Problem:
Change the Cartesian integral into an equivalent polar integral and
evaluate the polar integral:
\begin{eqnarray*}
\int_{0}^{2} \int_{0}^{\sqrt{1 - (1 - x)^2}} \frac{x + y}{x^2+y^2} \, dy \, dx \\
\end{eqnarray*}
Answer:
In this case, $x$ is going from $0$ to $2$. $y$ is going from $0$ to
$1$.
\begin{eqnarray*}
\int_{0}^{2} \int_{0}^{\sqrt{1 - (1 - x)^2}} \frac{x + y}{x^2+y^2} \, dy \, dx
&=& \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sqrt{3}}
r\Big( \frac{r \cos{\theta} + r\sin{\theta} }{r^2} \Big) \, dr d\theta \\
\int_{0}^{2} \int_{0}^{\sqrt{1 - (1 - x)^2}} \frac{x + y}{x^2+y^2} \, dy \, dx
&=& \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sqrt{3}}
\cos{\theta} + \sin{\theta} \, dr d\theta \\
\int_{0}^{2} \int_{0}^{\sqrt{1 - (1 - x)^2}} \frac{x + y}{x^2+y^2} \, dy \, dx
&=& \int_{0}^{\frac{\pi}{2}} ( \sin{\theta} + \cos{\theta} ) r \Big|_{0}^{\sqrt{3}} d\theta \\
\int_{0}^{2} \int_{0}^{\sqrt{1 - (1 - x)^2}} \frac{x + y}{x^2+y^2} \, dy \, dx
&=& \sqrt{3} \int_{0}^{\frac{\pi}{2}} (\sin{\theta} + \cos{\theta}) d\theta \\
\int_{0}^{2} \int_{0}^{\sqrt{1 - (1 - x)^2}} \frac{x + y}{x^2+y^2} \, dy \, dx
&=& \sqrt{3} ( -\cos{\theta} + \sin{\theta} ) \Big|_{0}^{\frac{\pi}{2}} \\
\int_{0}^{2} \int_{0}^{\sqrt{1 - (1 - x)^2}} \frac{x + y}{x^2+y^2} \, dy \, dx
&=& \sqrt{3} ( -(0) + 1 - ( -1 + 0 ) ) \\
\int_{0}^{2} \int_{0}^{\sqrt{1 - (1 - x)^2}} \frac{x + y}{x^2+y^2} \, dy \, dx
&=& 2 \sqrt{3}
\end{eqnarray*}
However, the book's answer is $\frac{\pi}{2} + 1$. I would like to know where
I went wrong and my guess is that I have the wrong limits when I convert to polar coordinates.
Thank you,
Bob