Lecture 15 : Monday May 5th
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15.1 Areas and Volumes
One of the fundamental applications of definite integrals is computing areas and volumes.
We have already seen that if f (x) is a continuous function defined on interval [a, b], and
f (x) ≥ 0 on [a, b]
Z
b
Area between f and the x−axis =
f (x) dx.
a
In this section, we extend this to finding volumes of various shapes. This will be done
in a similar way to finding areas with Riemann sums, where we sliced up the area into
rectangles and took the limit to get an integral. For computing volumes of simple shapes,
we also slice the volume up into pieces whose volume we know, and then let the thickness
of the pieces tend to zero. For example, suppose we want the volume of a cone of height
h whose base is a circle of radius ρ. This is shown below for h = ρ = 1:
0
0.2
0.4
0.6
-1
-0.5
0.8
1
-1
0
-0.5
0
0.5
0.5
1
1
Cone of radius and height 1
The main fact we use here is that if a cylinder has radius r and height h, then the volume
of the cylinder is πr2 h. If we slice the cone horizontally into n slices of thickness h/n,
then the ith slice is at height ih/n from the base of the cone, since the cone has height h.
If n is large, then the ith slice is almost a (very flat) cylinder. To estimate the volume of
this slice, we require the radius of the cone at height ih/n. This requires a little geometry:
if we take a vertical cross section through a diameter of the base of the cone, we get a
triangle whose base has length 2ρ and height is h. The radius of the ith slice is precisely
the width of this triangle at height ih/n – denoted r in the picture below.
1
Radius r
Vertical cross section of the cone
To find r, note that the triangles ABC and ADE are similar. Therefore
BC
AC
=
.
DE
AE
Since AE = h and AC = h − ih/n and DE = ρ, we get BC = ρ(1 − i/n). So the radius
of the slice at height ih/n is ρ(1 − i/n). The volume of this slice is approximately
πρ2 h ³
i ´2
1−
n
n
using our formula for a cylinder of radius ρ(1 − i/n) and height h/n (since the slices all
have thickness h/n). The approximate volume of the cone is therefore
n−1
X
πρ2 ih ³
i ´2
1−
.
n
n
i=0
This is immediately recognizable as a Riemann sum for the function f (x) = πρ2 h(1 − x)2
on the interval [0, 1]. If n → ∞, it follows that
Z 1
1
Volume of cone =
πρh(1 − x)2 dx = πρ2 h.
3
0
In our next example, we do the same for the sphere of radius ρ.
As a second example, we can compute the volume of a sphere of radius ρ. It is enough to
find the volume of a hemisphere. We slice the hemisphere into n horizontal slices, each
of which is approximately a cylinder if n is large and each of which has thickness ρ/n.
To find the radius r of the ith slice, which is at height iρ/n, we use geometry again. A
vertical slice through a diameter of the base of the hemisphere gives a semicircle (shown
in the figure below – the cross section of the ith slice is a shaded rectangle). At height
2
i/n in this semicircle, we draw a horizontal line, half of whose length is the radius r we
want. Since the sphere has radius ρ, the right-angled triangle ABC in the picture below
allows us to compute
r2 + (iρ/n)2 = ρ2
p
and therefore r = ρ 1 − i2 /n2 .
Circular cross sections
So the approximate volume of the ith slice is
³
i2 ´ ρ
πρ2 1 − 2 · .
n
n
Summing over h, we get the Riemann sum
³
i2 ´ ρ
πρ 1 − 2 · .
n
n
h=0
n−1
X
2
This is recognizable as a Riemann sum for f (x) = πρ3 (1 − x2 ) for 0 ≤ x ≤ 1. Therefore
the volume is
Z 1
2
Volume of hemisphere =
πρ3 (1 − x2 ) dx = πρ3 .
3
0
It follows that the volume of the whole sphere is 43 πρ3 .
It is not necessarily the case that the slices should be approximated by cylinders. For
example, if we take a square pyramid, then the slices do not look anything like cylinders.
It makes much more sense to approximate the slices by boxes. For example, if we have a
3
pyramid of height 1 whose base is a square with side length 2, then slicing the pyramid
into n horizontal slices of thickness 1/n, we get that the ith slice is approximately a box
whose base is a square with side length equal to r = 2(1 − i/n) (check this). So the
approximate volume of the ith slice must be
r2 ·
³
1
i ´2 1
=4 1−
· .
n
n
n
Adding up the approximate volumes of these slices, the volume of the pyramid is approximated by
µ
¶2
n−1
X
4
i
1−
.
n
n
i=0
As n → ∞, this Riemann sum becomes
Z 1
4
4
(1 − x)2 dx = .
3
0
So the volume of the pyramid is 4/3.
The method of approximating volume by slices and Riemann sums given above hinges on
the cross sections of the shape being fairly simple. In general, there is no way to compute
the volume of an arbitrary shape using this method. So just how far does this method go?
In the next section, we will consider special types of shapes, called solids of revolution,
whose volumes can be computed in a similar way.
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