chooke said:
ok now i have one last question and then im finished my assessment. how do i do this question;
The region in the first quadrant bounded by the graphs of f(x)=1/8 x^3 and
g(x)=2x is rotated about the y-axis. Find the volume of the solid so formed.
Hello again. ^^
To find point of intersection of both curves,
let f(x)=g(x)
x
3/8=2x
x
3/8-2x=0
x(x
2/8-2)=0
x(x/sqrt8 + sqrt2)(x/sqrt8 - sqrt2)=0 , by difference of 2 squares
x=0, x=(-sqrt2)(sqrt8), x=(sqrt2)(sqrt8)
x=0, 4, -4
The one we take is 4 because only this one is in the first quadrant.
So now the volume is given by:
4
/
| pi{[g(x)]
2 - [f(x)]
2 dx}= (see next line)
/
0
4
/
| pi{4x
2 - x
6/64 dx}= (see next line)
/
0
pi{4x
3/3 - x
7/448]} from 0->4 = pi[256/3 - 16384/448 - 0]
= 65536pi/1344 units
3
If the question doesn't ask for any rounding, then leave it in this exact form.
No need to say good luck since you've already done it.