# Math Help - Volume of water in a hemispheical bowl

1. ## Volume of water in a hemispheical bowl

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

2. Originally Posted by anonymous_maths
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 $x$ be the variable depth and $R$ be the variable radius.

$(r-x)^2+R^2=r^2$

$r^2+x^2-2rx+R^2=r^2$

$R^2=2rx-x^2$

$dV=\pi R^2dx$

= $\pi(2rx-x^2)dx$

$V=\int_{0}^{h}\pi(2rx-x^2)dx$

= $\pi(rx^2-\frac{x^3}{3})\big|_0^{h}$

= $\pi(\frac{3rx^2-x^3}{3})\big|_0^{h}$

= $\pi(\frac{3rh^2-h^3}{3})$

= $\frac{\pi h^2}{3}(3r-h)$

3. Originally Posted by anonymous_maths
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, and the method of disks looks like the easiest method.

CB