Calculus Of One Real Variable – By Pheng Kim Ving

12.7 
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1. Surfaces Of Revolution 
Consider a sphere of radius r. See Fig. 1.1. We wish to find the surface area
of this sphere. For this purpose we set up
an xycoordinate
system in such a way that its origin is at the centre of the sphere. In Fig.
1.1 we add the zaxis
to get
Fig. 1.1 Surface Area Of A Sphere. 
the xyzcoordinate
system to make it clear that the sphere is in the 3D space. The sphere
intersects the xyplane
in a
circle. Clearly we can consider the surface of the sphere as generated by
revolving the upper semicircle about the xaxis.
We can also consider it as generated by revolving the lower semicircle about
the xaxis. Its
area is calculated in Example
5.1. The upper and lower semicircles are graphs of functions.
When the graph of a function is revolved (rotated) about the
xaxis, it generates a
surface, called a surface of revolution.
See Fig. 1.2. In general, when a plane curve is revolved about a line in the
plane of the curve, it generates a surface
called a surface of revolution. In this section we'll find areas of
surfaces of revolution.
Fig. 1.2
Graph of f, when revolved about xaxis,
generates a surface of revolution.

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2. Revolution About The xaxis 
Fig. 2.1
Area of element of surface of revolution
swept

the circumference of revolution at x_{i}_{–1}) and P_{i}_{–1 }P_{i} is the length of
chord P_{i}_{–1 }P_{i}.
Now r(x_{i}_{–1}) =  f(x_{i}_{–1}); the absolute
value is necessary because a radius must be positive or 0 and we don't require
the restriction that f(x) be positive or 0 on
[a, b], so that general f(x) may be sometimes negative as well as sometimes
positive. Wherever f(x) is negative, we
take its absolute value for the radius of the circle generated by the point (x, f(x));
see Examples 2.1 and 2.2 below. By
Section
12.6 Part 2, there's c_{i}
in [x_{i}_{–1}, x_{i}]
such that:
In Section
12.6 Parts 1 and 2,
we approximated the length of the graph by the sum P_{0 }P_{1} +P_{1 }P_{2} + ... +
P_{n}_{–1 }P_{n} of the chords. So
here we of course approximate the area S
generated by the graph over [a,
b] by the sum S_{1} +
S_{2} + ... + S_{n} of the areas generated
by the chords. Consequently:
We've just shown that:
The area S of the surface of revolution for y = f(x) from x = a to x = b about the xaxis is:

Let s
be the arc length of the graph of f
over [a, b]. As discussed in Section
12.6 Part 3, the arc length element or
differential of arc length for f
at x is the
length ds, as
shown in Fig. 2.2, of a short tangent line segment at x, given by:
So Eq. [2.1] can be written as:
Fig. 2.2

The area S can be regarded as the integral of the differential area dS:

The fact that S
is the integral of dS
is consistent with the fact that s
is the integral of ds.
If you forget Eq. [2.1], try to
obtain it by going thru Eqs. [2.2], [2.3], and [2.4].
Find the area of the surface of revolution obtained by
revolving the graph of y
= f(x) = 2x from x
= –3 to x = –1
about
the xaxis.
Incorrect Solution
Let S be the required area. Then:
EOS
Of course the solution above is incorrect, since an area
can't be negative. The culprit is the incorrect absolute value. It
must be that 2x = –2x,
as 2x < 0 on
[–3, –1]. The correct solution is given below.
Correct Solution
Let S be the required area. Then:
EOS
A sketch of the graph of f
for help doesn't have to be drawn if not asked for. Simply recall the
generalsituation Fig. 2.2
mentally, apply Eq. [2.1], and get the correct absolute values.
Calculate the area of the surface generated by revolving the arc of y = x^{3} from x = –1 to x = 2 about the xaxis.
Let S be the desired area. Then:
EOS
If f(x) (or y) changes sign over [a, b],
we must consider subintervals of [a,
b] where it's positive and
where it's negative
in order to get its correct absolute values, then add up the integrals over
those subintervals.
Again a sketch of the graph of f for help doesn't have to be drawn if not asked
for. Simply recall the generalsituation Fig.
2.2 mentally, apply Eq. [2.1], and get the correct absolute values.
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3. Revolution About The yaxis 
Suppose the graph of a continuously differentiable function f over [a, b]
is revolved about the yaxis.
See Fig. 3.1. The
graph generates a surface of revolution. We're going to find the area S of this surface. Let x be an arbitrary point in [a,
b]. The tangent line segment
at x whose
length is the differential arc length ds
describes the differential area dS
of S.
The area S is the
integral of the differential area dS.
We have:
the absolute value being required since there are x < 0 if a < 0. So:
The area S of the surface of revolution for y = f(x) from x = a to x = b about the yaxis is:

Fig. 3.1
Differential surface area dS is described by differential arc

Remark that in Eq. [3.1], the integration is still along the
xaxis, not along the yaxis, although the
revolution is this time
about the yaxis!
Let S be the required area. Then:
A sketch of the graph of f
for help doesn't have to be drawn if not asked for. Simply recall the
generalsituation Fig. 3.1
mentally, apply Eq. [3.1], and get the correct absolute values.
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4. General Case 
Fig. 4.1
Area of revolution is integral of
differential area. Radius of

Fig. 4.2
Area of revolution is integral of
differential area. Radius of

In summary:
The area S of the surface of revolution for f(x) from x = a to x = b:

Find the area of the surface generated by revolving the line segment y = x + 2 from y = 2 to y = 5 about the line y = 4.
The xcoordinate
of the point of intersection of y
= x + 2 and y = 4 is x where x
+ 2 = 4 or x = 2. From
y = x + 2 we
have x = y – 2. When y increases from 2 to 5 along
the line y = x + 2, x increases from 2 – 2 = 0 to 5 – 2 = 3. Let S be
the desired area. Then:
EOS
Because the formulas for areas of surfaces of revolution
presented in this section correspond to integration along the
xaxis, we must change any yinterval to its
corresponding xinterval
in order to get the xlimits
of integration to be
utilized in these formulas.
A sketch of the graph of f
for help doesn't have to be drawn if not asked for. According to the case at
hand, simply recall
the generalsituation either Fig. 4.1 or Fig. 4.2 mentally, apply either Eq.
[4.1] or Eq. [4.2], and get the correct absolute
values. Eq. [4.3] can be used in any case.
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5. Applications 
Fig. 5.1 Graph For Example 5.1. 
EOS
Problems & Solutions 
1. Find the area of the surface of revolution obtained by revolving the graph of:
from x = 1 to x = 4 about the xaxis.
Let S be the required area. Then:
2. Calculate the area of the surface of revolution obtained by revolving the curve:
Let S be the desired area. Then:
3. Compute the area of the surface generated by revolving the arc y = ln x from x = 1 to x = e about the yaxis.
Solution
Let S be the required area. Then:
4. Find the area of the surface obtained by revolving
the line segment x = y + 3 from y = –5 to y = 1 about the line
x
= 2.
From x
= y + 3 we
have y = x – 3. When y increases from –5 to 1 along
the line x = y + 3, x increases from – 5 + 3 =
–2 to 1 + 3 = 4. Let S
be the desired area. Then:
The equation of the line segment (0, h)–( r,
0) in the xycoordinate
system with its origin at the center of the circular
base of the cone is:
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