insane0hflex said:Homework Statement
A cathedral dome is designed with three semi circular supports of radius r so that each horiontal cross section is a regular hexagon. Show that the volume of the dome is r^3 * sqrt(3)
an accompanying figure - http://imgur.com/3fSqh
Homework Equations
Vdisk=∫pi*(f(x))^2 dx
LCKurtz said:That dome is not a solid of revolution since the cross sections are not circles. What you need to do is figure out the area of the hexagonal cross section at height h and do the volume by integrating cross section areas for h from 0 to r.
insane0hflex said:Okay, that helps a little bit. So I would derive an expression for the area of the base, then take the integral from 0 to h? (since h = r correct)?
The formula for calculating the volume of a cathedral dome using the volumes of revolution, disk method is V = π∫ba(f(x))2dx, where a and b are the limits of the base of the dome and f(x) is the equation of the curve used to create the dome.
Some of the assumptions made when using the volumes of revolution, disk method to calculate the volume of a cathedral dome include: the base of the dome is circular, the dome is symmetrical, and the dome is created by rotating a curve around a central axis.
The height of the dome does not affect the calculated volume using the volumes of revolution, disk method. This is because the method only considers the shape of the base and does not take into account the height of the dome.
Other methods that can be used to calculate the volume of a cathedral dome include: the shell method, the washer method, and the cross-section method. Each method may be more suitable for different dome shapes and can provide more accurate results in certain cases.
The volumes of revolution, disk method can be applied to any structure that can be created by rotating a curve around a central axis. This includes structures such as towers, silos, and water tanks. The method can also be used to find the volume of objects such as vases and bottles with curved shapes.