# calculus - Double integral with polar coordinates

$$\int\int_D xdxdy, D = \{(x, y) \in R^2: y=3x, y=x, x^2+y^2 = 1\}$$

If I sketch the domain and switch to polar coordinates it seems like I'm supposed to compute this integral:

$$\int_\frac{\pi}{3}^\frac{\pi}{4} \int_0^1 \rho^2cos(\theta)d\rho d\theta$$.

Is this right?

It's almost correct – the only mistake is that $$\int_{\pi/3}^{\pi/4}$$ should be $$\int_{\pi/4}^{\tan^{-1}3}$$, where $$\tan^{-1}3$$ is the slope of the line $$y=3x$$.