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:
$$\frac{1}{\sqrt 3}\int_{0}^{\tan^{-1}(\sqrt 3)}\int_0^{\sec(\theta)}r^3\,\cos(\theta)\,dr\,d\theta$$
