## Introduction to Double Integration

1. Consider ${z=f(x,y)}$. What is the volume of the region
(1)$\displaystyle S = \{ (x,y,z)\in{\mathbb R}^{3}\mid z=f(x,y),\; a\leq x\leq b,\; c\leq y\leq d\}$
The first thing we do: consider the rectangle ${R=[a,b]\times[c,d]}$ and form a partition of ${[a,b]}$ into ${M-1}$ segments and ${[c,d]}$ into ${N-1}$ segments. This gives us a mesh of rectangles ${R_{ij} = [x_{i-1},x_{i}]\times[y_{j-1},y_{j}]}$ as specified by the following diagram:

Observe the area of ${R_{ij}}$ is
(2)$\displaystyle \Delta A = \Delta x\Delta y.$
We can approximate the volume ${V}$ of ${S}$ by a sort of Riemann sum, picking points ${(x_{ij}^{*}, y_{ij}^{*})}$ in ${R_{ij}}$ and taking
(3)$\displaystyle \sum^{M}_{i=1}\sum^{N}_{j=1}f(x^{*}_{ij},y^{*}_{ij})\Delta A\approx V.$
However, as with Riemann sums, we recover the exact volume when we take the limits ${M,N\rightarrow\infty}$:
(4)$\displaystyle V = \lim_{M,N\rightarrow\infty}\sum^{M}_{i=1}\sum^{N}_{j=1}f(x^{*}_{ij},y^{*}_{ij})\Delta A$
if the limit exists.

2. Definition. We define the “Double Integral” of ${f}$ over the integral ${R}$ as
(5)$\displaystyle \iint_{R}f(x,y)\,\mathrm{d} A = \lim_{M,N\rightarrow\infty}\sum^{M}_{i=1}\sum^{N}_{j=1}f(x^{*}_{ij},y^{*}_{ij})\Delta A$
if this limit exists.

Note this sum is called a double Riemann sum, just as a for the single integral we had a Riemann Sum.

3. Properties of Integrals. We won’t prove, but note, there are three important properties double integrals satisfy. Two can be grouped together as linearity:
(6)$\displaystyle \iint_{R}\bigl[f(x,y)+g(x,y)\bigr]\,\mathrm{d} A=\iint_{R}f(x,y)\,\mathrm{d} A+\iint_{R}g(x,y)\,\mathrm{d} A$
and
(7)$\displaystyle \iint_{R}cf(x,y)\,\mathrm{d} A = c\iint_{R}f(x,y)\,\mathrm{d} A$
where ${f}$ and ${g}$ are continuous on ${R}$, and ${c\in{\mathbb R}}$ is a constant.

If ${f(x,y)\geq g(x,y)}$ for all ${(x,y)\in R}$, then
(8)$\displaystyle \iint_{R}f(x,y)\,\mathrm{d} A\geq\iint_{R}g(x,y)\,\mathrm{d} A.$

4. Problem: How do we compute ${\iint_{R}f(x,y)\,\mathrm{d} A}$?

### Iterated Integrals

5. Fubini’s Theorem. If ${f}$ is continuous on the integral ${R=[a,b]\times[c,d]}$, then
(9)$\displaystyle \iint_{R}f(x,y)\,\mathrm{d} A=\int^{b}_{a}\left(\int^{d}_{c}f(x,y)\,\mathrm{d} y\right)\mathrm{d} x =\int^{d}_{c}\left(\int^{b}_{a}f(x,y)\,\mathrm{d} x\right)\mathrm{d} y.$
We won’t prove it (that’s what real analysis discusses!), but the parenthetic terms are functions of a single variable and describe the area of a “sheet”. By stacking “sheets” we can compute the volume of the region.

6. This is fine for rectangular regions, but over arbitrary regions what do we do? We bound the region ${D}$ by a pair of curves ${g_{1}(x)\leq y\leq g_{2}(x)}$:

In these situations, we write
(10)$\displaystyle \iint_{D}f(x,y)\,\mathrm{d} A=\int^{b}_{a}\int^{g_{2}(x)}_{g_{1}(x)}f(x,y)\,\mathrm{d} y\,\mathrm{d} x.$
Please note the order of integration, and corresponding boundary of integration!

7. We can likewise bound the region by curves ${h_{1}(y)\leq x\leq h_{2}(y)}$, for example:

The trick lies with writing these integrals as
(11)$\displaystyle \iint_{D}f(x,y)\,\mathrm{d} A=\int^{d}_{c}\int^{h_{2}(y)}_{h_{1}(y)}f(x,y)\,\mathrm{d} x\,\mathrm{d} y.$
Again, note the order of integration!

Example 1. Lets consider the domain
(12)$\displaystyle D = \{(x,y)\mid -1\leq x\leq1,\; 2x^{2}\leq y\leq 1+x^{2}\}.$
Evaluate the integral
(13)$\displaystyle I = \iint_{D}(x-y)\,\mathrm{d} A.$

Solution:\quad\ignorespaces We first doodle this domain, shade it in red:

Now we use Fubini’s theorem writing
(14)$\displaystyle I = \int^{1}_{-1}\int^{1+x^{2}}_{2x^{2}}(x-y)\,\mathrm{d} y\,\mathrm{d} x.$
Observe the order of integration, and the bounds of integration are determined by specifying ${y\geq 2x^{2}}$ and ${y\leq1+x^{2}}$ for the ${\int\mathrm{d} y}$ quantity. Similarly, since ${-1\leq x\leq1}$, we determine the bounds of integration for the ${x}$ integral.

Using linearity, we can write
(15)$\displaystyle I = I_{1}+I{2}$
where
(16)$\displaystyle I_{1}=\int^{1}_{-1}\int^{1+x^{2}}_{2x^{2}}x\,\mathrm{d} y\,\mathrm{d} x$
and
(17)$\displaystyle I_{2}= \int^{1}_{-1}\int^{1+x^{2}}_{2x^{2}}-y\,\mathrm{d} y\,\mathrm{d} x$
Now we can evaluate each of these separately.

We see for ${I_{1}}$, there are no ${y}$ expressions, so we have
(18)$\displaystyle \int^{1}_{-1}\int^{1+x^{2}}_{2x^{2}}x\,\mathrm{d} y\,\mathrm{d} x= \int^{1}_{-1}x\int^{1+x^{2}}_{2x^{2}}\,\mathrm{d} y\,\mathrm{d} x.$
Performing this inner integral
(19)$\displaystyle \int^{1+x^{2}}_{2x^{2}}\mathrm{d} y = \bigl(1+x^{2}\bigr)-(2x^{2})=1-x^{2}.$
We plug this back in:
(20)$\displaystyle \int^{1}_{-1}x\left(\int^{1+x^{2}}_{2x^{2}}\,\mathrm{d} y\right)\mathrm{d} x =\int^{1}_{-1}x\bigl(1-x^{2}\bigr)\mathrm{d} x.$
This can be solved quickly as
(21)\displaystyle \begin{aligned} \int^{1}_{-1}x\bigl(1-x^{2}\bigr)\mathrm{d} x &= \left.\frac{x^{2}}{2}-\frac{x^{4}}{4}\right|^{1}_{-1}\\ &= \left(\frac{(1)^{2}}{2}-\frac{(1)^{4}}{4}\right) -\left(\frac{(-1)^{2}}{2}-\frac{(-1)^{4}}{4}\right)\\ &=0. \end{aligned}
Thus
(22)$\displaystyle I_{1}=0.$
Moreover this implies ${I=I_{2}}$.

What to do about ${I_{2}}$? Recall
$\displaystyle I_{2}= \int^{1}_{-1}\int^{1+x^{2}}_{2x^{2}}-y\,\mathrm{d} y\,\mathrm{d} x$
We first perform the inner integral
(23)\displaystyle \begin{aligned} \int^{1+x^{2}}_{2x^{2}}y\,\mathrm{d} y &= \left.\frac{y^{2}}{2}\right|^{1+x^{2}}_{2x^{2}}\\ &=\frac{(1+x^{2})^{2} - (2x^{2})^{2}}{2}\\ &=\frac{1+2x^{2}+x^{4}-4x^{4}}{2} = \frac{1+2x^{2}-3x^{4}}{2}. \end{aligned}
We plug this back in
(24)$\displaystyle \int^{1}_{-1}-\left(\int^{1+x^{2}}_{2x^{2}}y\,\mathrm{d} y\right)\mathrm{d} x =-\int^{1}_{-1}\frac{1+2x^{2}-3x^{4}}{2}\mathrm{d} x.$
We can calculate this out directly as
(25)\displaystyle \begin{aligned} \int^{1}_{-1}(1+2x^{2}-3x^{4})\,\mathrm{d} x &=\left.x+\frac{2x^{3}}{3}-\frac{3x^{5}}{5}\right|^{1}_{-1}\\ &=\left(1+\frac{2}{3}-\frac{3}{5}\right)-\left(-1-\frac{2}{3}+\frac{3}{5}\right)\\ &=2+\frac{4}{3}-\frac{6}{5}=\frac{32}{15}. \end{aligned}
But we forgot the coefficient ${-1/2}$ in front of the integral! Thus our integral is
(26)$\displaystyle I_{2}=\frac{-32}{30}$
giving us our final result
(27)$\displaystyle I = 0 - \frac{32}{30},$
which concludes our example.