Convert $\int_0^{\pi/6} \int_{0}^{2/\cos\theta} r \ dr \ d\theta$ to cartesian coordinates?

Question

$$\int_0^{\pi/6} \int_{0}^{2/\cos\theta} r \ dr \ d\theta$$

I'm not sure how to sketch out the region of this while it's still in polar coordinates (in particular, I don't know how to interpret $r=2/ \cos\theta$ or what that would look like on the cartesian plane).

So I first try to convert it. At some point $r = 2/\cos\theta$, so $r \cos\theta = x = 2$.

Also, the original outer integrals tells us we're dealing with the right angle triangle within the unit circle corresponding to angle $\pi/6$. This triangle has a height of $1/2$ and a width of $\sqrt3/2$. I use these two numbers to calculate the slope of the triangle's diagonal and find that that diagonal can be described with the line $y=(\sqrt3/4)x$ .

So the integral I arrive at is $$\int_0^2 \int_0^{(\sqrt3/4)x} dy \ dx$$

But apparently that's wrong. I'm not sure how else to go about this, any help is appreciated!

Answer 1

Your triangle has horizontal leg =2 (since $x=r\cos\theta=2$). Since the angle opposite to the vertical leg is $\pi/6$, the height is $2\tan(\pi/6)=2\sqrt{3}/3$, and the equation of the hypotenuse is $y=(\sqrt{3}/3)x$. Your integral in Cartesian coordinates is $$ \int\limits_0^2\int\limits_0^{\sqrt{3}x/3}\,dy\,dx $$ since $rdrd\theta=dxdy$.

Answer 2

The integral represents the shaded area and the correct conversion is

$$\int_0^{\pi/6} \int_{0}^{2/\cos\theta} r dr d\theta= \int_0^2 \int_0^{x/\sqrt3 }dy dx$$

The triangle has the width $2$ and the height $2/\sqrt3$.