General method

If $R := \{ (x,y) ~\vert~ a \leq x \leq b, g(x) \leq y \leq h(x) \}$ then

$$\int \int_R f(x,y) dx dy = \int_a^b (\int_{g(x)}^{h(x)} f(x,y) dy) dx$$

To compute the iterated integral, first find $F(x,y)$ for which $\frac{\partial F}{\partial y} = f(x,y)$ so that

$$\int_{g(x)}^{h(x)} f(x,y) dy = F(x,y) \vert_{y = g(x)}^{y = h(x)} = F(x,h(x)) - F(x,g(x))$$

Then find $\Phi(x)$ for which $\Phi'(x) = F(x,h(x)) - F(x,g(x))$ and we find

$$\int_{a}^b (F(x,h(x)) - F(x,g(x))) dx = \Phi(b) - \Phi(a)$$

Putting these together, we have

$$\int \int_R f(x,y) dx dy = \Phi(b) - \Phi(a)$$

Example

Let $R = \{ (x,y) ~\vert~ 1 \leq x \leq 2, 1 \leq y \leq e^x \}$. Compute

$$\int \int_R \frac{x}{y} dx dy$$

Solution

$$\begin{eqnarray*} \int \int_R \frac{x}{y} dx dy & = & \int_{1}^2 (\int_1^{e^x}) ... ...x=2} \\ & = & \frac{8}{3} - \frac{1}{3} \\ & = & \frac{7}{3} \end{eqnarray*}$$

Another example

Let $R = \{ (x,y) ~\vert~ 1 \leq x \leq 3, \ln(x) \leq y \leq 4 \}$. Compute

$$\int \int_R x e^y dx dy$$

Solution

$$\begin{eqnarray*} \int \int_R e^{x+y} dx dy & = & \int_1^3 (\int_{\ln(x)}^4 x e^... ...- \frac{1}{3} + e^4 \frac{1}{2} \\ & = & \frac{26}{3} - 4 e^4 \end{eqnarray*}$$

Changing variables

If the region $R$ is defined by a constant bound on $y$ with the bound on $x$ varying as a function of $y$, then one may compute a double integral over $R$ as an iterated integral where one first integrates with respect to $x$ and then with respect to $y$.

Example

Let $R = \{ (x,y) ~\vert~ y \leq x \leq 2 y, 1 \leq y \leq 2 \}$. Compute

$$\int \int_R e^{x + y} dx dy$$

$$\begin{eqnarray*} \int \int_R e^{x + y} dx dy & = & \int_1^2 (\int_{y}^{2y} e^{x... ...1}{3} e^9 - \frac{1}{2} e^6 - \frac{1}{3} e^3 + \frac{1}{2} e^2 \end{eqnarray*}$$

Rectangles and Fubini's Theorem

If the region $R$ is a rectangle with horizontal and vertical sides relative to the coordinate axes, the one may compute a double integral over $R$ as an iterated integral by first integrating with respect to $x$ and then with respect to $y$ or vice versa.

Example

Let $R = \{ (x,y) ~\vert~ 0 \leq x \leq 2, 2 \leq y \leq 5 \}$. Compute

$$\int \int_R \frac{x^2}{y} dx dy$$

Solution

$$\begin{eqnarray*} \int \int_R \frac{x^2}{y} dx dy & = & \int_0^2 (\int_{2}^5 \fr... ...3} x^3] \vert_{x = 0}^2 \\ & = & \frac{8}{3} \ln(\frac{5}{2}) \end{eqnarray*}$$

Another solution

$$\begin{eqnarray*} \int \int_R \frac{x^2}{y} dx dy & = & \int_2^5 (\int_0^2 \frac... ...8}{3} (\ln(5) - \ln(2)) \\ & = & \frac{8}{3} \ln(\frac{5}{2}) \end{eqnarray*}$$