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I have to calculate the volume under the surface $$f(x,y)=e^{-(x^2+y^2)}$$

I have no clue what to do, but I'm just using the integration I learned.

I take a little slice, and the area of that slice should be $$\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\,dx$$

Then, I have to evaluate this slice for each $y$, so I have to take the integral of this integral, if it makes sense or something...

So I got $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\,dx\,dy$$

I have no clue how to evaluate the integral, but I punched this exact expression into a calculator and got 3.1415926..., so I think I did something meaningful...

Does what I did work, find the area of a slice for some random $y$ and integrate that slice area expression from $y=-\infty$ to $y=+\infty$??

Update: How do I switch this into polar coordinates?

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    switch to polar coordinates – 2017-02-26
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    how do I do that @WillJagy – 2017-02-26
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    Have you considered converting to polar coordinates? – 2017-02-26
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    i don't know how to do that @amd – 2017-02-26
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    The translation from $(x, y)$ to $(r, \theta)$ is to plug in $r\cos\theta$ for $x$ everywhere you see it, $r\sin\theta$ for $y$ every time you see it, and $dx\,dy=r\,dr\,d\theta$. You have to figure out the bounds for $r$ and $\theta$, but in this case, the radius goes from $0$ to $\infty$ and the angle goes from $0$ to $2\pi$. – 2017-02-27
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    Note that $r^2 = x^2+y^2$ and $dxdy = rdrd\theta$. Proving the latter fact is not trivial, but not overly difficult. – 2017-02-27

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Denote $x=r\cos\theta$ and $y=r\sin\theta$, thus $$ J = \det\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos \theta \end{vmatrix} =r\cos^2\theta + r \sin^2 \theta = r(\cos^2\theta+\sin^2\theta)=r. $$ Now, as you don't have any restrictions, your integration region is given by $$ D=\{(r\cos\theta, r\sin \theta) | 0 \le \theta \le 2\pi, 0 \le r < \infty\},$$ $$ \int \int e^{-(x^2+y^2)}dxdy = \int_{0}^{2\pi} \int_{0}^\infty e^{-r^2}rdrd\theta= -\frac{1}{2}\int_{0}^{2\pi} (e^{-r^2}|^{\infty}_{0} d\theta =\frac{1}{2}\int_{0}^{2\pi} d\theta = \pi, $$ where in the third step, one can see that $\frac{\partial}{\partial r}e^{-r^2}=-2re^{-r^2}$, thus $$\int e^{-r^2} r dr = -1/2\int(-2) e^{-r^2} rdr = -\frac{1}{2}e^{-r^2}+C. $$

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    Where did that -1/2 come from?? – 2017-02-27
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    Hi, pls see the edited answer. – 2017-02-27