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

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.

• Why not directly compute each integral $$\int_y^1 x \sqrt{x^2+3y^2} dx$$ since $$f(x)=x \sqrt{x^2+3y^2}$$ is an easy derivative? – Did May 19 '17 at 21:00
• I think the Jacobian of your transformation is $\;\frac1{\sqrt3}\;$ ... – DonAntonio May 19 '17 at 21:02
• In either order you choose to integrate you get at least one ugly integral to solve: something involving a hyperbolic sine and stuff... – DonAntonio May 19 '17 at 21:04
• Neither the limits of integration after the initial transformation nor the Jacbian is correct. – Mark Viola May 19 '17 at 21:07
• Thanks for the tips. I noticed my mistake on the Jacobian and the limits after the first transformation. Even though I've come to know of other (easier) ways of evaluating it, I'm curious as how it would be done if I followed the path I took originally. I'll edit the question. – bnoite May 19 '17 at 21:47

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

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.

• Still the hyperbolic sine is there...but that seems to be as simple as it gets. Changing the integration order also yields a similar integral and, perhaps, is a little easier. – DonAntonio May 19 '17 at 21:20
• Thank you for the hint. It really made things a lot easier. I'm still trying to wrap my mind around the way I was doing originally, though. I'll change the original question to be more clear. – bnoite May 19 '17 at 21:36