Volume of solid bounded above by $z=e^{-(x^2+y^2)}$ and below by xy plane 
I need to find the volume of solid bounded above by $z=e^{-(x^2+y^2)}$ and below by xy plane


Answer given: 

Why is the height $e^{-x^2}$, where did the $y$ goto? 
 A: The first part (i) is just a substitution of the cartesian coordinates by polar coordinates. Basically what you do is instead of integrating over all of $x$ and all of $y$, you integrate over all rings ($\phi \in [0,2\pi)$ ) with radius $r$. The infinitesimal unit of integration is thus $dx\,dy = r\,dr\,d\phi$ (if you have never seen this before let me know in the comments and I will try to include a picture. Otherwise look at substitution rules)
$$\begin{align}
x^2+y^2&=r^2 \\
dx\,dy&=r\,dr\,d\phi \\
\Rightarrow \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}\,dx\,dy &= \int_{0}^{2\pi}\!\!\!d\phi \int_0^\infty \!\!\!dr\, r e^{-r^2} \\
&= 2\pi\cdot\left[-\frac{1}{2}e^{-r^2}\right]_{r=0}^\infty \\
&=\pi
\end{align} $$
In the second (ii) part they only argue, that this must be the square of the integral $\int_{-\infty}^\infty e^{-x^2}\,dx$.
$$\begin{align}
B &= \int_{-\infty}^\infty e^{-x^2}\,dx \\
B^2 &= \int_{-\infty}^\infty e^{-x^2}\,dx \cdot \int_{-\infty}^\infty e^{-y^2}\,dy \\
&= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}\,dx\,dy \\
&= \pi
\end{align} $$
Therefore $B=\sqrt{\pi}$
