A water tank in the form of an inverted frustum of a cone has an altitude of 8 ft., and upper and lower radii of 6 ft. and 4 ft., respectively. Find the volume of the water tank and the wetted part of the tank if the depth of the water is 5 ft.

Question

anonymous · Answer

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anonymous · Answer

$\Huge V=\frac{\pi h} {3}( R^2+Rr+r^2)$
such that
$V$ is the volume
$R$ the big raduis
$r$ the small raduis

anonymous · Answer

* and $h$ is the hight

$\Huge V=\frac{8\pi}{3}(6^2+6×4+4^2)$

anonymous · Answer

$\Huge V=  \frac{608\pi}{3}$

anonymous · Answer

i hope that u get it , let me know if not :D