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## Homework Statement

Let a cone with height [itex]h[/itex] and base area [itex]A[/itex] have the density [itex]\rho (x) = \rho_{0} \frac{3x^{2} + 2xh}{h^{2}}, 0 \leq x \leq h[/itex]

the relation between cone radius [itex]r[/itex] and distance from cone apex [itex]x[/itex] is given by:

[itex]r = (\frac{B}{\pi h^{2}})^{\frac{1}{2}}x[/itex]

Find the total mass [itex]M[/itex] of the cone.

## Homework Equations

## The Attempt at a Solution

Ok, so i've seen problems where the density varies, but this is the first one i've looked at where both the eare and the density vary.

I take a small slice of the cone, [itex]V \approx A \Delta x[/itex] (this works for a rod, but hopefully this is fine when I take the limit of delta x)

Mass of the slice, [itex]m \approx \rho (x) A \Delta x[/itex]

Total mass of the cone, is the sum of the slices [itex]= \Sigma A \Delta x \rho (x)[/itex]

[itex]lim \Delta x \rightarrow 0[/itex]

[itex]M = \int^{h}_{0} A \rho (x) dx[/itex]

The area of a thin slice is given by [itex]A = \frac{r^{2} \pi h^{2}}{x^{2}}[/itex]

So my integral becomes [itex]M = \int^{h}_{0} \frac{r^{2} \pi h^{2}}{x^{2}} \rho (x) dx[/itex]

Subbing in the density funcion,

[itex]M = \int^{h}_{0} \frac{r^{2} \pi h^{2}}{x^{2}} \rho_{0} \frac{3x^{2} + 2xh}{h^{2}} dx[/itex]

[itex]M = \int^{h}_{0} \frac{r^{2}\pi h^{2} \rho_{0} (3x^{2} + 2xh)}{x^2 h^2} dx[/itex]

Firstly, is my method correct?

and secondly, how do I go about performing this integral?

thanks!