A Big, Weird Mountain

We can reduce the number of variables in the expression to one:

$H_0e^{-kr}$

The volume of this 'mountain' can be approached by using concentric cyclindrical 'shells', with 'height' $H_0e^{kr}$ , 'width' $2\pi r \Delta r$ , and 'volume' $2\pi H_0re^{-kr}\Delta r$ . As $\Delta r \to \infty$ , this becomes the integral:

$2\pi H_0\int_0^{\infty} re^{-kr} dr$

Evaluate the antiderivative using integration by parts:

$\int re^{-kr} dr = -\frac{1}{k}re^{-kr}-\int -\frac{1}{k}e^{-kr} dr =-\frac{1}{k}re^{-kr}-\frac{1}{k^2}e^{-kr}$

At $r=0$ , this is simply $-\frac{1}{k^2}$ . At $x\to\infty$ , it is

$\lim_{x\to\infty} -\frac{1}{k}re^{-kr}-\frac{1}{k^2}e^{-kr}$

While the second term obviously approaches zero, the first one requires to be solved by l'Hopital's Rule:

$\lim_{x\to\infty}-\frac{r}{ke^{kr}}=\lim_{x\to\infty}-\frac{1}{k^2e^{kr}}=0$

Thus, the volume is $2\pi H_0(0-(-\frac{1}{k^2}))=\frac{2\pi H_0}{k^2}$ , and when our values of $k$ and $H_0$ are inputted, we obtain $\sim1,691,655.24 m^3$ . The volume of the mineral deposit is the volume of the mountain multiplied by $\frac{79}{1,000,000}$ , or $133.640764 m^3$ . Finally, we multiply by the density, $34 \frac{kg}{m^2}$ , to obtain our answer of

$\large\sim4543.786$