Double Integrals#

Double Riemann Sum#

Let \(f\) be a continuous function on a rectangle \(R=\{(x, y)\) : \(a \leq x \leq b, c \leq y \leq d\}\). A double Riemann sum for \(f\) over \(R\) is created as follows.

Partition the interval \([a, b]\) into \(m\) subintervals of equal length \(\Delta x=\frac{b-a}{m}\). Let \(x_{0}, x_{1}, \ldots, x_{m}\) be the endpoints of these subintervals, where \(a=x_{0}<\) \(x_{1}<x_{2}<\cdots<x_{m}=b\).
Partition the interval \([c, d]\) into \(n\) subintervals of equal length \(\Delta y=\frac{d-c}{n}\). Let \(y_{0}, y_{1}, \ldots, y_{n}\) be the endpoints of these subintervals, where \(c=y_{0}<\) \(y_{1}<y_{2}<\cdots<y_{n}=d\).
These two partitions create a partition of the rectangle \(R\) into \(m n\) subrectangles \(R_{i j}\) with opposite vertices \(\left(x_{i-1}, y_{j-1}\right)\) and \(\left(x_{i}, y_{j}\right)\) for \(i\) between 1 and \(m\) and \(j\) between 1 and \(n\). These rectangles all have equal area \(\Delta A=\Delta x \cdot \Delta y\).
Choose a point \(\left(x_{i j}^{*}, y_{i j}^{*}\right)\) in each rectangle \(R_{i j}\). Then, a double Riemann sum for \(f\) over \(R\) is given by

\[ \sum_{j=1}^{n} \sum_{i=1}^{m} f\left(x_{i j}^{*}, y_{i j}^{*}\right) \cdot \Delta A . \]

Double Integral#

Let \(R\) be a rectangular region in the \(x y\)-plane and \(f\) a continuous function over \(R\). With terms defined as in a double Riemann sum, the double integral of \(f\) over \(R\) is

\[ \iint_{R} f(x, y) d A=\lim _{m, n \rightarrow \infty} \sum_{j=1}^{n} \sum_{i=1}^{m} f\left(x_{i j}^{*}, y_{i j}^{*}\right) \cdot \Delta A . \]

Interpretations#

\(\iint_{R} f(x, y) d A\) represents the ‘signed’ volume of the solid that lies under the surface \(f(x, y)\) above the region \(R\) in the \(x y\)-plane. (That is, the volume of the solid(s) the surface \(f\) bounds above the \(x y\)-plane minus the volume it bounds below the \(x y\)-plane on \(R\).)
\(A=\iint_{R} d A\) represents the area of the region \(R\) in the \(x y\)-plane.
\(f_{a v}=\frac{1}{A} \iint_{R} f(x, y) d A\) represents the average value of the function \(f(x, y)\) on the region \(R\).

Properties of Double Integrals#

Let \(f\) and \(g\) be continuous functions on a rectangle \(R=\{(x, y): a \leq x \leq b, c \leq y \leq d\}\), and let \(k\) be a constant. Then

\(\iint_{R}(f(x, y)+g(x, y)) d A=\iint_{R} f(x, y) d A+\iint_{R} g(x, y) d A\).
\(\iint_{R} k f(x, y) d A=k \iint_{R} f(x, y) d A\).
If \(f(x, y) \geq g(x, y)\) on \(R\), then \(\iint_{R} f(x, y) d A \geq \iint_{R} g(x, y) d A\).

Fubini’s Theorem#

If \(z=f(x, y)\) is continuous on \(R=[a, b] \times[c, d]\) then

\[ \iint_{R} f(x, y) d A=\int_{a}^{b}\left(\int_{c}^{d} f(x, y) d y\right) d x=\int_{c}^{d}\left(\int_{a}^{b} f(x, y) d x\right) d y . \]

These are called iterated integrals. They can be found by evaluating two single integrals.
\(\int_{c}^{d} f(x, y) d y\) means \(x\) is held fixed (treated as a constant) and \(f\) is integrated with respect to \(y\) from \(y=c\) to \(y=d\). This procedure is called partial integration. The result is a function of \(x\).
The second step is to integrate that function of \(x\) with respect \(x\) from \(a\) to \(b\).
Similarly, \(\int_{c}^{d}\left(\int_{a}^{b} f(x, y) d x\right) d y\) means integrate \(f\) with respect to \(x\) with \(y\) held fixed. This gives a function of \(y\) which is then integrated with respect to \(y\).
Iterated integrals are usual written without the brackets. (You should understand the \(\int \ldots d x\) or \(\int \ldots d y\) to act like a pair of brackets.)

Separable Integrals#

Consider the special case \(f(x, y)=g(x) h(y)\) on a rectangular region \(R=[a, b] \times[c, d]\).

\[ \iint_{R} f(x, y) d A=\int_{a}^{b}\left(\int_{c}^{d} g(x) h(y) d y\right) d x . \]

In finding the inner integral, \(g(x)\) is treated as a constant so we can write

\[ \iint_{R} f(x, y) d A=\int_{a}^{b} g(x)\left(\int_{c}^{d} h(y) d y\right) d x . \]

The inner integral also yields a constant so we can write

\[ \iint_{R} f(x, y) d A=\left(\int_{a}^{b} g(x) d x\right) \cdot\left(\int_{c}^{d} h(y) d y\right) . \]

I.e. in this special case, the double integral can be written as the product of two single integrals.