a hemisphere dome on top of a silo has a diameter of 40 feet, find the volume of the dome r=6cm
v=4/3 times 3.14 times radius6cubed
I find the explanation of the problem quite unclear. Here is one guess:
If you mean the volume of the material used, then if the outside radius is R, and the inside radius is r, the volume of material is
2π/3 (R^3-r^3)
Assuming the outside radius is 40ft and the thickness is 6cm, we have (since 1 ft = 30.48 cm),
2π/3 [(40*30.48)^3 - (40*30.48-6)^3] cm^3
Evaluate that. I also suspect a units typo. But you can fix that, right?
To find the volume of a hemisphere dome on top of a silo with a diameter of 40 feet, you can use the formula for the volume of a sphere, since a hemisphere is half of a sphere.
The formula for the volume of a sphere is: V = (4/3) * π * r^3, where V is the volume and r is the radius.
In this case, the diameter of the dome is 40 feet, so the radius would be half of the diameter, which is 40/2 = 20 feet.
To convert the radius to centimeters (cm), you need to multiply it by 30.48 (since there are 30.48 cm in 1 foot). So, the radius would be 20 feet * 30.48 cm/foot = 609.6 cm.
Now you can substitute the radius into the formula for the volume of a sphere:
V = (4/3) * π * (609.6 cm)^3
Calculating this would give you the volume of the hemisphere dome in cubic centimeters.