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## Homework Statement

Show that the moment of inertia of a hollow cone of mass M, radius R, and height h about its base is ##\frac{1}{4}M(R^2+2h^2)##

## Homework Equations

##I=\int r^2dm##

where r is the perpendicular distance from the axis

Surface Area of a cone ##= \pi R (R^2+h^2)^{1/2}##

## The Attempt at a Solution

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Mass density ##\sigma = \frac{M}{SA} = \frac{M}{\pi R(R^2+h^2)^{1/2}}##

Now ##I = \int r^2dm = \int r^2\sigma dA = \sigma \int r^2dA##

Now the distance from its base is given by ##r = r(x) = h(1-x/R)## and from the image I made it appears the area element is simply ##dA = 2\pi xds##, and ##ds = \sqrt{dx^2+dy^2} = \sqrt{1+y'(x)^2}dx##. Trivially ##y'(x)^2 = (h/R)^2## giving ##dA = 2\pi x \sqrt{1+(h/R)^2}dx## and $$I = \sigma \int r^2dA =\sigma \int_0^R h^2\left(1-\frac{x}{R}\right)^22\pi x \sqrt{1+\left(\frac{h}{R}\right)^2}dx \\ I = 2\pi\sigma h^2\sqrt{1+\left(\frac{h}{R}\right)^2}\int_0^R \left(1-\frac{x}{R}\right)^2 xdx$$

which clearly isn't going to give the answer since there's no way ##\sqrt{1+(h/R)^2}## will be cancelled