Week 2 Discussion Notes

Table of Contents

• Double Integrals

Double Integrals

In 1D, the integral represented a (signed) area:

This interpretation of double integrals is basically the same. Instead of area, we have volume:

You can already see that unlike in 1D, the region you integrate over can get very complicated. In 1D, you mostly integrated over intervals, but in 2D, $\mathscr{D}$ can be a rectangle, a disk, a star, etc. However, despite this, Fubini's theorem tells you that a double integral can be (in a nutshell) calculated as two 1D integrals:

Theorem (Fubini)

$$\begin{aligned} \iint_\mathscr{D} f\p{x, y} \,\diff{A} &= \iint_\mathscr{D} f\p{x, y} \,\diff{x} \,\diff{y} \\ &= \iint_\mathscr{D} f\p{x, y} \,\diff{y} \,\diff{x} \end{aligned}$$

In other words, Fubini's theorem tells you that you can calculate a double integral as an iterated integral, and that the order of the variables you integrate in doesn't matter.

Example 1.

Integrate $f\p{x, y} = \p{x + y}^2$ over $\mathscr{R} = \br{0, 2} \times \br{-1, 1}$.

Solution.

When given a rectangle, the first interval always corresponds to $x$ and the second interval always corresponds to $y$. This gives us the following bounds:

$$\begin{aligned} \iint_\mathscr{R} f\p{x, y} \,\diff{A} &= \int_{-1}^1 \int_0^2 \p{x + y}^2 \,\diff{x} \,\diff{y} \\ &= \int_{-1}^1 \left. \frac{\p{x + y}^3}{3} \right\rvert_{x=0}^{x=2} \,\diff{y} \\ &= \int_{-1}^1 \frac{\p{y + 2}^3 - y^3}{3} \,\diff{y} \\ &= \left. \frac{\p{y + 2}^4 - y^4}{12} \right\rvert_{-1}^1 \\ &= \boxed{\frac{20}{3}}. \end{aligned}$$

To illustrate Fubini's theorem, I'll calculate the integral in the other order too:

$$\begin{aligned} \iint_\mathscr{R} f\p{x, y} \,\diff{A} &= \int_0^2 \int_{-1}^1 \p{x + y}^2 \,\diff{y} \,\diff{x} \\ &= \int_0^2 \left. \frac{\p{x + y}^3}{3} \right\rvert_{y=-1}^{y=1} \,\diff{x} \\ &= \int_0^2 \frac{\p{x + 1}^3 - \p{x - 1}^3}{3} \,\diff{x} \\ &= \left. \frac{\p{x + 1}^4 - \p{x - 1}^4}{12} \right\rvert_0^2 \\ &= \frac{20}{3}. \end{aligned}$$

Example 2.

Integrate $f\p{x, y} = e^x$ over the region $\mathscr{D}$ bounded by $y = x + 1$, $y = x$, $x = 0$, and $x = 1$.

Solution.

Unlike the previous example, $\mathscr{D}$ is slightly more complicated than a rectangle. For example, for each $x$ in the interval $\br{0, 1}$, the interval that $y$ lives in changes:

$$\begin{aligned} x = 0 &\implies 0 \leq y \leq 1 \\ x = 1 &\implies 1 \leq y \leq 2. \end{aligned}$$

This means that the bounds of $y$ are a function of $x$:

For each $x$ in $\br{0, 1}$, $y$ will range over $\br{x, x+1}$, so the integral is

$$\begin{aligned} \iint_\mathscr{D} e^x \,\diff{A} &= \int_0^1 \int_x^{x+1} e^x \,\diff{y} \,\diff{x} \\ &= \int_0^1 \Bigl. ye^x \Bigr\rvert_{y=x}^{y=x+1} \,\diff{x} \\ &= \int_0^1 e^x \,\diff{x} \\ &= \boxed{e - 1}. \end{aligned}$$