LetRR
Df(x, y)dx dybe a given double integral. We want to describe how to convert the integral to an equivalent one with a different domain of integration and a correspondingly different integrand.
The setup is to assume that there is a function T: D∗ → D, where D∗ is a subset of R2. We think of T as transforming D∗ onto D, so we assume that T(D∗) = D and that T is one-to- one, except possibly along the boundary of D∗. This last part means that whenever a and b are distinct points of D∗, not on the boundary, then the values T(a) and T(b) are distinct, too:
a 6= b ⇒ T(a) 6= T(b). (For the purposes of integration, anomalous behavior confined to the boundary can be allowed typically without messing things up. See the remarks on page 130.)
The new integral will be an integral over D∗. It helps keep things straight if the coordinates in the domain and codomain have different names, so we think of T as a transformation from the uv-plane to thexy-plane and write T(u, v) = (x(u, v), y(u, v)).
Thus we want to convert the given integral over D with respect to x and y to an integral over D∗ with respect touandv. We outline the main idea. It is similar to what we did in Section5.5to define surface integrals with respect to surface area using a parametrization, though of course there is a big difference between motivating a definition and justifying a theorem. Another difference is that, in the interim, we have learned about the derivative of a vector-valued function.
Figure 7.2: Change of variables: x andy as functions of u andv via transformation T To begin, subdivide D∗ into small subrectangles of dimensions 4ui by 4vj, as in Figure 7.2.
The transformationT sends a typical subrectangle of area4ui4vj in theuv-plane to a small curvy quadrilateral in the xy-plane. Let 4Aij denote the area of the curvy quadrilateral. If we choose a sample point pij in each curvy quadrilateral, we can form a Riemann sum-like approximation of the integral over D:
Z Z
D
f(x, y)dx dy ≈X
i,j
f(pij)4Aij. (7.1)
To turn this into a Riemann sum overD∗, we need to relate the areas of the subrectangles and the curvy quadrilaterals.
To do so, we use the first-order approximation of T. Choose a point aij in the (i, j)th subrect- angle of D∗ such that T(aij) = pij. Then the first-order approximation says that, for points u in
7.1. CHANGE OF VARIABLES FOR DOUBLE INTEGRALS 175 D∗ nearaij:
T(u)≈T(aij) +DT(aij)ã(u−aij)
≈pij+DT(aij)ã(u−aij)
≈ pij −DT(aij)ãaij
+DT(aij)ãu. (7.2)
It may be difficult to get a foothold on the behavior ofT itself, but describing how the approximation transforms the (i, j)th subrectangle is quite tractable.
For this, we bring in some linear algebra. Note that the derivative DT(aij) is a 2 by 2 ma- trix. As shown in Chapter 1, any 2 by 2 matrix A = a b
c d
determines a linear transformation L: R2 → R2 given by matrix multiplication: L(x) = Ax. This transformation sends the unit square determined bye1 and e2 to the parallelogram determined byL(e1) = [ac] andL(e2) =b
d
, which are the columns of A (Figure7.3). We saw in Proposition 1.13 of Chapter 1 that the area
Figure 7.3: A linear transformationLsends the square determined bye1ande2to the parallelogram determined byL(e1) and L(e2).
of this parallelogram is |det
L(e1) L(e2)
|, where
L(e1) L(e2)
is the matrix whose rows are L(e1) and L(e2). This is precisely the matrix transpose At. Thus the area of the parallelogram is
|det(At)|=|detA|, where we have used the general fact that a matrix and its transpose have the same determinant (Proposition 1.14, Chapter1). In particular,Lhas changed the area by a factor of |detA|.
This principle extends easily to a couple of slightly more general cases.
• Given nonzero scalars`and w, letR be the|`|by |w|rectangle determined by`e1 and we2. Then L sends R to the parallelogram P determined by L(`e1) = ` L(e1) and L(we2) = wL(e2). Compared to the original case, the areas of both rectangle and parallelogram have changed by a common factor of|`w|, hence they still differ by a factor of |detA|.
• Suppose that we translate the rectangle R in the previous case by a constant vectora. This results in another|`|by|w|rectangleR0consisting of all points of the forma+x, wherex∈R.
By linearityL(a+x) =L(a)+L(x), soLtransformsR0to the translation of the parallelogram L(R) =P byL(a). Let’s call the translated parallelogramP0. Since translations don’t affect areas,R0 and its image P0 again differ by a factor of |detA|.
This shows the following.
Lemma 7.2. A linear transformationL:R2 →R2 represented with respect to the standard bases by a 2 by 2 matrix Asends rectangles whose sides are parallel to the coordinate axes to parallelograms and alters the area of the rectangles by a factor of |detA|.
Returning to the first-order approximation (7.2) of T, we conclude that multiplication by DT(aij) transforms the subrectangle of area 4ui4vj that contains aij to a parallelogram of area
|detDT(aij)| 4ui4vj. The determinant of the derivative, detDT(aij), is sometimes referred to as theJacobian determinantataij. Hence, after additional translation, the first-order approxima- tion (7.2) sends the subrectangle containingaij to a parallelogram that:
• approximates the curvy quadrilateral of area 4Aij that containspij and
• has area |detDT(aij)| 4ui4vj.
Thus the Riemann sum approximation (7.1) becomes:
Z Z
D
f(x, y)dx dy≈X
i,j
f(pij)ã 4Aij ≈X
i,j
f(T(aij))ã |detDT(aij)| 4ui4vj.
Letting 4ui and 4vj go to zero, this becomes an integral over D∗, giving the following major result.
Theorem 7.3(Change of variables theorem for double integrals). LetDandD∗ be bounded subsets ofR2 and letT:D∗ →Dbe a smooth function such that T(D∗) =Dand that is one-to-one, except possibly on the boundary of D∗. If f is integrable onD, then:
Z Z
D
f(x, y)dx dy= Z Z
D∗
f(T(u, v))|detDT(u, v)|du dv.
In other words,xandy are replaced in terms ofuandv in the functionf using (x, y) =T(u, v) and dx dy is replaced by |detDT(u, v)|du dv =
det ∂x
∂u
∂x
∂y ∂v
∂u
∂y
∂v
du dv.
By comparison, in first-year calculus, if f(x) is a real-valued function of one variable and x is expressed in terms of another variable u, say as x =T(u), then the method of substitution says that:
Z T(b) T(a)
f(x)dx= Z b
a
f(T(u))T0(u)du.
In this case, it seems simplest to understand the result in terms of antiderivatives using the chain rule and the fundamental theorem of calculus rather than through Riemann sums, though we won’t present the details of the argument. In particular, substituting for x in terms ofu includes the substitution dx = T0(u)du = dudxdu. The substitution dx dy = |detDT(u, v)|du dv in the previous paragraph is the analogue in the double integral case. We say more about the principle of substitution in the next section.
Returning to double integrals, in the case of polar coordinates, the substitutions for x and y are given by (x, y) =T(r, θ) = (rcosθ, rsinθ). Then DT(r, θ) =
cosθ −rsinθ sinθ rcosθ
and:
|detDT(r, θ)|=|rcos2θ+rsin2θ|=|r|=r.
By the change of variables theorem, this means that “dx dy =r dr dθ” and that polar coordinates transform integrals as follows.
Corollary 7.4. In polar coordinates:
Z Z
D
f(x, y)dx dy= Z Z
D∗
f(rcosθ, rsinθ)r dr dθ, where D∗ is the region of the rθ-plane that describes Din polar coordinates.