Does Trivial Actually Mean Tedious?

This installment in the ‘It’s Obvious. Not!’ series relates to the second edition of the book “div grad curl and all that” by H.M. Schey, published by W. W. Norton.

Near the end of the example I referenced here, the author of “div grad curl and all that” states that the following integral is ‘trivial’ and results in an answer of 1/6 pi, (specifically, this falls on page 26 of the second edition). As far as I can tell, the solution is more tedious than it is trivial. I’m hoping there really is a trivial solution. If you know it, please add it to the comments below. I’m posting two versions of the ‘tedious’ solution here.

The integral in question:

The author suggests switching to polar coordinates before solving the integral using the following substitutions:

The substitution that’s not mentioned is:

So, now to solve the ‘trivial’ integral, first use the substitutions mentioned above:

Factoring out the -r squared term in square root:

Using the trigonometry identity

we get:

Which is a little more readable as:

To arrive at the answer, first, we need to take the anti-derivative with respect to r and evaluate the integral over the limits of r stated in the book’s example: 0 to 1. Then, we’ll take the anti-derivative over theta and evaluate over the limits 0 to pi/2.

There are two ways to arrive at the anti-derivative with respect to r, the ‘clever’ way and the ‘go through all the steps’ way.

The ‘clever’ way:

Use the integration tables from Wikipedia. There, you’ll see a table of integrals involving:

Be careful here. u, x, and a are just symbolic notations chosen by the authors of the integration table. They have nothing to do with coordinate systems. a is always considered to be a constant, and x is the variable of the function in question. u is used as a substitution to make the tables easier to read. We want the integral:

So, our anti-derivative over (our) r is:

Evaluating over r gives:

The anti-derivative with respect to theta is:

which evaluates to pi/6. The result promised in the ‘trivial’ evaluation of the integral.

The ‘go through all the steps’ way:

Suppose you didn’t have the integration table, or didn’t think of it. Then, you could perform the integration out step-by-step using the ‘substitution rule’.

For our substitution, we’ll chose

Then, using the 'chain rule' of differentiation:

simplifying and multiplying both sides by dr

and

Now, if we substitute u into our original integral, we get

using the value we found for rdr gives

but,

so, we have

The anti-derivative for this is:

just as we found using the tables. Now you can return to the ‘clever’ solution above and proceed with evaluating this anti-derivative over the limits of r and then finding and evaluating the anti-derivative with respect to theta.