Course:MATH200/IntegrationOnSimple

Integration over Simple Regions

Simple Regions

Definition. A region D in R² is said to be type I if there is an interval [a,b] in R and there are two continuous function g,h:[a,b]→ R such that g(x)≤ h(x) for all x∈ [a,b] and

 $D=\{(x,y)\in {\mathbf{𝐑}}^2:x\in [a,b],g(x)\le y\le h(x).$

A region D is said to be type II if the region E={(x,y): (y,x)∈ D} is type I. In other words D is type II if there is an interval [a,b] in R and there are two continuous function g,h:[a,b]→ R such that g(y)≤ h(y) for all y∈ [a,b] and

 $D=\{(x,y)\in {\mathbf{𝐑}}^2:y\in [a,b],g(y)\le x\le h(y).$

Remark on Simple Regions

Suppose D is an type I region, let V denote the set of real numbers c such that the line x=c intersects D non-trivially. Let H dentoe the set of real numbers c such that the line y=c intersects D non-trivially. A region D is type I only if V is an interval [a,b] and, for each c in V, the intersection of the vertical line x=c with D is also an interval, [g(c),h(c)]. If the resulting functions g and h are continuous, then D is type I.

Integration on Simple Regions

Theorem. Suppose f:D→R is a continuous function with D in R^2. If D is type I as above, we have

 $\int \underset{D}{\int }f(x,y)dA=\underset{a}{\overset{b}{\int }}\underset{g(x)}{\overset{h(x)}{\int }}f(x,y)dydx.$

Similarly, if D is type II we have

 $\int \underset{D}{\int }f(x,y)dA=\underset{a}{\overset{b}{\int }}\underset{g(y)}{\overset{h(y)}{\int }}f(x,y)dxdy.$

Remark on the Theorem

The integrals in the theorem are sometimes called iterated integrals because one does the inner integral first and then iterates this procedure to do the outer integral.

We are not going to prove the theorem. (For this, see the text.) However, we remark that it is essentially the same as the method of calculating volumes of regions by cross-sections used in 1-variable calculus.

Example with Switching Order of Integration

Example. Compute $\underset{0}{\overset{1}{\int }}\underset{\sqrt{x}}{\overset{1}{\int }}{e}^{y^3}dydx.$

Solution. The integral is written as the sort of iterated integral that would come out of an type I region:

 $D=\{(x,y):x\in [0,1],\sqrt{x}\le y\le 1\}.$

However, we cannot easily compute the integral as written because the inner integral cannot be written in terms of elementary functions. (This follows from a theorem of Liouville.) However, D is also type II: we have

 $D=\{(x,y):y\in [0,1],0\le x\le y^2\}.$

The integral then becomes

 $\begin{array}{rl}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{y^2}{\int }}{e}^{y^3}dxdy& =\underset{0}{\overset{1}{\int }}[x{e}^{y^3}{]}_0^{y^2}dy\\ & =\underset{0}{\overset{1}{\int }}{y}^2{e}^{y^3}dy=(e-1)/3.\end{array}$

Remark. The technique above is known as switching the order of integration.