Serious Help Needed (1 Viewer)

[chooke]

i am going to repeat this question in hope others have caught on with the desperato tone i have in my title.

The region in the first quadrant bounded by the graphs of f(x)=1/8x^3 and
g(x)=2x is rotated about the y-axis. Find the volume of the solid so formed

 

[Riviet]

There is no need to start another thread asking the same question, as someone (like me) will eventually help you out. I understand it's urgent since it's due tomorrow, but still, plenty of people visit this forum and i'm sure there'll be someone to help you asap.

 

[Mountain.Dew]

okay, this is a bit tricky for 2U students, but i'll do the best i can.

steps to do:
1) draw curves
2) find intersection(s)
3) identify which curve, when rotated, will give a larger volume
4) rotate around x or y axis? change limits if have to.
5) do the intergral.

1) draw the curves y = (1/8)x^3 and y=2x on the same number plane. they are going to intersect at some pts.
2) consider 1/8 x^3 - 2x = 0, solve for x

x^3 - 16x = 0 (times by 8)
x(x^2 - 16) = 0 (factorise), we know x = 0 is one solution
(x - 4)(x + 4) = 0, we know x= 4, -4. BUT we only consider 1st quad, so
we only work with x=0 and x=4

3) the curve y = 1/8x^3, when rotated, has larger volume than y=2x when rotated. essentially, we are taking one volume off the other to find the bounded volume.

4) curves rotate around y-axis, so we have to change the limits. sub x= 0 into y=2x, y =0. sub x=4 into y=2x, y = 8.

5) use the general formula:

----------------/b---------------/d
volume = (pi)| ( x1 )^2 dy - | (x2)^2 dy , x1 and x2 are the two curves
---------------/a----------------/c

so, make x the subject for each of the two curves:

y = (1/8)x^3 --> x = 2y^(1/3)

y = 2x --> x = 1/2y

SO the final volume is THUS:

------------------/8---------------------/8
volume =( (pi) | ( 2y^(1/3) )^2 dy - | (1/2y)^2 dy ) Units ^2
-----------------/0---------------------/0

if ur still confused, PM me or reply.

 

[angmor]

is this a question found in 2 unit?

 

[Mountain.Dew]

mmmmmmm this is prolly a hard 2U, but a medium 3U.

 

[Riviet]

This question would be worth several marks if found in a 2 unit paper.

Anyway, I already posted the solution in another thread and chooke has already handed the assignment in long ago too.

This was my solution:

To find point of intersection of both curves,

let f(x)=g(x)
x³/8=2x
x³/8-2x=0
x(x²/8-2)=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{4x² - x⁶/64 dx}= (see next line)
/
0

pi{4x³/3 - x⁷/448]} from 0->4 = pi[256/3 - 16384/448 - 0]

= 65536pi/1344 units³

If the question doesn't ask for any rounding, then leave it in this exact form.