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.

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Integral, Multiple Integrals and tagged . Bookmark the permalink.

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