# Double and Triple Integrals

When I was first taught integrals, I was told they are like area. More specifically,

$$int_{0}^{1} f(x) d x$$

represents the infinite sum of the area of rectangles given by height (f(x)) times width (dx), taken on the interval from 0 to 1. Double and triple integrals are an extension of this same concept, but now we have multiple parameters that can vary.

$$int_{0}^{1} int_{0}^{1} f(x, y) d x d y$$

This represents the infinite sum of the volume of rectangular prisms given by height (f(x,y)) times base (dxdy), over the square region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

This is all well and good, but often we use integrals to compute things that have nothing to do with area or volume. In these cases it is vitally important to keep in mind what an integral really represents – an infinite sum of products. If you can understand this, then the number of dimensions don’t really matter. You just need to figure out how to express the bounds of the region and what one infinitesimal piece (for 2 dimensions its a rectangle and for 3 a rectangular prism) looks like. The rest is just computation. But enough blathering, let’s get into an example.

For this example, we will find the volume of space contained in the region bounded by z = siny, y = x, and the line y = π. The region looks something like this:

The best way to approach this problem is to consider tiny cubes of volume dV, given by dxdydz. We simply sum these cubes over the dimensions of x, y and z (the region R) to find the total volume.

$$V=iiint_{R} d V$$

The first part is to figure out the bounds / order of integration. This is arguably the hardest part, and should be done with care. You can’t hope to get the right answer if you mess up the setup!

Always try to consider the simplest order in which to calculate an integral. There is no ‘right way’, because the order of integration doesn’t affect the result, but there are certainly good ways and bad ways. Good ways reduce complexity, and help simplify our problem.

The obvious choice for me was to evaluate the z component first, and then to consider the double integral

$$V=iint_{A} z d x d y$$

The z bound is quite straightforward as it is given by the problem — z goes from 0 to siny.

So we have:

$$iint_{A} int_{0}^{text { siny }} d z d y d x, int_{0}^{text { siny }} d z=sin y, V=iint_{A} sin y d y d x$$

Where A is the area of the shape’s base in the x-y plane.

Now we have a choice — to integrate with respect to y first, or x. Here it is better to compute the integral with respect to y before x, because the height (z value) is given in terms of y. y is lower bounded by the line the y = x, and upper bounded by the line y = π. x goes from 0 to π.

$$int_{0}^{pi} int_{x}^{pi} sin y d y d x$$

Now, just crunch the numbers.