Need to prove that the formula for the volume of water V in a hemispherical bowl of radius r when the depth of water is h is given by:
V = ((pi*h^2)/3) * (3r-h)
I assume using integration. Solids of revolution maybe?
Any help appreciated.
Thanks
Need to prove that the formula for the volume of water V in a hemispherical bowl of radius r when the depth of water is h is given by:
V = ((pi*h^2)/3) * (3r-h)
I assume using integration. Solids of revolution maybe?
Any help appreciated.
Thanks
Draw a diagram. Let $\displaystyle x$ be the variable depth and $\displaystyle R$ be the variable radius.
$\displaystyle (r-x)^2+R^2=r^2$
$\displaystyle r^2+x^2-2rx+R^2=r^2$
$\displaystyle R^2=2rx-x^2$
$\displaystyle dV=\pi R^2dx$
= $\displaystyle \pi(2rx-x^2)dx$
$\displaystyle V=\int_{0}^{h}\pi(2rx-x^2)dx$
= $\displaystyle \pi(rx^2-\frac{x^3}{3})\big|_0^{h}$
= $\displaystyle \pi(\frac{3rx^2-x^3}{3})\big|_0^{h}$
= $\displaystyle \pi(\frac{3rh^2-h^3}{3})$
= $\displaystyle \frac{\pi h^2}{3}(3r-h)$