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BOAS
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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!