$\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 $y=3x$.

  • $\begingroup$ Thank you very much. $\endgroup$ – Frost832 May 25 at 9:05

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