# Thread: 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 $\displaystyle x$ be the variable depth and $\displaystyle R$ be the variable radius.

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

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

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

$\displaystyle dV=\pi R^2dx$

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

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

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

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

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

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

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

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# volume of water in a hemispherical bowl

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