26.3. METHODS FOR TRIPLE INTEGRALS 527
12. Find∫ 4
0∫ 2
y/21x e2 y
x dxdy. You might need to interchange the order of integration.
13. Find∫ 8
0∫ 4
y/21x e3 y
x dxdy.
14. Find∫ 1
3 π
0∫ 1
3 π
xsiny
y dydx.
15. Find∫ 1
2 π
0∫ 1
2 π
xsiny
y dydx.
16. Find∫
π
0∫
π
xsiny
y dydx
17. ∗ Evaluate the iterated integral and then write the iterated integral with the order ofintegration reversed.
∫ 3−3∫ x−x x2 dydx
You should get∫ 0
3
∫ −y
−3x2 dxdy+
∫ −3
0
∫ y
−3x2 dxdy+
∫ 3
0
∫ 3
yx2 dxdy+
∫ 0
−3
∫ 3
−yx2 dxdy
This is a very interesting example which shows that iterated integrals have a life oftheir own, not just as a method for evaluating double integrals.
18. ∗ Evaluate the iterated integral and then write the iterated integral with the order ofintegration reversed.
∫ 2−2∫ x−x x2 dydx.
26.3 Methods for Triple Integrals26.3.1 Definition of the IntegralThe integral of a function of three variables is similar to the integral of a function of twovariables. In this case, the term: “mesh” refers to a collection of little boxes which coversa given region in R.
Definition 26.3.1 Let R be a bounded region in the R3 and let f be a boundedfunction defined on R. We say f is Riemannn integrable if there exists a number, denotedby∫
R f dV and called the Riemannn integral such that if ε > 0 is given, then whenever oneimposes a sufficiently fine mesh enclosing R and considers the finitely many boxes whichintersect R, numbered as {Qi}m
i=1 and a point (xi,yi,zi) ∈ Qi, it follows that∣∣∣∣∣∫
Rf dV −∑
if (xi,yi,zi)volume(Qi)
∣∣∣∣∣< ε
Of course one can continue generalizing to higher dimensions by analogy. By exactlysimilar reasoning to the case of integrals of functions of two variables, we can consideriterated integrals as a tool for finding the Riemannn integral of a function of three or morevariables.
26.3.2 Iterated IntegralsAs before, the integral is often computed by using an iterated integral. In general it isimpossible to set up an iterated integral for finding
∫E f dV for arbitrary regions, E but
when the region is sufficiently simple, one can make progress. Suppose the region E overwhich the integral is to be taken is of the form E = {(x,y,z) : a(x,y)≤ z ≤ b(x,y)} for(x,y) ∈ R, a two dimensional region. This is illustrated in the following picture in whichthe bottom surface is the graph of z = a(x,y) and the top is the graph of z = b(x,y).