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Double Integrals

# Double Integrals over Rectangular Regions

Page 1
Problems 1-2
Page 2
Problems 3-5

Let $$R$$ be a rectangular region$$\left[ {a,b} \right] \times \left[ {c,d} \right]$$ of the $$xy$$-plane. Then using the Fubini’s theorem we can write the double integral in this region through the iterated integral:

${\iint\limits_R {f\left( {x,y} \right)dxdy} } = {\int\limits_a^b {\left( {\int\limits_c^d {f\left( {x,y} \right)dy} } \right)dx} } = {\int\limits_c^d {\left( {\int\limits_a^b {f\left( {x,y} \right)dx} } \right)dy} .}$

The region $$R$$ here is simultaneously the region of type $$I$$ and type $$II,$$ so that we have a free choice as to whether to integrate $$f\left( {x,y} \right)$$ with respect to $$x$$ or $$y$$ first. It is usually better to evaluate the easier integral first.

In the special case where the integrand $$f\left( {x,y} \right)$$ can be written as the product of two functions $$g\left( {x} \right) h\left( {y} \right),$$ we have

${\iint\limits_R {f\left( {x,y} \right)dxdy} } = {\iint\limits_R {g\left( x \right)h\left( y \right)dxdy} } = {\left( {\int\limits_a^b {g\left( x \right)dx} } \right)\cdot}\kern0pt{ \left( {\int\limits_c^d {h\left( y \right)dy} } \right)}$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Evaluate the double integral $$\iint\limits_R {xydxdy}$$ over the rectangular region $${R =}$$ $${ \left\{ {\left( {x,y} \right)|\;2 \le x \le 5,\; }\right.}$$ $${\left.{ 0 \le y \le 1} \right\}.}$$

### ✓Example 2

Calculate the double integral $$\iint\limits_R {x{y^2}dxdy}$$ over the region $${R =}$$ $${ \left\{ {\left( {x,y} \right)|\;1 \le x \le 5,\; }\right.}$$ $${\left.{ 0 \le y \le 2} \right\}.}$$

### ✓Example 3

Calculate the integral $$\iint\limits_R {\large\frac{{dxdy}}{{{{\left( {x + y} \right)}^2}}}\normalsize}$$ over the rectangular region $${R =}$$ $${ \left\{ {\left( {x,y} \right)|\;0 \le x \le 2,\; }\right.}$$ $${\left.{ 1 \le y \le 2} \right\}.}$$

### ✓Example 4

Evaluate the integral $$\iint\limits_R {\cos \left( {x + y} \right)dxdy}$$ over the region $${R =}$$ $${ \left\{ {\left( {x,y} \right)|\;0 \le x \le {\large\frac{\pi }{2}\normalsize},\;}\right.}$$ $${\left.{ 0 \le y \le {\large\frac{\pi }{2}\normalsize}} \right\}.}$$

### ✓Example 5

Evaluate the integral $$\iint\limits_R {\left( {x – {y^2}} \right)dxdy}$$ over the region $${R =}$$ $${ \left\{ {\left( {x,y} \right)|\;2 \le x \le 3,\;}\right.}$$ $${\left.{ 1 \le y \le 2} \right\}.}$$

### Example 1.

Evaluate the double integral $$\iint\limits_R {xydxdy}$$ over the rectangular region $${R =}$$ $${ \left\{ {\left( {x,y} \right)|\;2 \le x \le 5,\; }\right.}$$ $${\left.{ 0 \le y \le 1} \right\}.}$$

#### Solution.

We see that the integrand $$f\left( {x,y} \right)$$ is the product $$g\left( {x} \right) h\left( {y} \right).$$ Then we have

${\iint\limits_R {xydxdy} = \int\limits_2^4 {xdx} \cdot \int\limits_0^1 {ydy} } = {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_2^4 \cdot \left. {\left( {\frac{{{y^2}}}{2}} \right)} \right|_0^1 } = {\left( {8 – 2} \right)\left( {\frac{1}{2} – 0} \right) }={ 3.}$

### Example 2.

Calculate the double integral $$\iint\limits_R {x{y^2}dxdy}$$ over the region $${R =}$$ $${ \left\{ {\left( {x,y} \right)|\;1 \le x \le 5,\; }\right.}$$ $${\left.{ 0 \le y \le 2} \right\}.}$$

#### Solution.

Since the integrand $$f\left( {x,y} \right)$$ is the product $$g\left( {x} \right) h\left( {y} \right),$$ we can write

${\iint\limits_R {x{y^2}dxdy} } = {\int\limits_1^5 {xdx} \cdot \int\limits_0^2 {{y^2}dy} } = {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_1^5 \cdot \left. {\left( {\frac{{{y^3}}}{3}} \right)} \right|_0^2 } = {\left( {\frac{{25}}{2} – \frac{1}{2}} \right)\left( {\frac{8}{3} – 0} \right) }={ 64.}$
Page 1
Problems 1-2
Page 2
Problems 3-5