Multivariate Calculus - Bruce E. Shapiro

LECTURE 16. DOUBLE INTEGRALS OVER RECTANGLES 131
Next, we evaluated each of the two inner integrals:
∫ ( (
6 ( ) ∣ ))
R f(x, y)dA = x y| 4 1 ∣∣∣
4
0
+
0
2 y2 dx
0
∫ 6
(
= x (4 − 0) +
=
0
∫ 6
0
(4x + 8)dx
( 1
2 (16) − 1 2 (0) ))
dx
After the inside integrals have been completed what remains only depends on the
other variable, which is x. There is no more y in the equation:
∫ 6
R f(x, y)dA = (4x + 8)dx
0
= (2x 2 + 8x) ∣ ∣ 6 0
= 2(36) + 8(6) − 2(0) − 8(0) = 72 + 48 = 120
Usually the Riemann sum will only yield a good approximation, but not the exact
answer, with the accuracy of the approximation improving as the squares get smaller.
It is only a coincidence that in this case we obtained the exact, correct answer.
Example 16.3 Evaluate
∫ 7
4
(∫ 5
1
)
(3x + 12y)dx dy
Solution. We evaluate the inner integral first. Since the inner integral is an integral
over x, within the inner integral, we can treat y as a constant:
∫ 7
(∫ 5
) ∫ 7
( ∫ 5 ∫ 5
)
(3x + 12y)dx dy = 3 xdx + 12y dx dy
4 1
4 1
1
∫ ( (
7
∣ )
1 ∣∣∣
5 (
= 3
4 2 x2 + 12y x| 1) )
5 dy
1
∫ 7
( ( 25
= 3
2 − 1 )
)
+ 12y (5 − 1) dy
2
=
4
∫ 7
4
(36 + 48y) dy
After the integration over x is completed, there should not be any x’s left in the
equation – they will all have been “integrated out” and what remains will only
depend on y.
∫ 7
(∫ 5
) ∫ 7
(3x + 12y)dx dy = (36 + 48y) dy
4
1
4
= (36y + 24y 2 ) ∣ ∣ 7 4
= [(36)(7) + 24(49) − (36)(4) − 24(16)]
= 252 + 1176 − 144 − 384 = 900
Math 250, Fall 2006 Revised December 6, 2006.