Show that $$\iint_R\left(x+y\right)\mathrm{d}x\mathrm{d}y=1$$ where $R$ is the triangle with vertices at $\left(0,0\right)$, $\left(1,0\right)$ and $\left(0,2\right)$.
Having trouble here with this question, am I supposed to just use polar coordinates with $x=\cos\theta$, $y=\sin\theta$, $\mathrm{d}x\mathrm{d}y=r\mathrm{d}r\mathrm{d}\theta$ with limits being $0\leq x\leq1$ and $0\leq y\leq2$ ? I tried writing it out but it doesn't make sense because of the integration factors.
Try drawing the triangle on paper. Can you work out the range of $x$ within this triangle for a given value of $y$?
Once you've done that, you should be able to express the integral directly with Cartesian coordinates where the limits for the inner integral depend on $y$.
Polar coordinates are not useful here. We have by Fubini's theorem $$\iint_R\left(x+y\right)\mathrm{d}x\mathrm{d}y=\intop_{x=0}^{1}\left(\intop_{y=0}^{2\left(1-x\right)}\left(x+y\right)\mathrm{d}y\right)\mathrm{d}x=\intop_{x=0}^{1}\left[xy+\frac{y^2}{2}\right]_{y=0}^{y=2\left(1-x\right)}\mathrm{d}x$$ $$=\intop_{x=0}^{1}\left(2x\left(1-x\right)+2\left(1-x\right)^2\right)\mathrm{d}x=2\intop_{x=0}^{1}\left(1-x\right)\mathrm{d}x=2\left[x-\frac{x^2}{2}\right]_{x=0}^{x=1}=1.$$
