Integration. (1 Viewer)

[crazyymonkey]

need help with this question.

1) the region, enclosed by the parabola y^2 = 4ax and the line x=a, is rotated about the x-axis. Find the volume of the solid formed.

please do working out if possible because i need to get how to do it.

 

[3unitz]

the volume for any curve rotated about the x-axis from a to b is:

V = pi I y^2 dx (limits a,b)

so our volume for this question:

V = pi I (4ax) dx (limits 0,a) ... since y^2 = 4ax, and we're going from 0 to a

4a is just a constant so we can pull him out of the integral:

V = pi.4a I x dx (limits 0,a) ... now all we gotta do is integrate x

V = pi.4a [x^2/2] (0,a)

V = pi.4a (a^2/2 - 0^2/2)

V = pi.2a^3 units^3

 

[tommykins]

Sideways parabola.

v = pi int y² dx from a->0
= pi int 4ax dx from a->0

if it's wrong, i apologise, no handy paper nearby.

damn you 3u.

 

[3unitz]

haha, nice foot work.