# Calculus

## Double Integrals # Double Integrals over Rectangular Regions

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

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

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Evaluate the double integral $\iint\limits_R {xydxdy}$ over the rectangular region

$R = \left\{ {\left( {x,y} \right)|\;2 \le x \le 5,\; 0 \le y \le 1} \right\}.$

### Example 2

Calculate the double integral $\iint\limits_R {x{y^2}dxdy}$ over the region

$R = \left\{ {\left( {x,y} \right)|\;1 \le x \le 5,\; 0 \le y \le 2} \right\}.$

### Example 1.

Evaluate the double integral $\iint\limits_R {xydxdy}$ over the rectangular region

$R = \left\{ {\left( {x,y} \right)|\;2 \le x \le 5,\; 0 \le y \le 1} \right\}.$

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

$R = \left\{ {\left( {x,y} \right)|\;1 \le x \le 5,\; 0 \le y \le 2} \right\}.$

Solution.

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

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

See more problems on Page 2.