Double Integrals

What I would like to do with the ideas that follow is show you how to compute double integrals by using calculs you already know (from Math 3B) with just a couple new ideas. These were methods that helped me demystify these problems as a student, and I hope they help you, too!

1) Single Integrals

We can't do double integrals without single integrals. So let's quickly refresh what these are.

A definite integral finds the signed area under the curve between two points, usually referred to as $a$ and $b$.

Let's look at an example: $$\int_0^1 f(x) \ dx.$$

This integral measures the area under the "curve" $y=2x$ between $x=0$ and $x=1$.

2) Approximating with Riemann Sums

Riemann sums are a nice way to think about integrals. They approximate area under the curve, and regular integrals are just limits of these sums as $n$, the number of subdivisions, approaches $\infty$.

Riemann Sum Review: To estimate the area under the curve, we chop up our interval into $n$ equal pieces, pick sample points from each of those pieces, and add up the areas of those rectangles.

3) The Ruse

There was a hidden purpose in giving you the previous example. That integral actually calculates a double integral. Each height on the graph $f(x)=2x$ represents the area of a slice of the function $g(x,y) = 4xy$.

You see, the slices we were adding had a value of $2x$ for each $x$ in the interval $[0,1]$, but this number $2x$ was really meant to represent the area of each of the triangluar slices you see below. Take a look at the slices, and then we'll go through the calculations.

Area Calculation: Each slice has an area that depends on the $x$-value for that slice. The height of the surface at a point $(x,y)$ is $4xy$, so the area of the triangle is given by the integral

4) Riemann Sum Version

Here is the Riemann Sum version if you would like to see it.

Conclusion: Finding the volume under a surface when you have a rectangular region of integration is just a question of setting up two integrals.

For example, if we wish to compute the area under the surface $z = f(x,y) = 4xy$ then we can set up the double integral $$ \int_0^1 \int_0^1 4xy \ dy \ dx$$ and then solve: $$ \int_0^1 \left( \int_0^1 4xy \ dy \right) \ dx = \int_0^1 \left( 2x \right) \ dx = 1.$$

5) Integrating Over Other Regions

The previous strategy works great, but only for rectangular regions. So what if we wanted to integrate over the region between the curves $y=x^2$ and $y=\sqrt{x}$?

Strategy: Use what you know from single-variable calculus and extend it in the same way we extended our last integral to a double integra. I'll walk you through the steps.

First, let's set up an integral to express the area of the region. $$ \int_0^1 \sqrt{x} - x^2 \ dx. $$ This integral expresses the area of the region we would like to integrate over. Let's look at slices again.

Now express these heights as a second integral based on the $x$-value in question: $$ \sqrt{x} - x^2 = \int_{x^2}^{\sqrt{x}} 1 \ dy. $$

Do you agree that these are equal? You might ask why I'm over-complicating a simple integral. The reason lies in interpreting what this double integral means. Since $$ \int_0^1 \sqrt{x} - x^2 \ dx = \int_0^1 \left( \int_{x^2}^{\sqrt{x}} 1 \ dy \right) \ dx, $$ let's take a look at the meaning of this double integral on the right hand side.

For each value of $x$, what was the height $\sqrt{x} - x^2$ is now the area $1 \cdot (\sqrt{x} - x^2)$ of a volume slice under the surface with a constant height of 1. The reason this is helpful, is because you can leverage what you just did to find the volume under any surface, not just the simple surface $z=1$. I'll show you.

Example: Compute the volume under the surface $z = f(x,y) = 4xy$ over the region bounded by the curves $y=x^2$ and $y=\sqrt{x}$.

The setup is the same as before, only we replace the constant height $z=1$ with the function in question. $$ \int_0^1 \int_{x^2}^{\sqrt{x}} 4xy \ dy \ dx $$ Then we just compute.