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

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.

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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}

V = 2077.66 cc ----------------answer.