-
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the line x + y =1.
Solution. We recognise an area as numerically equal to the volume of a solid of height 1,
so if R is the region described, the area is
"
1 y= 1-x2 1
1 dx dy = dy dx = 1 - x2 - (1 - x) dx = . . . .
R 0 y=1-x 0
And we also find that Fubini provides a method for actually calculating integrals; some-
times one way of doing a repeated integral is much easier than the other.
9.7. Example. Sketch the region of integration for
1 1
exy
x2 dx dy.
0 y
Evaluate the integral by reversing the order of integration.
Solution. The diagram in Fig.9.2 shows the area of integration.
Interchanging the given order of integration, we have
1 1 1 x
exy exy
x2 dx dy = x2 dy dx
0 y 0 0
1
exy
x2 x
= dx
x
1 1
2
ex
= x dx - xdx
0 0
1 2 x2 1 1
ex e
= - = ( -2) .
2 2 2
9.8. Exercise. Evaluate the integral
"
( x - y2) dy dx,
R
where R is the region bounded by the curves y = x2 and x = y4.
9.3. CHANGE OF VARIABLE THE JACOBIAN 97
9.3 Change of Variable the Jacobian
Another technique that can sometimes be useful when trying to evaluate a double (or triple
etc) integral generalise the familiar method of integration by substitution.
Assume we have a change of variable x = x(u, v) and y = y(u, v). Suppose that the
region S in the uv - plane is transformed to a region S in the xy - plane under this
transformation. Define the Jacobian of the transformation as
"x "x
"(x, y)
J(u, v) = "u "v = .
"y "y
"(u, v)
"u "v
It turns out that this correctly describes the relationship between the element of area dx dy
and the corresponding area element du dv.
With this definition, the change of variable formula becomes:
f(x, y) dx dy = f(x(u, v), y(u, v)) |J(u, v)| du dv.
S S
Note that the formula involves the modulus of the Jacobian.
9.9. Example. Find the area of a circle of radius R.
Solution. Let A be the disc centred at 0 and radius R. The area of A is thus dx dy. We
A
evaluate the integral by changing to polar coordinates, so consider the usual transformation
x = r cos ¸, y = r sin ¸ between Cartesian and polar co-ordinates. We first compute the
Jacobian;
"x "y "x "y
=cos ¸, =sin ¸, = -r sin ¸, = r cos ¸.
"r "r "¸ "¸
Thus
"x "x
"r "¸ cos ¸ -r sin ¸ = r(cos2 ¸ +sin2 ¸) =r.
J(r, ¸) = "y "y =
sin ¸ r cos ¸
"r "¸
We often write this result as
dA = dx dy = rdrd¸
Using the change of variable formula, we have
R 2À
R2
dx dy = |J(r, ¸)| dr d¸ = rdrd¸ =2À .
2
A A 0 0
We thus recover the usual area of a circle.
Note that the Jacobian J(r, ¸) = r > 0, so we did indeed take the modulus of the
Jacobian above.
9.10. Example. Find the volume of a ball of radius 1.
98 CHAPTER 9. MULTIPLE INTEGRALS
Solution. Let V be the required volume. The ball is the set {(x, y, z) | x2 + y2 + z2 d" 1}.
It can be thought of as twice the volume enclosed by a hemisphere of radius 1 in the upper
half plane, and so
V =2 1 -x2 -y2 dx dy
D
where the region of integration D consists of the unit disc {(x, y) | x2 + y2 d" 1}. Although
we can try to do this integration directly, the natural co-ordinates to use are plane polars,
and so we instead do a change of variable first. As in 9.9, if we write x = r cos ¸, y = r sin ¸,
we have dx dy = rdrd¸. Thus
V =2 1 -x2 -y2 dx dy = 2 ( 1 - r2) rdrd¸
D D
2À 1
= 2 d¸ 1 - r2 dr
0 0
1
(1 - r2)3/2
= 4À -
3
4À
= .
3
Note that after the change of variables, the integrand is a product, so we are able to do the
dr and d¸ parts of the integral at the same time.
And finally, we show that the same ideas work in 3 dimensions. There are (at least) two
co-ordinate systems in R3 which are useful when cylindrical or spherical symmetry arises.
One of these, cylindrical polars is given by the transformation
x = r cos ¸, y = r sin ¸, z = z,
and the Jacobian is easily calculated as
"(x, y, z)
= r so dV = dx dy dz = rdrd¸dz.
"(r, ¸, z)
The second useful co-ordinate system is spherical polars with transformation
x = r sin Æ cos ¸, y = r sin Æ sin ¸, z = r cos Æ.
The transformation is illustrated in Fig 9.3.
It is easy to check that Jacobian of this transformation is given by
dV = r2 sin ÆdrdÆd¸ = dx dy dz.
9.11. Example. The moment of inertia of a solid occupying the region R, when rotated
about the z - axis is given by the formula
I = (x2 + y2)ÁdV.
R
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