Tutorial 4
1. Volume
Generalizing last week idea of volume between two curves, we can think of solids of revolutions obtained by rotating a disk or a washer in 3D space. To get a volume, we set up an
integral as a function of a variable, say x, expressing a slice element in terms of area.
The volume can be though of as a approximation by taking slabs of the solid S, with thickness
∆x, and then taking the limit of this approximation as ∆x → 0. We get in an analogous fashion
to the area between curve the volume, this time expressed as a function of the slices area of S,
namely
V ( S) =
Z
I
A( x)dx
1.1 Volume: the disk method
The first method is to consider a typical element of S to be a disc of area A( x) = π R( x)2 with
thickness dx.
Remark
The above problem can also easily be solved as a triple integral. If you are in a polar-cylindrical
referential, we have variables θ ∈ [0, 2π ], a radius r and a height z; such integral is of the form,
after a change of variable x = r cos(θ ), y = r sin(θ ), z = z, one gets
ZZZ
dxdydz =
ZZZ
rdrdzdθ =
Z 2π Z zu Z f ( z)
zl
0
0
rdrdzdθ = 2π
Z zu
f ( z)2
zl
2
dz
using the fact that our integration with respect to z, r does not depend on θ, we get the 2π
factor. Then, our integral expresses the stretching of a point element to a line r, rotating it
around a circle and revolving this into the space given a height function that governs the
shape of our solid S.
1
Finding the volume of a given solid is usually not a problem of integration (at this stage), but
of setup. Given functions and restrictions in the two-dimensional space ( X, Y ), we need to
determine
◦ With respect to which variable (either x or y) we take slices;
◦ Around which line y = a (where x-axis corresponds to y = 0) or x = b (where the y-axis is
a special case which is equivalent to x = 0) we are revolving, and whether our solid touches
the line.
◦ Whether we have set an annulus (between two functions) or a full disk.
◦ Remember to consider the radius as a function of the distance between the axis of rotation,
as either f ( y) − a or f ( x) − b
1. Find the volume of the solid obtained by rotating the region bounded by the given curves
about the specified line. Sketch the region, the solid, and a typical disk.
(a) y = 1/ x, x = 1, x = 2 about the x-axis.
Solution. We are revolving a full circle around the x-axis with radius 1/ x; the setup for the
integral is thus
Z 2
1
A( x)dx =
Z 2
1
h π i2
1
1
= π 1−
π 2 dx = −
x
x 1
2
1
0.8
0.5
0.6
1
0.5
0.4
−1
−0.5
x
1
1.5
2
0.2
0
0
0.5
1
1.5
2
(b) y = sec( x), y = 1, x = −1, x = 1; about the x-axis.
2
Solution. Using the linearity, we see that we have an annulus, the outside region is bounded
by sec( x) [the outer radius] , while the inner cylinder has radius 1 [the inner radius].
1.5
1
1
−2
−0.5
x
1
2
−1
−1
0.5
−1
−1
1
0
0.5
1
Thus, we get A( x) = π ( Ro ( x))2 − π ( Ri ( x)) = π (sec2 ( x) − 1). The bounds for the integral
are with respect to x; thus
Z 1
−1
A( x)dx = π
Z 1
−1
sec2 ( x)dx − 2π = π (tan(1) − tan(−1)) − 2π = 2π (tan(1) − 1)
2
(c) y = x 3 , x = 1, y = 0; about the x-axis (i.e. y = 0).
Solution. This time, we are revolving around the y-axis, so little more care is needed.
Carefully looking at the picture, you might realize that what we are revolving does not touch
the y-axis all along, but rather is between the cylinder of radius 1 around the y-axis and the
curve. Our washer is an annulus, and we thus want to have the radius in terms of R( y) (the
3
inverse function being x = y 2 ), we setup as usual, with the bounds y = 0, y = 1, the latter
3
can be found by plug-in since (1) 2 = 1.
Z 1
0
A( y)dy =
Z 1
0
πdy −
Z 1
0
h y i1
3
π y3 dy = π − π
= π
4 0
4
(d) y = x2 , y = 4; about y = 4
3
1
1
0.8
0.5
0.6
−1
0.4
1
−1
0.2
0.2
0.4
0.6
0.8
1
x
1
Solution. This is the first example in which we are not rotating around the y or x axis
directly. Our solid S does touch the axis y = 4, and the cross-section has thickness dx; Note
that our cross-section has radius 4 when x = 0 and 0 at x = ±2; in fact, the radius is 4 − x2 (a
picture might help here).
8
4
6
3
4
2
2
1
−2
−2
−1
0
1
−2
2
−1
1
2
2
The setup for A( x) = π (4 − x2 )2 = π (16 − 8x2 + x4 ), which we integrate from x ∈ [−2, 2]
Thus,
8 3 1 5 2
512
V=
A( x)dx = π
(16 − 8x + x )dx = π 16x − x + x =
π
3
5
15
−2
−2
−2
Z 2
Z 2
2
4
4
Notice that we have two choices; either we take twice the value of the integral, from x = 0 to
x = 2 using a symmetry argument, or we refrain from this and take absolute values (otherwise, you will get a volume of zero, and you should go back to see why this does not make
sense).
(e) y = x2 , x = y2 ; about x = −1
Solution. We may note that we are rotating a solid far from its edges with q
slices dy; here
2
2
the inner radius of the annulus is y − (−1) = y + 1 and the outer radius is ( y) + 1. The
curve goes from x = 0 to x = 1 and the integral for the volume can be expressed as
Z 1
0
Z
√
2
2
2
π ( y + 1) − ( y + 1) dy = π
1
0
√
π y + 2 y + 1 − y4 − 2y2 − 1 dy
2 3
y5
2
= π y + 2 · y2 −
− y3
3
5
3
29
=
π
30
2
1
2.5
1
2
0.8
0
0.6
1.5
0.4
1
0.2
0.5
−1
−0.5
−4
0
0.5
1
−2
2
x
−2
1.5
2. Find the volume of a frustum of a pyramid with square base of side b, square top of side a
and height h
Solution. What we want is the equation of the curve (linear equation) that governs the radius
5
Figure 1: Frustum of a pyramid with bases length b, a and frustum height h
a
b
of the pyramid. It is not hard to see that by centering the axes so that the base is at the origin
and the missing vertex on the y-axis. We get the intercept on the y axis using similar triangles.
If the height of the missing pyramid is g, then
b
a
=
g
h+g
⇒g=
and the height of the pyramid with base b is h +
ah
b−a
ah
. The slope is
b−a
h−0
2h
y2 − y1
= a b =
x2 − x1
a−b
2 − 2
we thus have the equation for the curve as y =
x=
2h
ah
x+
or equivalently,
a−b
b−a
b
a−b
y+
2h
2
Since the slice corresponds to squares with dimension A( y) = 4x2 (why?), replacing in the
above formula for x in terms of y gives
Z h
0
Z h
a−b
b
A( y)dy = 4
y+
dy
2h
2
0
Z h
( a − b)2 2 2b(b − a)
2
=
y
+
y
+
b
dy
h2
h
0
h
( a − b)2 3 b(b − a) 2
2
=
y +
y +b y
3h2
h
0
6
=
1 2
( a + ab + b2 )h
3
Remark
We are accustomed to working with y as a function of x; finding the equation of the curve in
terms of x directly would have been faster; there is nothing wrong in doing so.
3. Find the volume of a tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4 cm and 5 cm.
Solution. Without loss of generality, say the height is 5 cm, and we have a right angle triangle
for base, with perpendicular sides 3 cm and 4 cm, with area 6. We want to express the integral
as a function of height, with cross element being the triangle similar to the one of dimension
3 × 4. We can judge the area of the base triangles by considering similar pyramids, with height
5 − z; using similar triangles with sides
b
3
= 1
5
5−z
or simply scaling by factor of
Z 5
0
5−z
5
2
b
= 2
5
5−z
2
the area of the base. That gives
Z
1 5
A( z)dz =
25 − 10z + z2 dz
25 0
5
1
1
=
25z − 5z2 + z3 = 10cm3
25
3
0
4. Set up an integral for the volume of a solid torus with radii r and R and by interpretin the
integral as an area, find the volume of the torus.
Solution. The torus consists of a circle centered at ( R, 0) with radius r, rotated 2π radians
around the y axis. A slice would consist of the circle, which can be described by the equations
( x − R)2 + y2 = r2
7
q
We can decompose this into two equations, corresponding to x = R ± r2 − y2 . The range of
y is (−r, r) and setting this as an area between
q
q two functions, the outer function f ( y) = R +
r2 − y2 and the inner function g( y) = R −
Z r
−r
r2 − y2 , we have
Z r
f ( y)2 − g( y)2 dy
0
Z r
q
=
π4R r2 − y2 dy
A( y)dy = 2
π
−r
What we have corresponds to a half circle of radius r, the area of which is
the final answer for the volume of the torus as 2Rπ 2 r2
1 2
πr , which gives
2
5. (?) A hole of radius r is bored through the center of a sphere of radius R > r. Find the
volume of the remaining portion of the sphere.
 Do not try to simply subtract the volume of the cylinder from that of the sphere.
1.2 Volume: the shell method
It is not always easy to apply the disk method, for functions such as cos( x). Another approach
that can be used is called the shell method, and consist in reparametrizing the problem as to
integrate (see Remark) to integrate first the height and expressing it in terms of the radius.
This gives a factor 2πr (corresponding to the circumference), along with a radius element dr
and a height function h(r), possibly the difference between the functions.
The volume element is V = 2πrh∆r. The reference manual gives a good way to remember
the formula; think of unwinding the shell; one get then a “box” of height f ( x), circumference
2π x and thickness dx.
6. Use cylindrical shells to find the volume generated by rotating the region bounded by the
given curves about the specified axis. Sketch the region.
(a) y = x2 − 6x + 10, y = − x2 + 6x − 6; about the y-axis;
Solution. The two curves intersect at the points (2, 2) and (4, 2); since we are rotating about
the y-axis, we take circumference to be 2π x and the height of a box will be the difference
between f 1 ( x) = − x2 + 6x − 6 and f 2 ( x) = x2 − 6x + 10. The shell method gives then
8
volume
Z 4
2
V ( x)dx =
Z 4
2
2π x − x2 + 6x − 6 − ( x2 − 6x + 10) dx
Z 4
−2x3 + 12x2 − 16x dx
2
4
1
= 4π − x4 + 2x3 − 4x2
4
2
= 2π
= 16π
3
2
2
1
5
1
−4
0
1
2
3
−2
2
x
4
−5
4
√
(b) x = y, x = 0, y = 1; about the x-axis.
Solution. The axis is different, so this time we want to have areas in terms of integrals as a
function of y. The curve touches the y axis and we integrate thus
Z 1
0
√
2π y ydy = 2π
Z 1
0
3
y 2 dy =
4
π
5
(c) y = 4x − x2 , y = 8x − 2x2 ; about x = −2
Solution. This is an interesting solid of revolution; see the picture. Note that in this example,
9
1
1
1
0.8
−1
0.6
0.4
x
1
−1
−1
0.2
0
0.2
0.4
0.6
0.8
1
we keep the curves, but the radius element will be 2 + x; we have
Z 4
0
2
2
Z 4
2π (2 + x)(4x − x2 )dx
4
1
2
= 2π 4x2 + x3 − x4
3
4
0
256
π
=
3
2π (2 + x) (8x − 2x ) − (4x − x ) dx =
0
8
8
6
6
4
4
2
2
x
−8 −6 −4
−2
0
1
2
3
2
−5
4
4
10
(d) y = x2 , x = y2 ; about y = −1
Solution. Although the axis is not the same, due to symmetry we already have found the
answer in Problem 1e. See the graph; we have radius y + 1 for the shell and the value of y
varies between 0 and 1; the volume is given by
V=
Z 1
0
2π ( y + 1)
Z 1
√
y − y2 dy
√
y − y3 − y2 dy
0
1
2 3
1
1
2 5
y 2 + y 2 − y4 − y3
= 2π
5
3
4
3
0
29
=
π
30
= 2π
3
y2 +
7. Describe the volume represented by
Z π /2
0
2π ( x + 1) cos( x)dx
Solution. We rotate the curve cos( x) bounded by the x-axis (y = 0) from 0 to π /2 around
the axis x = −1. This gives the resulting solid.
8. Use cylindrical shells to find the volume of the torus of Problem 4.
Solution. Circumference will be 2π xq, element slice corresponds to dx for thickness, and
finally the height is, using symmetry, 2 r2 − ( x − R)2 . When x = R, we get the full radius. x
ranges from R − r to R + r, the volume integral will be
V=
=
=
Z R +r
R −r
Z r
−r
Z r
−r
2π x2
q
r2 − ( x − R)2 dx
4π (u + R)
4πu
p
p
r2 − u2
r2 − u2 +
Z r
−r
4π R
p
r2 − u2
and the first integral vanishes because it is an odd function. The second amounts back to what
we had in the last step of the solution to Problem 4
11