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_{R} f (x, y) d x d y = \int_{a}^{b} (\int_{c}^{d} f (x, y) d y) d x = \int_{c}^{d} (\int_{a}^{b} f (x, y) d x) d y .

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 $f (x, y)$ can be written as the product of two functions $g (x) h (y),$ we have

\iint_{R} f (x, y) d x d y = \iint_{R} g (x) h (y) d x d y = (\int_{a}^{b} g (x) d x) \cdot (\int_{c}^{d} h (y) d y)

Solved Problems

Example 1.

Evaluate the double integral $\iint_{R} x y d x d y$ over the rectangular region

R = {(x, y) | 2 \leq x \leq 4, 0 \leq y \leq 1} .

Solution.

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

\iint_{R} x y d x d y = \int_{2}^{4} x d x \cdot \int_{0}^{1} y d y = {(\frac{x^{2}}{2}) |}_{2}^{4} \cdot {(\frac{y^{2}}{2}) |}_{0}^{1} = (8 - 2) (\frac{1}{2} - 0) = 3.

Example 2.

Calculate the double integral $\iint_{R} x y^{2} d x d y$ over the region

R = {(x, y) | 1 \leq x \leq 5, 0 \leq y \leq 2} .

Solution.

Since the integrand $f (x, y)$ is the product $g (x) h (y),$ we can write

\iint_{R} x y^{2} d x d y = \int_{1}^{5} x d x \cdot \int_{0}^{2} y^{2} d y = {(\frac{x^{2}}{2}) |}_{1}^{5} \cdot {(\frac{y^{3}}{3}) |}_{0}^{2} = (\frac{25}{2} - \frac{1}{2}) (\frac{8}{3} - 0) = 32.

See more problems on Page 2.