• May 18th 2006, 12:02 AM
LilDragonfly
I really need someone to help me understand this very clearly:

A hemispherical bowl has an internal radius of 13 cm, and contains water to a maximum depth of 8 cm. Find the volume of water in the bowl. Show all working.

Thank you!
• May 18th 2006, 02:53 AM
ticbol
Quote:

Originally Posted by LilDragonfly
I really need someone to help me understand this very clearly:

A hemispherical bowl has an internal radius of 13 cm, and contains water to a maximum depth of 8 cm. Find the volume of water in the bowl. Show all working.

Thank you!

The bowl then is a hollow hemisphere. Water level is 8cm deep at the center.

One way to get the volume of water is by established formulas. One formula is that of a spherical cap,
V = pi(h/6)[3r^2 +h^2] ---------(i)
where
h = height of cap == depth of water here = 8cm
r = radius of base of the cap.

If the center of the circle is at (0,0), the equation of the great circle of the sphere of the inside of the bowl is
x^2 +y^2 = 13^2. -------(ii)
The y of the water level then is -5cm.
So the x of the water level is
x^2 +(-5)^2 = 13^2
x^2 = 169 -25 = 144
x = +,-12cm.
Hence, the radius of the base of the cap here, r = 12cm.

Therefore,
V = pi(8/6)[3(12^2) +8^2] = 661.33pi = 2077.64 cu.cm. ----answer.

-----------------------------------
If you are into integration, one way is to place the bottom of the water at (0,0), whereby the center of the great circle of the sphere of the inside of the bowl is at at (0,13), thus, the equation of that circle is
(x-0)^2 +(y-13)^2 = 13^2
x^2 +(y-13)^2 = 169 -----(iii)

Still one way to get the volume of the water in the bowl here is by disc method:
dV = pi(x^2)dy -----where x = radius of horizontal disc, and dy = thickness of the disc.
So,
V = INT.(0-->8)[pi(x^2)]dy -----------(iv)

We express the x^2 into y^2, because of the dy.
From (iii),
x^2 = 169 -(y-13)^2
x^2 = 169 -(y^2 -26y +169)
x^2 = 26y -y^2 -------------(iiia)
Substitute that into (iv),
V = (pi)*INT.(0->8)[26y -y^2]dy
V = (pi)[(26y^2)/2 -(y^3)/3](0->8)
V = (pi)[13y^2 -(y^3)/3](0->8)
V = (pi){[13(8^2) -(8^3)/3] -[0]}
V = 661.33pi = 2077.64 cu.cm. --------------answer.

Another way, using the setup when the center of the great circle is at (0,0).
Same horizontal circular discs.
dV = pi(x^2)dy
Integrate this dV from (y = -13) to (y = -5)

x^2 +y^2 = 13^2
x^2 = 169 -y^2

V = (pi)INT.(-13 --> -5)[169 -y^2]dy
V = pi[169y -(y^3)/3](-13 --> -5)
V = pi{[169(-5) -(1/3)(-5)^3] -[169(-13) -(1/3)(-13)^3]
V = pi{[-845 +41.67] -[-2197 +732.33]}
V = pi{661.34}
• May 19th 2006, 02:49 PM
LilDragonfly
Wow, that was a super response. Thank you for that :) Could I e-mail you with my original answer so that you can let me know what I was doing wrong? This would be a great help to me in the future because I will understand it better. I tried to attach it but the file is too large. Thanks again.
• May 19th 2006, 05:25 PM
ticbol
Quote:

Originally Posted by LilDragonfly
Wow, that was a super response. Thank you for that :) Could I e-mail you with my original answer so that you can let me know what I was doing wrong? This would be a great help to me in the future because I will understand it better. I tried to attach it but the file is too large. Thanks again.

Sorry, but I prefer that you show it here. I do not welcome private messages. I do not reply to private messages re Math.
If the file is too large, why not break it into smaller files that you can post separately.
• May 19th 2006, 09:50 PM
LilDragonfly
Ok that is completley understandable. It took a long while to cut down the attachments to size. I'm sorry that the equations appear squashed, I hope that you can still understand it. I have the first part of my working in this message, and the second part as attachments 1, 2, 3, 4 and 5. Thank you for your quick response.

x² + y² = 13² = 169 = y = √169 - x²
So,
when y = - 5, x² = 169 - (-5)² = 169 - 25 = 144 = x = 12.
Rotate the position of the quarter circle about the y axis between x = 0 and x = 12.
• May 19th 2006, 11:27 PM
ticbol
Quote:

Originally Posted by LilDragonfly
Ok that is completley understandable. It took a long while to cut down the attachments to size. I'm sorry that the equations appear squashed, I hope that you can still understand it. I have the first part of my working in this message, and the second part as attachments 1, 2, 3, 4 and 5. Thank you for your quick response.

x² + y² = 13² = 169 = y = √169 - x²
So,
when y = - 5, x² = 169 - (-5)² = 169 - 25 = 144 = x = 12.
Rotate the position of the quarter circle about the y axis between x = 0 and x = 12.

Ummm, okay. So you used vertical cylindrical shells as your dV. That is one way.
You got it all correct, except that you forgot that the height or depth of the water is 8 cm only. Your solution is for more than the water in the bowl.

Your dV = 2pi(x)[f(x)]dx is for the whole semispherical bowl. (If the integration is from x=0 to x=12, then a portion of the water volume will not be included, though, because above the water level at y = -5cm, the x is more than 12cm.)
dV = 2pi(x)[f(x) -5]dx ---the water level is at y = -5 cm only, remember.
dV = 2pi(x)[sqrt(169 -x^2) -5]dx
So,
V = (2pi)INT.(0-->12)[sqrt(169 -x^2) -5](x dx)
V = (2pi)INT.(0-->12){[sqrt(169 -x^2)](x dx) -[5x]dx}
Blah, blah, blah,
V = 4339.6 -(2pi)[(5x^2)/2](0-->12)
V = 4339.6 -5pi[x^2](0-->12)
V = 4339.6 -5pi[144 -0]
V = 4339.6 -2261.9
V = 2077.7 cc ---------------answer. Same as my findings.

That 2261.9 cc of water is the volume not filled up in the semispherical bowl.
------------------
Ooopps, No, rather, that 2261.9 cc is the volume of the solid cylinder above the water level up to the top of the bowl. I forgot that the dx was from x=0 to x=12 only.
The x for the lid or top of the bowl is 13cm.