Consider the line segment `c` defined by `y = 2t + 1` for `t` in [1,2] (a) Use geometry to calculate the area of the surface obtained by rotating `c` about the t-axis. (b) Use the surface integral...
Consider the line segment `c` defined by `y = 2t + 1` for `t` in [1,2]
(a) Use geometry to calculate the area of the surface obtained by rotating `c` about the t-axis.
(b) Use the surface integral formula to verify the answer that you got in (a).
a) Using geometry, the object we wish to find the surface area of is a cone with no base and no 'nose' (a 'conical fustrum').
The formula for the surface area of this cone with no base and no 'nose' (which is a smaller cone) is given by
`S = pi (r_2 l_2 - r_1l_1)` where `r_2` is the radius of the base and `l_2` is the length of the side of the cone with its 'nose', and `r_1` is the radius of the base of the 'nose' and `l_1` the length of the side of the cone that is the 'nose'. ` `` `
Here, `r_2 = 2t + 1` where `t = 2` so that `r_2 = 5`, and `r_1 = 2t+1` where `t=1` so that `r_1 = 3`. The length `l_2` is the length of the line segment defined by `f(t) = 2t + 1` for `t` in `[-1/2,2]` so that, using Pythagoras and letting `(x_0,y_0) = (-1/2,0)` (the apex of the cone) and `(x_1,y_1) = (2,f(2))` ,
`l_2 = sqrt((y_1-y_0)^2 + (x_1-x_0)^2) = sqrt(5^2 + 2.5^2) = sqrt(5^2 + (5/2)^2)`
`= 5sqrt(1 + 1/4) = 5sqrt(5/4)`
Similarly, `l_1 = sqrt(3^2 + 1.5^2) = sqrt(3^2 + (3/2)^2) = 3sqrt(1 + 1/4) = 3sqrt(5/4)` so that all in all the surface area of the geometric object of interest is given by
`S = pi(5(5sqrt(5/4)) - 3(3sqrt(5/4))) = pi((25 -9)sqrt(5/4)) `
`= pi(16sqrt(5/4)) = pi(8sqrt(5)) = 8sqrt(5)pi`
b) The surface integral formula says that the surface of the rotated line `f(t)` about the t-axis (in the segment `t` in `[1,2]`) is given by
`S= 2pi int_1^2 f(t) sqrt(1 + f'(t)^2) dt` (for each small increment `dt` in `t` the length of the line segment is `sqrt((dt)^2 + (dtf'(t))^2)` , the radius of the circle of thickness `dt` is `f(t)` and the circumference is then `2pi f(t)` . We integrate over `t`)
Here then we have that
`S = 2pi int_1^2 (2t+1) (sqrt(1 + 4)) dt`
`= 2sqrt(5)pi (t^2 + t)|_1^2 = 2sqrt(5)pi(6 - 2) = 8sqrt(5)pi`
as in a)
Answer: the surface area of the object formed by the line being rotated about the t-axis is `8sqrt(5)pi` . The object formed is called a` ` 'conical fustrum'