Evaluate the double integral $\int_0^1\int_0^x x \sqrt{x^2+3y^2} \,dy\,dx$ changing it to polar coordinates

Question

I'm asked to evaluate the integral using polar coordinates

$$\int_0^1\int_0^x x \sqrt{x^2+3y^2} \,dy\,dx$$

The approach I took was to change variables, letting $u = x$ and $v=\sqrt{3}y$, this way the integral becomes $$\frac{1}{\sqrt{3}}\int_0^1\int_0^{\sqrt{3}u} u \sqrt{u^2+v^2}\, dv\,du$$

and it gets pretty obvious to change variables again, this time to polar coordinates using $u = r \cos(\theta)$ and $v = r \sin(\theta)$

$$\frac{1}{\sqrt{3}}\int_?^?\int_?^? r^3\cos(\theta)\, d\theta \,dr$$

But the deal is I can't seem to get the limits of integration right. I've tried some approaches but I got to nowhere, basically, so it's not even worth writing them out here. Thanks for the help in advance.

Answer 1

Since the region of integration in u-v coordinates is still a triangle, it can be converted to polar coordinates. We can look at the bounds for r and $\theta$ separately.

In general with polar coordinates, it's easier when the inner integral is evaluated with respect to r ($\theta$ held constant), so we'll do that here.

What are the bounds for $\theta$? Looking at the following diagram, we can see that $\theta$ goes from 0 to approximately $\frac{\pi}{4}$.

To find $\theta$ exactly, we can use basic trigonometry. The triangle has legs of length $\sqrt{3}$ and 1. Therefore, $\tan(\theta) = \frac{\sqrt{3}}{1}$, from which we can take the inverse to find $\theta=\tan^{-1}(\sqrt 3)$. So, with the bounds of the outer integral, we're effectively doing a circular "scan" from $\theta = 0$ to $\theta = \tan^{-1}(\sqrt 3)$.

To find the bounds of the inner integral for r, imagine fixing a particular value of $\theta$ between 0 and $\tan^{-1}(\sqrt 3)$. Now, we need to figure out the line that r will traverse, i.e., where does r start and where does it stop, for a particular fixed value of $\theta$.

Such a line will look like the orange dotted lines in the following diagram (each orange line corresponds to a separate fixed $\theta$).

So then, r starts at r = 0, and ends at the line u = 1. The standard formulas for polar coordinates are $x = r\,\cos(\theta)$ and $y=r\,\sin(\theta)$, except that we are using u and v instead of x and y. Therefore, u = 1 implies that $r\,\cos(\theta) = 1$, which implies that $r = \frac{1}{\cos(\theta)} = \sec(\theta)$.

The final integral should look like this: $$\frac{1}{\sqrt 3}\int_{0}^{\tan^{-1}(\sqrt 3)}\int_0^{\sec(\theta)}r^3\,\cos(\theta)\,dr\,d\theta$$

Answer 2

hint

Put $$y=\frac {xt}{\sqrt {3} }$$

the integral becomes

$$\frac {1}{\sqrt {3}}\int_0^1 x^3(\int_0^\sqrt{3}\sqrt {1+t^2}dt)dx $$

with $t=\sinh(u) $, you can finish.