi)f(xi)Δx* *
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

Volume of V ≈
i=1
n
∑
Therefore the volume of V = lim
i=1
n
∑
n ∞
2πD(xi)f(xi)Δx* *
2πD(xi)f(xi)Δx* *
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

Volume of V ≈
i=1
n
∑
i=1
n
∑
n ∞
By the FTC, the volume of V = ∫x=a
b
2πD(xi)f(xi)Δx* *
2πD(xi)f(xi)Δx* *
2πD(x)f(x)dx
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*
radius* height

Volume of V ≈
i=1
n
∑
i=1
n
∑
n ∞
By the FTC, the volume of V = ∫x=a
b
2πD(xi)f(xi)Δx* *
2πD(xi)f(xi)Δx* *
2πD(x)f(x)dx
This known as the shell method.
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*
radius* height

Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x

Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2.
Example B.
y = x2
y = 2x

x = 0 to x = 2.
Example B.
y = x2
y = 2x
x=3

x = 0 to x = 2. Given a vertical cross–section at x,
its length is
Example B.
y = x2
y = 2x
x=3
x

its length is (2x – x2)
Example B.
y = x2
y = 2x
x=3
2x–x2
x

its length is (2x – x2) which is the height of the
cylindrical shell it forms.
Example B.
y = x2
y = 2x
x=3
2x–x2
x

cylindrical shell it forms.
Example B.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x

cylindrical shell it forms. The radius is D(x) =
Example B.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x

cylindrical shell it forms. The radius is D(x) = (3 – x).
Example B.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x
3–x

Hence the volume is
Example B.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x
3–x
∫x=0
2
2π(3 – x)(2x – x2)dx =

Hence the volume is
Example B.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x
3–x
∫x=0
2
2π(3 – x)(2x – x2)dx =16π
3

b. Revolve the area as shown around y = –1,
find the volume it formed by the shell method.
y = x2
y = 2x
y = –1

We are to track a horizontal
cross–section, parallel to the axis
of rotation, as it orbits around
y = –1.
y = x2
y = 2x
y = –1

y = –1. Hence we need to solve
each equation for x.
y = x2
y = 2x
y = –1

y = x2
y = 2x
y = –1
y = x2  x = f(y) = √y, and
y = 2x  x = g(y) = y/2.

y = x2
y = 2x
y = –1
y = x2  x = f(y) = √y, and
y = 2x  x = g(y) = y/2.
x = √yx = y/2

y = x2
y = 2x
y = –1
y = x2  x = f(y) = √y, and
y = 2x  x = g(y) = y/2.
x = √yx = y/2
The template spans from y = 0 to y = 4.

y = x2
y = 2x
y = –1
y = x2  x = f(y) = √y, and
y = 2x  x = g(y) = y/2.
x = √yx = y/2
The template spans from y = 0 to y = 4.
For a fixed y, the horizontal cross–section forms
a cylindrical shell as it revolves around y = –1.

y = –1
x = √y
x = y/2
y

y = –1
x = √y
x = y/2
y = –1
x = √y
x = y/2
y
y

At y, the horizontal cross–section
has length (√y – y/2).
y = –1
x = √y
x = y/2
y
y = –1
x = √y
x = y/2
y
√y – y/2

The radius of rotation is (y + 1).
y = –1
x = √y
x = y/2
y
y = –1
x = √y
x = y/2
y
√y – y/2
y + 1

The radius of rotation is (y + 1).
Therefore the volume of the solid is
∫y=0
4
2π(√y – y/2)(y + 1)dy = 104π
15
y = –1
x = √y
x = y/2
y
y = –1
x = √y
x = y/2
y
√y – y/2
y + 1

6 volumes of solids of revolution ii x

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