Calculus

Double Integrals

Double Integrals Logo

Iterated Integrals

  • 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 \(II\).

    Region of integration of type I
    Figure 1.
    Region of integration of type II
    Figure 2.

    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

    \[{R }={ \left\{ {\left( {x,y} \right)|\;u\left( y \right) \le x \le v\left( y \right),\;\;}\right.}\kern0pt{\left.{c \le y \le d} \right\}.}\]

    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 a problem to see the 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