Calculus

Double Integrals

Iterated Integrals

Page 1
Problem 1
Page 2
Problems 2-5

Regions of Type \(I\) and Type \(II\)

The most powerful tool for the evaluation of the double integrals is the Fubini’s theorem. It works not for a general region \(R\) but for some special regions which we call Regions of type \(I\) or type \(II\).

Definition \(1\).
A plane region \(R\) is said to be of type \(I\) if it lies between the graphs of two continuous functions of \(x\) (Figure \(1\)), that is
\[{R }={ \left\{ {\left( {x,y} \right)|\;a \le x \le b,\;\;}\right.}\kern0pt{\left.{p\left( x \right) \le y \le q\left( x \right)} \right\}.\;\;}\] Definition \(2\).
A plane region \(R\) is said to be of type \(II\) if it lies between the graphs of two continuous functions of \(y\) (Figure \(2\)), that is
\[{R }={ \left\{ {\left( {x,y} \right)|\;u\left( y \right) \le x \le v\left( y \right),\;\;}\right.}\kern0pt{\left.{c \le y \le d} \right\}.}\]

Region of integration of type I

Figure 1.

Region of integration of type II

Figure 2.

Fubini’s Theorem

Let \(f\left( {x,y} \right)\) is a continuous function over a type \(I\) region \(R\) such that
\[{R }={ \left\{ {\left( {x,y} \right)|\;a \le x \le b,\;\;}\right.}\kern0pt{\left.{p\left( x \right) \le y \le q\left( x \right)} \right\}.}\] Then the double integral of \(f\left( {x,y} \right)\) in this region is expressed in terms of the iterated integral:
\[
{\iint\limits_R {f\left( {x,y} \right)dA} }
= {\int\limits_a^b {\int\limits_{p\left( x \right)}^{q\left( x \right)} {f\left( {x,y} \right)dydx} } .}
\] For a region of type \(II\) we have the similar theorem:

If \(f\left( {x,y} \right)\) is a continuous function on a type \(II\) region \(R\) such that
\[{R }={ \left\{ {\left( {x,y} \right)|\;u\left( y \right) \le x \le v\left( y \right),\;\;}\right.}\kern0pt{\left.{c \le y \le d} \right\}.}\] then
\[
{\iint\limits_R {f\left( {x,y} \right)dA} }
= {\int\limits_c^d {\int\limits_{u\left( y \right)}^{v\left( y \right)} {f\left( {x,y} \right)dxdy} } .}
\] Thus, the Fubini’s theorem allows to calculate double integrals through the iterated ones.
To evaluate an iterated integral, we first find the inner integral and then the outer integral.

Solved Problems

Click on problem description to see solution.

 Example 1

Evaluate the iterated integral \(\int\limits_0^1 {\int\limits_1^2 {xydydx} } .\)

 Example 2

Find the iterated integral \(\int\limits_0^1 {\int\limits_y^{{y^2}} {\left( {x + 2y} \right)dxdy} }.\)

 Example 3

Calculate the iterated integral \(\int\limits_1^2 {\int\limits_0^y {x\sqrt {{y^2} + {x^2}} dxdy} }.\)

 Example 4

Evaluate the iterated integral \(\int\limits_0^1 {\int\limits_0^y {\ln \left( {{y^2} + 1} \right)dxdy} } .\)

 Example 5

Express the integral as an integral with the order of integration reversed: \(I = \int\limits_0^5 {\int\limits_{\frac{x}{2}}^x {f\left( {x,y} \right)dydx} } .\)

Example 1.

Evaluate the iterated integral \(\int\limits_0^1 {\int\limits_1^2 {xydydx} } .\)

Solution.

We first evaluate the inner integral and then the outer integral:
\[
{\int\limits_0^1 {\int\limits_1^2 {xydydx} } }
= {\int\limits_0^1 {\left[ {\int\limits_1^2 {xydy} } \right]dx} }
= {\int\limits_0^1 {\left[ {x\left. {\left( {\frac{{{y^2}}}{2}} \right)} \right|_1^2} \right]dx} }
= {\int\limits_0^1 {\frac{3}{2}dx} }
= {\frac{3}{2}\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1 }={ \frac{3}{4}.}
\]

Page 1
Problem 1
Page 2
Problems 2-5