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3.5 Change of Variables in Multiple Integrals 117
3.5 ChangeofVariables in Multiple Integrals
Given the difficulty of evaluating multiple integrals, the reader may be wondering if it is
possible to simplify those integrals using a suitable substitution for the variables. The an-
swer is yes, though it is a bit more complicated than the substitution method which you
learned in single-variable calculus.
Recall that if you are given, for example, the definite integral
! "
2
x3 x2−1dx,
1
then you would make the substitution
u=x2−1 ⇒ x2=u+1
du=2xdx
which changes the limits of integration
x=1 ⇒ u=0
x=2 ⇒ u=3
so that we get
!2x3"x2−1dx = !21x2·2x"x2−1dx
1 1 2
!31 #
= 0 2(u+1) udu
! # $
3
= 1 u3/2+u1/2 du , which can be easily integrated to give
2 0
#
= 14 3 .
5
Letustakeadifferentlookatwhathappenedwhenwedidthatsubstitution,whichwillgive
some motivation for how substitution works in multiple integrals. First, we let u = x2−1.
Ontheinterval of integration [1,2], the function x$→ x2−1 is strictly increasing (and maps
[1,2] onto [0,3]) and hence has an inverse function (defined on the interval [0,3]). That is,
on [0,3] we can define x as a function of u,namely
x = g(u) = #u+1.
Thensubstituting that expression for x into the function f(x)=x3#x2−1gives
3/2#
f (x) = f(g(u)) = (u+1) u,
118 CHAPTER3. MULTIPLEINTEGRALS
andweseethat
dx ′ ′
du = g (u) ⇒ dx = g (u)du
1 −1/2
dx = 2(u+1) du,
so since
g(0)=1 ⇒ 0=g−1(1)
g(3)=2 ⇒ 3=g−1(2)
then performing the substitution as we did earlier gives
! ! "
2 2
f (x)dx = x3 x2−1dx
1 1
!
3 #
= 1(u+1) udu , whichcanbewrittenas
0 2
!
3 #
3/2 u·1 −1/2
= 0 (u+1) 2(u+1) du , which means
!2f(x)dx = !g−1(2) f(g(u))g′(u)du.
1 g−1(1)
In general, if x= g(u) is a one-to-one, differentiable function from an interval [c,d] (which
youcanthinkofasbeingonthe“u-axis”)ontoaninterval[a,b](onthex-axis),whichmeans
that g′(u) ̸= 0ontheinterval(c,d), so that a = g(c)andb = g(d), then c = g−1(a)andd =
g−1(b), and
!bf(x)dx = !g−1(b)f(g(u))g′(u)du. (3.17)
a g−1(a)
This is called the change of variable formula for integrals of single-variable functions, and it
is whatyouwereimplicitlyusingwhendoingintegrationbysubstitution. Thisformulaturns
out to be a special case of a more general formula which can be used to evaluate multiple
integrals. We will state the formulas for double and triple integrals involving real-valued
functions of two and three variables, respectively. We will assume that all the functions
involved are continuously differentiable and that the regions and solids involved all have
2
“reasonable” boundaries. The proof of the following theorem is beyond the scope of the text.
2See TAYLOR and MANN,§15.32and§15.62forallthedetails.
3.5 Change of Variables in Multiple Integrals 119
Theorem3.1. ChangeofVariablesFormulaforMultipleIntegrals
Let x=x(u,v)andy=y(u,v)defineaone-to-onemappingofaregionR′ inthe uv-planeonto
aregionR in the xy-plane such that the determinant
% %
% ∂x ∂x%
% %
%∂u ∂v%
J(u,v) = % % (3.18)
%∂y ∂y%
% %
is never 0 in R′.Then %∂u ∂v%
&f(x,y)dA(x,y) = &f(x(u,v),y(u,v))|J(u,v)|dA(u,v) . (3.19)
R R′
We use the notation dA(x,y)anddA(u,v) to denote the area element in the (x,y)and(u,v)
coordinates, respectively.
Similarly, if x = x(u,v,w), y= y(u,v,w)andz= z(u,v,w) define a one-to-one mapping of
asolidS′ in uvw-space onto a solid S in xyz-space such that the determinant
% %
% ∂x ∂x ∂x%
% %
%∂u ∂v ∂w%
% %
%∂y ∂y ∂y%
J(u,v,w) = % % (3.20)
% %
%∂u ∂v ∂w%
% %
% ∂z ∂z ∂z %
% %
is never 0 in S′,then %∂u ∂v ∂w%
'f(x,y,z)dV(x,y,z)='f(x(u,v,w),y(u,v,w),z(u,v,w))|J(u,v,w)|dV(u,v,w) . (3.21)
S S′
The determinant J(u,v)informula(3.18)iscalledtheJacobian of x and y with respect
to u and v, and is sometimes written as
J(u,v) = ∂(x,y) . (3.22)
∂(u,v)
Similarly, the Jacobian J(u,v,w) of three variables is sometimes written as
J(u,v,w) = ∂(x,y,z) . (3.23)
∂(u,v,w)
Notice that formula (3.19) is saying that dA(x,y)=|J(u,v)|dA(u,v), which you can think of
as a two-variable version of the relation dx= g′(u)duin the single-variable case.
Thefollowing example shows how the change of variables formula is used.
120 CHAPTER3. MULTIPLEINTEGRALS
& x−y
Example3.9.Evaluate ex+y dA, where R={(x,y):x≥0,y≥0,x+y≤1}.
R
Solution: First, note that evaluating this double integral without using substitution is prob-
ably impossible, at least in a closed form. By looking at the numerator and denominator of
the exponent of e, we will try the substitution u = x− y and v= x+ y.Tousethechangeof
variables formula (3.19), we need to write both x and y in terms of u and v.Sosolvingfor
x and y gives x = 1(u+v)andy= 1(v−u). In Figure 3.5.1 below, we see how the mapping
2 2
x=x(u,v)= 1(u+v), y= y(u,v)= 1(v−u) maps the region R′ onto R in a one-to-one manner.
2 2
y v
x=1(u+v)
2 1
1
y=1(v−u)
x+y=1 2 R′
u=−v u=v
R x u
0 0
1 −11
Figure 3.5.1 The regions R and R′
Nowweseethat
% %
% ∂x ∂x% % %
% % % 1 1 % % %
%∂u ∂v% % 2 2 % 1 %1% 1
% % % %
J(u,v) = = % % = ⇒|J(u,v)|= = ,
%∂y ∂y% % 1 1 % 2 %2% 2
% % −2 2
%∂u ∂v%
so using horizontal slices in R′, we have
& x−y &
ex+y dA = f (x(u,v), y(u,v))|J(u,v)|dA
R R′
!1!v u 1
= ev 2 dudv
0 −v
!1# u %u=v $
= vev % dv
0 2 %u=−v
= !1v(e−e−1)dv
0 2
v2 %1 1( 1) e2−1
= (e−e−1)% = e− =
4 %0 4 e 4e
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