Integrating using polar co-ordinates.

I ran into some trouble with some integrals that use polar coordinates.

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Change the following cartesian integrals into an equivalent polar integral. Then evaluate the polar integral.

Sounds fair!(Clapping)

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1). $\displaystyle \int_0^6 \int_0^y x~dx~dy$

....(Angry)

The region for integration is a triangle which I thought was a little weird. I think the r limits start from 0 to something (I called it b in the integral below), but I think the $\displaystyle \theta$ limits are $\displaystyle \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$.

So, without the final r limit, I have $\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_0^b r^2cos(\theta)~dr~d \theta$.

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2). $\displaystyle \int_0^2 \int_0^{\sqrt{1-(x-1)^2}}\frac{x+y}{x^2+y^2}~dy~dx$

My problem with this one is that the region of integration is not a circle centred at the origin but rather the point $\displaystyle (1,0)$. I am aware that the region is a semi circle. For this reason, are the $\displaystyle \theta$ limits $\displaystyle 0 \leq \theta \leq \pi$?

Are the r limits $\displaystyle 0 \leq r \leq \sqrt{1-(rcos(\theta)-1)^2}$?

However, the integrand is going to be $\displaystyle \frac{r(rcos(\theta)+rsin(\theta))}{r^2}=cos(\thet a)+sin(\theta)$.

So overall, I get that the integral should be $\displaystyle \int_0^{\sqrt{1-(rcos(\theta)-1)^2}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta$.

Clearly the r limits are incorrect!

From here, I decided to put $\displaystyle \theta=\pi$ into the upper limit of the double integral:

$\displaystyle \int_0^{\sqrt{r^2+2r}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta=\int_0^{\pi} \int_0^{\sqrt{r^2+2r}} cos(\theta)+sin(\theta)~d \theta dr$.

Is this integral the correct one to evaluate?