My attempt:
I calculated density by substituting $x = 10$ in the given expression.
Used formula for volume of a cone to calculate the volume.
Worked out the mass using the formula M = D$\times$V.
My final answers for mass was approximately $261800kg$.
I doubt if this is the correct answer! I suspect that I am missing something. Can someone help, please?
$\endgroup$ 33 Answers
$\begingroup$Assuming that the base of the cone $C$ is located at the plane $z=0$, then, given a point $(x,y,z)$ of the cone, the density at that point is $z$, not $x$. So, the computation that you have to do is$$\int_C100(20-z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^5\int_0^{10-2z}100(20-z)r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta.$$
$\endgroup$ $\begingroup$Split the cone into a series of thin disks each with volume ${\rm d}V = \pi r^2\, {\rm d} x$ with $x = 0 \ldots h$ and $r = R \left(1-\tfrac{x}{h} \right)$.
$$ m = \int \rho(x) {\rm d}V = \pi R^2 \int \limits_0^h 100 (20-x) \left(1-\tfrac{x}{h} \right)^2 {\rm d}x $$
$\endgroup$ $\begingroup$Let the height be $x$. The variation of radius can be found out as follows. At $x=10,r=0 $ and $x=0,r=5$ thus $r=c+kx$ solving we have $r=5-0.5x$ then $dm=pdV=100(20-x)\pi(5-0.5x)^2dx$ Now integrate this from $x=0-10$. You will get the total mass.
$\endgroup$ 5
