524 CHAPTER 26. THE RIEMANNN INTEGRAL ON Rp
For each y between 0 and 4, the variable x, goes from y to 4.∫ 4
y
(x2y+ yx
)dx =
883
y− 13
y4 − 12
y3
Now ∫R
f dA =∫ 4
0
(883
y− 13
y4 − 12
y3)
dy =672
5.
Here is a similar example.
Example 26.1.4 Let f (x,y) = x2y for (x,y)∈ R where R is the triangular region defined tobe in the first quadrant, below the line y = 2x and to the left of the line x = 4. Find
∫R f dA.
x
y
4
R
Put the integral with respect to x on the outside first. Then∫
R f dA =∫ 4
0∫ 2x
0(x2y)
dydxbecause for each x ∈ [0,4], y goes from 0 to 2x. Then
∫ 2x0(x2y)
dy = 2x4. and so∫
R f dA =∫ 40(2x4)
dx = 20485 .
Now do the integral in the other order. Here the integral with respect to y will be on theoutside. What are the limits of this integral? Look at the triangle and note that x goes from0 to 4 and so 2x = y goes from 0 to 8. Now for fixed y between 0 and 8, where does x go? Itgoes from the x coordinate on the line y = 2x which corresponds to this y to 4. What is thex coordinate on this line which goes with y? It is x = y/2. Therefore, the iterated integral is∫ 8
0
∫ 4
y/2
(x2y)
dxdy.
Now∫ 4
y/2(x2y)
dx = 643 y− 1
24 y4 and so∫
R f dA =∫ 8
0( 64
3 y− 124 y4
)dy = 2048
5 the same an-swer.
A few observations are in order here. In finding∫
S f dA there is no problem in settingthings up if S is a rectangle. However, if S is not a rectangle, the procedure always isagonizing. A good rule of thumb is that if what you do is easy it will be wrong. Thereare no shortcuts! There are no quick fixes which require no thought! Pain and sufferingis inevitable and you must not expect it to be otherwise. Always draw a picture and thenbegin agonizing over the correct limits. Even when you are careful you will make lots ofmistakes until you get used to the process.
Sometimes an integral can be evaluated in one order but not in another.
Example 26.1.5 For R as shown below, find∫
R sin(y2)
dA.
x
8
4
R