Calculus- Early Transcendentals, 2021a

514 Multiple Integration
y
.
......
..
. ......... .
...
. . . ...........
. . ........
x
−1 0 1
Figure 14.18: The approximating parallelogram.
The area of this parallelogram is the length of the cross product:
i j k
〈−r 0 sinθ 0 ,r 0 cosθ 0 ,0〉dθ ×〈cosθ 0 ,sinθ 0 ,0〉dr =
−r 0 sinθ 0 r 0 cosθ 0 0
dθ dr
∣ cosθ 0 sinθ 0 0∣ = 〈0,0,−r 0 sin 2 θ 0 − r 0 cos 2 θ 0 〉dθ dr
= 〈0,0,−r 0 〉dθ dr.
The length of this vector is r 0 drdθ. So in general, for any values of r and θ, the area in the xy-plane
corresponding to a small rectangle anchored at (θ,r) in the rθ-plane is approximately rdrdθ. In other
words, “r” replacesthe“r(?)” in Equation 14.2.
In general, a substitution will start with equations x = f (u,v) and y = g(u,v). Again, it will be straightforward
to convert the function being integrated. Converting the limits will require, as above, an understanding
of just how the functions f and g transform the uv-plane into the xy-plane. Finally, the small
vectors we need to approximate an area will be 〈 f u ,g u ,0〉du and 〈 f v ,g v ,0〉dv. The cross product of these
is 〈0,0, f u g v − g u f v 〉dudv with length | f u g v − g u f v |dudv. The quantity | f u g v − g u f v | is usually denoted
∂(x,y)
∣∂(u,v)
∣ = | f ug v − g u f v |
and called the Jacobian. Note that this is the absolute value of the two by two determinant
∣ f ∣
u g u∣∣∣
,
f v g v
which may be easier to remember. (Confusingly, the matrix, the determinant of the matrix, and the absolute
value of the determinant are all called the Jacobian by various authors.)
Because there are two things to worry about, namely, the form of the function and the region of
integration, transformations in two (or more) variables are quite tricky to discover.
Example 14.17: Integral of an Ellipse
Integrate x 2 − xy + y 2 over the region x 2 − xy + y 2 ≤ 2.
Solution. The equation x 2 − xy + y 2 = 2 describes an ellipse as in Figure 14.19; the region of integration
is the interior of the ellipse. We will use the transformation x = √ 2u − √ 2/3v, y = √ 2u + √ 2/3v.