Iterated Integrals

The Riemann sum above is actually two sums, one in i and one in j:

$$\int_R f dA = \lim_{\Delta x, \Delta y \rightarrow 0} \sum_i \sum_j f(x_i,y_j)\Delta x \Delta y.$$

This is very similar to the way we do integrals in one variable:

$$\int_a^b f(x) dx = \lim_{\Delta x \rightarrow 0} \sum_i f(x_i) \Delta x.$$

We could apply this to the two variable case by breaking up the summation and the limits, as follows:

$$\int_R f dA = \lim_{\Delta x \rightarrow 0} \lim_{\Delta y \rightarrow 0} \sum_i \sum_j f(x_i,y_j)\Delta x \Delta y$$

$$= \lim_{\Delta x \rightarrow 0} \sum_i \left(\lim_{\Delta y \rightarrow 0} \sum_j f(x_i,y_j) \Delta y \right) \Delta x$$

$$= \lim_{\Delta x \rightarrow 0} \sum_i \left( \int_c^d f(x_i,y) dy \right) \Delta x$$

$$= \int_a^b \left( \int_c^d f(x,y) dy \right) dx.$$

This gives a clue as to how we should go about evaluating the double integral. We do it one variable at a time (surprise!) This type of integral is called an iterated integral since we integrate with respect to one variable and then another. As it turns out, we could have organized the summations differently and reversed the order of integration:

$$\int_R fdA = \int_a^b \left( \int_c^d f(x,y) dy \right) dx = \int_c^d \left( \int_a^b f(x,y) dx \right) dy.$$

Usually, reversing the order will take more work, but for a rectangular region such as this, it's pretty straightforward.

Here's an example: Integrate the function f(x,y) = x² + 2xy + y² over the rectangle, R, given by $-1 \le x \le 1, -3 \le y \le 3$.

$$\int_R f dA = \int_{-1}^1 \left( \int_{-3}^3 (x^2 + 2xy + y^2)dy \right) dx$$

$$= \int_{-1}^1 \left[ x^2y + xy^2 + \frac{1}{3}y^3 \right]_{-3}^3 dx$$

$$= \int_{-1}^1 ((3x^2 + 9x + 9) - (-3x^2 + 9x - 9)) dx$$

$$= \int_{-1}^1 (6x^2 + 18) dx = (2x^3 + 18x)_{-1}^1 = 40.$$

We could also evaluate this by integrating in the opposite order:

$$\int_R f dA =\int_{-3}^3 \left( \int_{-1}^1 (x^2 + 2xy + y^2) dx \right) dy$$

$$= \int_{-3}^3 \left[ \frac{1}{3}x^3 + x^2y + xy^2 \right]_{-1}^1 dy$$

$$= \int_{-3}^3 \left[ \left( \frac{1}{3} + y + y^2 \right) - \left(-\frac{1}{3} +y - y^2\right) \right] dy$$

$$= \int_{-3}^3 \left( \frac{2}{3} + 2y^2 \right) dy = \left(\frac{2}{3}y + \frac{2}{3}y^3 \right)_{-3}^3 = 40.$$