geometry
posted by CJ on .
PLEASE HELP ME! i have this homework due tomorrow and i only have 1 question left and am sooo desperate for an answer:
The diameter of the Dome of the Rock in Jerusalem is 20m. If it is plated with gold that is 5cm thick, find the cost of plating the dome in gold given that 1kg of gold costs $43,395 and 1g of gold is equal to 19.3cm>3
p.s cm>3 means cm cube

surface area of sphere = (4/3) pi r^2
so
surface area of hemisphere = (2/3)pi r^2
here r = 10 m
so
surface area = (200/3) pi meters^2
there are 10^4 cm^2 per meter^2
so
surface area = (200/3) pi*10^4 cm^2
volume = surface area * depth
= (200/3) pi *10^4 * 5 cm^3
grams = volume*density
= (200/3) pi *10^4 * 5 (1/19.3)
cost per gram = 43.395
so
cost = (200/3) pi *10^4 * 5 (1/19.3)*43.395 
The volume of the gold would be
V = (pi/4)D^2*(thickness) = 15.7 m^3
= 15.7*10^6 cm^3
The mass would be (density)*V = 303.2 *10^6 g = 303.2*10^3 kg
Multiply that by the cost per kg.
I get about 13 billion dollars 
I treated it as a 5 cm layer over a cylinder. Damon treated it as a hemisphere. He is correct.
http://www.historyofscience.com/G2I/timeline/images/dome_of_the_rock.jpg 
thank you sooo much damon :)