Section 7.7: Double integrals

If $R$ is some region in the plane and $f(x,y)$ is a (continuous) function taking positive values, then the volume of $\{ (x,y,z) ~\vert~ (x,y) \text{ in } R \text{ and } 0 \leq z \leq f(x,y) \}$ is the

integral

$$\int \int_R f(x,y) dx \, dy$$

More generally, if $f(x,y)$ is any continuous function, then $\int \int_R f(x,y) dx \, dy$ is the signed volume of the solid bounded by $f$.

Example

If $R := \{ (x,y) ~\vert~ 0 \leq x \leq 1, 0 \leq y \leq 1 \}$ is the unit square and $f(x,y) \equiv 1$, then the solid bounded by $f$ over $R$ is the unit cube. So, $\int \int_R f(x,y) dx \, dy = 1$.

Iterated integrals

If $R = \{ (x,y) ~\vert~ a \leq x \leq b, g(x) \leq y \leq h(x) \}$ (where $a \leq b$ are constants and $g$ and $h$ are continuous functions of $x$), then

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

Here, $F(x) = \int_{g(x)}^{h(x)} f(x,y) dy$ is itself a function of $x$.

Example

Suppose $f(x,y) = xy$ and $R = \{ (x,y) ~\vert~ 1 \leq x \leq 2, x \leq y \leq x^2 \}$. Compute

$$\int \int_R f(x,y) dx \, dy$$

$$\begin{eqnarray*} \int \int_R f(x,y) dx \, dy & = & \int \int_{ \{ (x,y) ~\vert~... ...rac{16}{8} - \frac{1}{12} + \frac{1}{8}) \\ & = & \frac{27}{8} \end{eqnarray*}$$

Another example

Compute

$$\int \int_{ \{ (x,y) ~\vert~ 0 \leq x \leq 1, 0 \leq y \leq e^x \}} e^x dx \, dy$$

Solution

$$\begin{eqnarray*} \int \int_{ \{ (x,y) ~\vert~ 0 \leq x \leq 1, 0 \leq y \leq e^... ...2} e^{2x}) \vert_{x=0}^{x = 1} \\ & = & \frac{1}{2} (e^2 - 1) \end{eqnarray*}$$

Yet another example

Compute

$$\int \int_{ \{ (x,y) ~\vert~ 1 \leq x \leq 4, \sqrt{x} \leq y \leq x^3 \} } \frac{x}{y^2} dx \, dy$$

Solution

$$\begin{eqnarray*} \int \int_{ \{ (x,y) ~\vert~ 1 \leq x \leq 4, \sqrt{x} \leq y ... ...16}{3} + \frac{1}{4} - \frac{2}{3} - 1 ) \\ & = & \frac{47}{12} \end{eqnarray*}$$