21. Double Integrals

Now that we can do derivatives involving $\large f(x,y)$, it's time to move on to integrals.

In single variable calculus, we approximate the area under $\large f(x)$ with areas of rectangles. The approximation is called a Riemann sum:

$$\large \int_a^b f(x)\, dx\approx \sum_{k=1}^Nf(x_k)\Delta x$$

The approximation improves as we use more rectangles, and the limit of the Riemann sum gives us the exact area under the curve:

$$\large \int_a^b f(x)\, dx =\lim_{N\to \infty} \sum_{k=1}^Nf(x_k)\Delta x$$

The Double Interal

Now, we have $\large z = f(x,y)$. Our goal is to find the volume under the surface above some region $\large R$:

DoubleInt1

Image from: https://www.geeksforgeeks.org/double-integral/

We call this volume the double integral of $\large f(x,y)$ on $\large R$, and we use the following notation:

$$\large \iint_R f(x,y)\, dA$$

The notation for a double integral consistes of two integral signs, an integrand ($\large f(x,y)$, the function we are integrating), and an area element.

Volume under a surface

To compute the double integral, we look to find the volume under $\large f(x,y)$. Our approach is quite similar to the single variable approach to area under a curve.

Our approach is similar. We break the region $\large R$ down into pieces (rectangles):

DoubleInt2

Image from: https://allen.in/jee/maths/double-integral

If $\large R$ is the rectangle $\large [a,b]\times[c,d]$, then we break $\large [a,b]$ down into $\large x_0, x_1,\dots,x_N$ with $\large \Delta x = \frac{b-a}{N}$. We similarly break down $\large [c,d]$ to get $\large y_0,\dots,y_M$ with $\large \Delta y = \frac{d-c}{M}$.

With these partitions, we can say the volume under the surface $\large f(x,y)$ and above one of these small rectangles is approximately $\large f(x_i,y_j)\Delta x\Delta y$ (this is a height times a width times a length to give the volume of a box). We add up the volumes of all the small boxes to approximate the volume under the surface:

$$\large \iint_R f(x,y)\, dA\approx \sum_{i=1}^N\sum_{j=1}^M f(x_i,y_j)\Delta x\Delta y$$

Like before, we can improve the approximation by using more little rectangles. The limit gives us the exact value of the double integral (the exact volume under the surface):

$$\large \iint_R f(x,y)\, dA= \lim_{M,N\to\infty} \sum_{i=1}^N\sum_{j=1}^M f(x_i,y_j)\Delta x\Delta y$$

The image below illustrates the limit convergence to the exact volume:

DOUBLE+INTEGRALS+12 1 Image from: https://slideplayer.com/slide/9376985/

Actual computation

If $\large R$ is a rectangle $\large [a,b]\times[c,d]$, we compute a double integral by simply integrating in $\large x$ first and then in $\large y$ (or vice versa):

$$\large \iint_R f(x,y)\, dA = \int_a^b\int_c^d f(x,y) \, dy dx = \int_c^d\int_a^b f(x,y) \, dx dy$$

The iterated integrals above are just a sequence of good, old fashioned, single variable integrals done back-to-back. The video below illustrates how this is done:

As you can imagine after that example, these double integrals can be precisely as difficult or easy as the single variable integrals they reduce to. We could see integrals that require substitution, integration by parts, trig identities, and partial fraction decompositions, just like we saw in single variable calculus.

Facts about double integrals

We'll end this section with some things to know about double integrals:

$\large \iint_R f(x,y)dA$ is the volume under $\large f$ and above $\large R$.
$\large \iint_R 1 dA$ is the area of the region $\large R$.
The average value of a function $\large f(x,y)$ on the region $\large R$ is $\large \frac{1}{Area(R)}\iint_R f(x,y) dA$.
If $\large \rho(x,y)$ is a mass density of a plate with shape $\large R$, then $\large \iint_R \rho(x,y) dA$ is a total mass.