Question

Volume Question

Answer 1

for the rod i am meant to assume it is a cylinder and for the oblate spheroid I am given v = (4/3)(ab^2)

Answer 2

@sillybilly123 @Nnesha @563blackghost

Answer 3

@Hero

Answer 4

The prolate ellipsoid described by \(\left( \dfrac{x}{b} \right)^2 + \left( \dfrac{y}{b} \right)^2 + \left( \dfrac{z}{a} \right)^2 = 1\), with \(a > b\), has volume \(V = \dfrac{4 \pi a b^2}{3}\)

tbc, \(b\) is the radius of the circle on the x-y plane.

You can get that by using a Volume of revolution....or just look it up

If the rod has the same volume then:


\(\dfrac{\pi d^2 L}{4} = \dfrac{4 \pi a b^2}{3} \implies L = \dfrac{16 a b^2}{3 d^2} \qquad \star\)

If they have the same length, ***and *** the rod is lying along the z-axis (which makes physics-wise, though a drawing is always paints a thousand words), then:

\(L = 2a \qquad \circ\)

And so:

\( \star, \circ \implies d = \dfrac{2 \sqrt 2 b}{\sqrt 3 } \)

And so:

\(\dfrac{L}{d} = \dfrac{2a}{ \dfrac{2 \sqrt 2 b}{\sqrt 3 }} = \sqrt{\dfrac{3}{2}}\dfrac{a}{b}\)