# the volume of a spherical segment with base radius, r and height, h, is given by the formula v= 1/6 pie*h (3r^2+h^2)a domed stadium is in the shape of a spherical segment with a base radius of 150...

the volume of a spherical segment with base radius, r and height, h, is given by the formula v= 1/6 pie*h (3r^2+h^2)

a domed stadium is in the shape of a spherical segment with a base radius of 150 m. the dome must contain a volume of 3 500 000 m^3. Determine the height of the dome at its centre to the nearest tenth of a metre.

*print*Print*list*Cite

V = (1/6)pi * h(3r^2 + h^2)

r = 150 m

v= 3,500,000 m^3

To find the height we will substitue in v"

V = (1/6) pi *h (3r^2 + h^2)

3,500,000 = (1/6)*pi *h (3(150^2) + h^2)

Multiply by 6/pi

==> 3,500,000*6/ (22/7) = h(3*(150^2) + h^2)

==> h(67,500 + h^2) = 21,000,000/ (22/7)

==:> h^2 + 67,500 h - 21,000,000*7/22 = 0

Multiply by 22:

==> 22h^2 + 1485,000h - 147,000,000 = 0

Divide by 2:

==> 11h^2 + 742,500 h - 73,500,000 = 0

==> h= [-742,500 + sqrt(742,500^2 + 4*11*73,500,00]/22

= 98.8 m

**Then the height h = 98.8 m **

The formula for the volume of the spherical dome is given:

v =(1/6)pi*h(3r^2+h)...(1),

Given base radius r = 150m and volume of the dome v = 3,500,000 m^3.

To determine the height , we solve for h and then put the values for r and v.

Multiply by 6/pi both sides of eq (1):

6v/pi = h(3r^2+h)

6v/pi = h^2+(3r^2)h.

We rearrange as a quadratic in h and then solve foh using formula for quadratic equation:

h^2 +(3r^2)h -6v/pi

Therefore h = {-3r^2 +sqrt[(3r^2)^2-4(-6v/pi)]}/2

h = {-3r^2+sqrt(9r^2 +24v/pi)}/2

h = {-3*150^2+sqrt(9*150^4+24*350000/pi)}/2

h = 98.8848802m