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Double Integrals

# Iterated Integrals

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Problem 1
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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
${R }={ \left\{ {\left( {x,y} \right)|\;u\left( y \right) \le x \le v\left( y \right),\;\;}\right.}\kern0pt{\left.{c \le y \le d} \right\}.}$

Figure 1.

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