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

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Answer given:

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Why is the height $e^{-x^2}$, where did the $y$ goto?

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    The $y$ got "integrated out" just like the $x$ did. The text is evaluating the volume in two different ways in order to produce the important result $\int_{-\infty}^\infty e^{-t^2}dt=\sqrt{\pi}$. I wish the book had used $re^{-r^2}$ in the first part.2012-04-30
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    Note that the "height" of the intermediate (one dimensional) integral is not $e^{-x^2}$ but rather $2\pi x e^{-x^2}$. Also you might want to (at least consider) changing the book. I have seen this explained much better several times. If such problems as this arise often while reading it you should definitely change the literature.2012-04-30
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    Totally agree @AndréNicolas, using x as the variable of integration in (i) is very misleading.2012-04-30

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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 $B=\sqrt{\pi}$