# How much paint is needed for truncated cone tower?

• miss_sana
In summary, the problem involves finding the amount of paint needed to cover a spherical tank with a flared supporting column. The tank holds 54,000 gallons of water and has a radius of 12 feet. The supporting column has a base that is one-third the radius of the tank and a height of 15 feet. The connection between the tank and the column has a base that is two-thirds the radius of the tank and a height of 10 feet. The paint must be applied to an average thickness of .16 cm. To solve this problem, the surface area of the truncated cone supporting column must be calculated and added to the surface area of the spherical tank.
miss_sana

## Homework Statement

So there's about 4 problems that iI just don't understand. The first one is called H20 in the S-K-Y.
Theres a drawing and it kind of looks like a graduated cylinder with a circle on top. It says 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 suporting 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 spehere flares out to a diameter equal to two-thirds the radius of the sphere and is 10 ft tall.
The paint must be applied to an average thickness of .16 cm over the entire structure. I need to calculate how much paint will be needed to paint the tower based on the details given.

## The Attempt at a Solution

now i know its a surface area problem, and I already converted the gallons into feet, and found that the radius of the circle is 12 ft long. Also, I know that the surface area of what would be the circle is 576pi, but the thing is, it's not whole because it's connected to the supporting column. if anyone could help, that'd be great! =)

Last edited:
I take it you mean "sphere" rather than circle. The base sounds like a truncated cone. You can find a formula for its surface area here:
http://pagesperso-orange.fr/jean-paul.davalan/calc/cone/cone-en.html

## What is the formula for finding the volume of a truncated cone?

The formula for finding the volume of a truncated cone is V = (1/3) * π * h * (r^2 + r * R + R^2), where h is the height of the cone, r is the radius of the smaller base, and R is the radius of the larger base.

## How do you find the slant height of a truncated cone?

The slant height of a truncated cone can be found using the Pythagorean theorem. It is given by l = √(h^2 + (r - R)^2), where h is the height of the cone, r is the radius of the smaller base, and R is the radius of the larger base.

## What is the relationship between the volume of a cone and a truncated cone?

The volume of a truncated cone is equal to the volume of a cone with the same height and base radius, minus the volume of the smaller cone that was cut off. This can be represented by the formula V = (1/3) * π * h * (R^2 + R * r + r^2), where h is the height of both cones and R is the radius of the larger base of the truncated cone.

## How do you find the lateral surface area of a truncated cone?

The lateral surface area of a truncated cone can be found by using the formula A = π * l * (r + R), where l is the slant height of the cone and r and R are the radii of the smaller and larger bases, respectively.

## What is the difference between a cone and a truncated cone?

A cone is a three-dimensional shape with a circular base and a curved surface that tapers to a single point, known as the apex. A truncated cone, on the other hand, is a cone that has had its top cut off by a plane that is parallel to the base. This results in a smaller circular base and a frustum-shaped surface.

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