A problem in Integration?

Question

using integration, how can i find the volume of a frustum of a right circular cone with height h, lower base radius R and top radius r?

Also, how can i find the volume of a frustum of a pyramid with square base of side b, square top of side a and height h?

Answer 1

do this by ythe method of slices: that is find the area of the cross section in terms of height and multiply by dy ( this is the slice) then add the slices (integrate from 0 to h).

assume h means the height of the frustrum rather than the original height of the cone the radius at any height y is proportional the the change in radius:

radius = R - y(R-r)/h and area = pi(radius)^2
a= pi( Rh -yR +yr)^2/h^2

then volume = integral ady = integral[pi(Rh-yR=yr)^2/h^2]dy
integrate between the limits of 0 to h

the pyramid can be done the same way by finding area in terms of vertical height

Answer 2

Volume of a frustrum pyramid w/ square base:

V= (1/3)(h)[B+b+((B*b)^0.5)], where B=AREA of base1, and b= AREA of base2.

Volume of Cone frustrum: (not sure what you mean by "right")

V= (0.262h)(D^2 + d^2 + Dd), where D= diam. of base1, and d= diam. of base2.

i think to integrate this, you would need to use double integration, but not sure...at least you have the formulas, and a cool website
they ask for a password, but ignore it.

Answer 3

you can do a double integral in polar coordinates. Model the object using a function with 2 polar variables and then integrate over the limits

Answer 4

crucial by using areas is: ?udv=uv-?vdu so after substitution and placing u=x and dv=F'(x), ?xF'(x)dx=xF(x)-?F(x) and because ?F(x)=G(x) because of fact F(x)=G'(x) ?xF'(x)dx=xF(x)-G(x).

Answer 5

http://mathworld.wolfram.com/ConicalFrustum.html