### 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

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

### 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 or tap 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}.}

\]