# Thread: another solid revolution / intergration T_T

1. ## another solid revolution / intergration T_T

a hemispherical bowl of radius a units is filled with water to a depth of a/2 units.Findby intergration, the volume of the water.

2. Originally Posted by chibiusagi
a hemispherical bowl of radius a units is filled with water to a depth of a/2 units.Findby intergration, the volume of the water.
When r=a, the d=a, as the bowl is hemisphere, d=r.
Therefore, when d=0.5a, r=0.5a.

Volume of a hemisphere
$=\frac{2}{3}\pi r^3$
$=\frac{\pi a^3}{12}$

I only know this method.
I don't know the method by integration, sorry.

3. You should get $V=\pi\int_{\frac{a}{2}}^a{(a^2-x^2)}\,\mathrm{d}x$.

4. When r=a, the d=a, as the bowl is hemisphere, d=r.
Therefore, when d=0.5a, r=0.5a.

Volume of a hemisphere
=\frac{2}{3}\pi r^3
=\frac{\pi a^3}{12}
If you cut a hemisphere in half the two parts are not hemispheres so this method doesn't work.

I would use the equation for a circle $y = \sqrt {a^2-x^2}$ and rotate the region between a/2 and a around the x-axis.