Evaluating double intergral of $\int_{0}^{1} \int_{0}^{2} \sqrt{5x^2+4y^2+1}\;dy\;dx$

Question

Are there practical methods for integrating something like this (i.e. something where it's not a disk so you cannot simply use polar coordinates) without using a calculator?

$$\int_{0}^{1} \int_{0}^{2} \sqrt{5x^2+4y^2+1}\;dy\;dx$$

Answer 1

In this particular instance, you can do it without too much trouble. There are no completely general ways to go about nasty integrals, so I don't know how much can be said about that aspect of the question. However, this particular "type" of problem deals with quadrics, and things that look like parabaloids, and for those, we can do it by hand. The basic idea is that you have already identified that this function is "almost" amenable to polar coordinates, but the pesky constants make the transformation not work out. So we first make a linear change of coordinates which eliminate those constants, and then the trick works.

The trick is to make the change of coordinates $u = \sqrt{5}x, v = 2y$. Then you can rewrite your integral as $$\int_a^b \int_c^d \sqrt{u^2 + v^2 + 1} \ (\det J) \ dv du$$

where $a,b,c,d$ are the new bounds of integration from the change of variables, and $\det J$ is the Jacobian determinant. We see that $a = 0, b = \sqrt{5}, c = 0, d = 4$, and the Jacobian matrix is

$$J = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y}\\ \end{bmatrix} = \begin{bmatrix} \sqrt{5} & 0 \\ 0 & 2\\ \end{bmatrix} $$

so $\det J = 2 \sqrt{5}$.

We have obtained the integral $$2 \sqrt{5} \int_0^\sqrt{5} \int_0^4 \sqrt{u^2 + v^2 + 1} \ dv du$$

which is amenable to using polar coordinates now. We write $ u = r \cos \theta$ and $v = r \sin \theta$ and we for now ignore the bounds, obtaining the indefinite integral:

$$2 \sqrt{5} \iint r\sqrt{r^2 + 1} \ dr d\theta$$

which is easily evaluated to be $\frac{2 \sqrt{5}}{3} \theta (r^2+1)^{3/2}$. Now change back to $u,v$ coordinates and evaluate.