Hoping someone can let me know if I am correct with this solution.
Given $ \int \!\!\! \int_D e^{-x^2-y^2} \, dA $
where D is bounded by $x= \sqrt{4-y^2} $ and the $y$ axis
I need to Identify the integral in Cartesian coordinates then convert to cylindrical coordinates?
This is my attempt to find the x bounds $y \le x\le \sqrt{4-y^2} $
to find the $y$ bounds
$0\le y\le\sqrt{4} $
so am I correct to write the Cartesian integral as
$ \int_{0}^{2} \! \int_{y}^{\sqrt{4-y^2}}e^{-x^2-y^2} \,dx\,dy $
To convert to Cylindrical coordinates this is bound by the y axis in quadrant 1 &4 so I get
$ \int_{ -\pi /2}^{\pi/2} \! \int_{0}^{2} e^{-r^2} \,rdrd \theta $
The cylindrical coordinates integral is OK. The Cartesian coordinates formula does not look fine to me.It all starts with your first inequality. Why is $y\le x$? Bounded by $y$ axis means $x=0$. Your integration domain $D$ is the half disk of radius 2, centered at 0, in quadrants 1 & 4. You can describe this domain either in first coordinate $x$ or first coordinate $y$. In the $x$ case, $0\le x\le 2$ and $-\sqrt{4-x^2}\le y \le \sqrt{4-x^2}$. If you start integrating $y$ first, you have $-2 \le y \le 2$ and $0\le x \le \sqrt{4-y^2}$