Volume of cathedral dome (Using volumes of revolution, disk method)

[insane0hflex]

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]

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

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]

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.

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)?

 

[LCKurtz]

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)?

No, that isn't correct. You don't start with the area of the base. You want the area of the upper hexagon in the picture, which will depend on h. h is the distance from the base to the plane of that hexagon, in other words the variable distance between the two red dots.