Volume question...

[Boba_Fett_0]

A bowl is shaped like a hemisphere with diameter 30cm. A ball with diameter 10cm is placed in the bowl and water is poured into the bowl to a depth of h centimeters. Find the volume of water in the bowl.

I'm not really sure where to begin, though I think there will be two separate equations to solve it, one in which 0 < h < 10 and another in which h > 10

 

[Deleted member 4993]

A bowl is shaped like a hemisphere with diameter 30cm. A ball with diameter 10cm is placed in the bowl and water is poured into the bowl to a depth of h centimeters. Find the volume of water in the bowl.

I'm not really sure where to begin, though I think there will be two separate equations to solve it, one in which 0 < h < 10 and another in which h > 10

Follow the method shown at:

http://www.freemathhelp.com/forum/viewtopic.php?t=25763

find the volume of the larger cap (r = 15) for height 'h'..................................(1)

find the volume of the smaller cap (r = 5) for height 'h'..................................(2)

Subtract (2) from (1)

 

[galactus]

Consider if the ball is partially submerged and when it is totally submerged.

If it is partially submerged, \(\displaystyle \L\\0\leq{h}<10\)

If it is totally submerged, \(\displaystyle \L\\10\leq{h}<15\)

The radius of the bowl is 15 cm and the ball has radius 5 so points on the sections satisfy the equations:

\(\displaystyle \L\\x^{2}+y^{2}=225\)

and

\(\displaystyle \L\\x^{2}+(y+10)^{2}=25\)

Find for the two separate cases.

For \(\displaystyle \L\\0\leq{h}<10\):

\(\displaystyle \L\\V={\pi}\int_{-15}^{h-15}\left[(225-y^{2})-(25-(y+10)^{2})\right]dy\)

For \(\displaystyle \L\\10\leq{h}<15\)

\(\displaystyle \L\\V={\pi}\int_{-15}^{-10}\left[(225-y^{2})-(25-(y+10)^{2})\right]dy+{\pi}\int_{-10}^{h-15}(225-y^{2})dy\)

I hope I typed this out right!.

 

[Boba_Fett_0]

Thanks guys, I think I understand it now