# Thread: H20 in the S-K-Y

1. ## H20 in the S-K-Y

Essentially, I need to find the surface area of a given water tower for a Calculus AB project:
• The spherical top holds a little over 54,000 gallons of water, the base of the top is 60 feet above the ground and the diameter of the supporting column is one-third the radius of the sphere. The base of the supporting column flares out like a cone so that the diameter of the bottom is equal to the radius of the sphere; this bottom section is 15 feet tall. Likewise, the connection between the support and the sphere flares out to a diameter equal to two-thirds the radius of the sphere and is 10 feet tall.
Because no diameters or radii are given, I am having trouble finding a good starting point. Any help or insights would be greatly appreciated...
-Simon

2. I would convert the gallons into cubic metres, then take a look at V = (4/3)(pi)(r^3). That should give you a radius to work with. It's always the simple stuff .

3. I had thought of this; although, the problem states that the volume of the water is approximately 54,000 gallons (a little over). Any ideas on how to create a more precise estimate?

4. I don't think there's any way to get a more accurate measurement from a given number. All the other measurements refer back to the sphere, I don't think there's any way to go about it other than getting a radius from the sphere to start things off.