Engineering Math

14.3.2 Iterated Integrals

As before, the integral is often computed by using an iterated integral. In general it is impossible to set up an iterated integral for finding ∫ _EfdV for arbitrary regions, E but when the region is sufficiently simple, one can make progress. Suppose the region E over which the integral is to be taken is of the form E =

for ∈ R, a two dimensional region. This is illustrated in the following picture in which the bottom surface is the graph of z = a and the top is the graph of z = b.

Then

∫ ∫ ∫ b(x,y) E fdV = R a(x,y) f (x,y,z)dzdA

It might be helpful to think of dV = dzdA. Now ∫ _a

b

(x,y)

fdz is a function of x and y and so you have reduced the triple integral to a double integral over R of this function of x and y. Similar reasoning would apply if the region in ℝ³ were of the form or .

In this case, R is the region shown.

Thus the region E is between the plane z = 1 −

on the top, z = 0 on the bottom, and over R shown above. Thus

∫ ∫ ∫ 1−(x+y) ∫ 1∫ 1−x∫ 1−(x+y) 1 1dV = dzdA = dzdydx =- E R 0 0 0 0 6

Of course iterated integrals have a life of their own although this will not be explored here. You can just write them down and go to work on them. Here are some examples.

The inside integral yields ∫ _3y^x

dz = x² − 4xy + 3y². Next this must be integrated with respect to y to give ∫ ₃^xdy = −3x² + 18x − 27. Finally the third integral gives

∫ 3∫ x∫ x ∫ 3( ) (x− y)dzdydx = − 3x2 + 18x − 27 dx = − 1. 2 3 3y 2

The inside integral is ∫ ₀^y+z cos

dx = 2cosz siny cosy + 2sinz cos²y − sinz − siny. Now this has to be integrated.

= − 1− 16cos5y +20 cos3y − 5cosy− 3(sin y)y+ 2 cos2y.

Finally, this last expression must be integrated from 0 to π. Thus

∫ π ∫ 3y ∫ y+z cos (x + y)dxdzdy 0 0 0

∫ π = (− 1− 16cos5y+ 20cos3y− 5 cos y− 3(sin y)y+ 2cos2y)dy = − 3π 0

The inside integral is just a function of x and y. (In fact, only a function of x.) The order of the last two integrals must be interchanged. Thus the iterated integral which needs to be done in a different order is

∫ 2∫ 3− 32x f (x,y) dydx. 0 0

As usual, it is important to draw a picture and then go from there.

Thus this double integral equals

∫ 3 ∫ 23(3−y) f (x,y)dxdy. 0 0

Now substituting in for f

,

∫ 3∫ 23(3−y)∫ x2 dzdxdy. 0 0 0

In the yz plane, the first of the following pictures corresponds to x = 0.

Therefore, the outside integrals taken with respect to z and y are of the form ∫ ₀

∫ ₀

23

−ydzdy, and now for any choice of in the above triangular region, x goes from 0 to 16 − y². Therefore, the iterated integral is

∫ 23∫ 23−y∫ 16−y2 860 0 0 0 dxdzdy = 243

The first listed cylinder intersects the xy plane in the disk, x² + y² ≤ 1. What is the volume of the three dimensional region which is between this disk and the two surfaces, z =

and z = −? An iterated integral for the volume is

∫ 1 ∫ √1−-x2∫ √1−x2 √---- √ ----dzdydx = 16. −1 − 1−x2 − 1− x2 3

Note that I drew no picture of the three dimensional region. If you are interested, here it is.

One of the cylinders is parallel to the z axis, x² + y² ≤ 1 and the other is parallel to the y axis, x² + z² ≤ 1. I did not need to be able to draw such a nice picture in order to work this problem. This is the key to doing these. Draw pictures in two dimensions and reason from the two dimensional pictures rather than attempt to wax artistic and consider all three dimensions at once. These problems are hard enough without making them even harder by attempting to be an artist.