Double Integrals over Rectangular Regions
Let R be a rectangular region [a, b] × [c, d] 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 (x, y) 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
Example 1.
Evaluate the double integral \[\iint\limits_R {xydxdy}\] over the rectangular region
\[R = \left\{ {\left( {x,y} \right)|\;2 \le x \le 4,\; 0 \le y \le 1} \right\}.\]
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
\[\iint\limits_R {x{y^2}dxdy} = \int\limits_1^5 {xdx} \cdot \int\limits_0^2 {{y^2}dy} = \left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_1^5 \cdot \left. {\left( {\frac{{{y^3}}}{3}} \right)} \right|_0^2 = \left( {\frac{{25}}{2} - \frac{1}{2}} \right)\left( {\frac{8}{3} - 0} \right) = 32.\]
See more problems on Page 2.
