Needs help with a volume question (2 Viewers)

[hon1hon2hon3]

Heres is a volume questions . . starts off really easy , then the second part pisses me off. . .

1. i) Find the volume generated when the area bounded by y=sin x and the x - axis, for 0 to pie, is rotated about the x-axis.

This is easy , i got pie square on 2.

ii) This area described in (i) is now rotated about the line x=2pie, find the volume of the solid formed.

How do you do this ? Help >_> . . . i know that the slice area will became pie times ( B square - A square ) where B is the longer radius of the circle and A is the shorter one . . i also know that the sum of B and A is pie .

I finally got up to an equation of 2 pie square S 2sin^-1 y - pie .dy . . . limits from 0 to pie/2 . . . can any one tell if i am right ? . . . and if soo how do i solve it . . . if not . . . . please post the correct solution >_> thx in advance have fun doing it

I got a answer of -pie cube . . since volume has to be possitive thus its pie cube . . . . but i dont think its right , i used a whole lot of things , intergration by parts , how sin behave . . making it to inverse . . . cutting the area into two . . since its symmertrical about x=pie/2 . . . . all this tells me i am wrong = =. . . can some one please work it out +O+ ZOMG ZOMG OMZG 4 unit

 

[conics2008]

step 1 ) sketch y=sinx

step 2 ) draw line x= 2pi which is in the frist quad where it makes its first revoultion.

step 3) consider the part of the graph from pi - 0 and use shells to find the voulme. sorry i thought it was the whole graph.

step 4) take care in your working out.

step 5) praise your self.

step 6 ) if you need help just let me know and ill do it on paper hopefully =]

 

[hon1hon2hon3]

its just from 0 to pie . . therefore only have to worry about the area above the a-axis . . and please do it >_> . . . i have been on this question for half an hours= =. . . and i finally got a answer of pie cube , dont know if its right . . . dang it, dont have the solution ="= i swear to god never do 4 unit maths with out the solution ="= . . . speachless. . .

 

[conics2008]

ok im postin up the solution now..

step 1) identify your outer radius and your inner radius

R (outer) = 2pi-x r(inner)= 2pi-x-dx

step 2) d V = pi { (R-r )(R+r) } * y where y=sinx as your height of the shell

sub in the outer and inner and simplify

you should get at the end

D V = lim dx > 0 sigam from pi to 0 2pi { 2pi sin x - xsinx }

keep in mind you already done sinx from part i the only thing new you have to do is use IBP on xsinx from pi to 0

it looks long integration thats why take care with your working out..

i stopped at this step

2pi S 2pi sin(x) << part i + S -xsinx

 

[conics2008]

what your answer for part i, it looks simple but cbf

 

[conics2008]

shit sorry wait.. ill give u the dam solution..

 

[conics2008]

answer i get is 6pi^2

 

[hon1hon2hon3]

i am trying to follow your solution , hold up i guess >_>

 

[hon1hon2hon3]

OMG , i think i forgot to tell you one important things. . . u gotta do it with slice method . . . yours looks like cylindrical shell method yeah @@?

 

[conics2008]

well the first part ive done with slice and second part is with shells ???

whats the problem ??

 

[hon1hon2hon3]

mmm, well they told us, we need to use slice to do the whole questions . . soo . . gotta use slice >_>

 

[conics2008]

here is the working out for part ii ) lol im soo hopeless xD

we know the outter radius is 2pi-x and inner is 2pi-x-dx

use this as total voulme

dV = pi { R^2-r^2 } = pi(R-r)(R+r)

dv= pi { (2pi-x+2pi-x-dx)(2pi-x-2pi+x+dx)[ sinx ] }

collect and simplify

you get 4pi^2sinx - 2pi xsinx

hence total voulme is

S pi > 0 4pi^2 sinx + S 2pi -xsinx << use IBP for this and u should get either 10pi^2 or 6pi^2 ( but im pretty sure its 6pi^2 ) i might stuffed up for the 10 ?

 

[hon1hon2hon3]

you call yourself hopeless= = then what am i ZOMG ="=. . . i am trying to make sense out of your solution ="=

 

[conics2008]

i had no hope with slices.. its too dam hard to find the roots in terms of y ?? how did u do it ?

 

[conics2008]

if you draw a typical slice p.d to x=2pi you would need to find the roots which is kind of impossible with sin(x) ?/

I know these types of question with functions only ??

 

[hon1hon2hon3]

For your solution . . delta V bit. . . how come u suddenly have a sin x ? The way i did it was way too complicated to type it out ="= and i bet the way i did was wrong . . there must be some easier way , dam it >_< ZOMG i am noob

 

[conics2008]

wtf// that was the height of the cylinder ??

the dam slice method is not working out here buddy.. im pretty sure its too dam hard for slice to work in this case..

because if you think about it there is no end points ?? only two roots

 

[hon1hon2hon3]

well it is not impossible . . i will try to type it out for you the way i did it.

i used that pie times ( r2 square - r1 square ) . . . you get what i mean yeah . . . then i let the long radius be X1 and the shorter be X2 . . . and from using the difference between two square . . pie times ( 2 pie - X1 + 2 pie - X2 ) ( 2 pie - X1 - ( 2 pie - X2) . . . then u get pie ( 4 pie - X1 - X2) (X2 - X1)

i will post this first . in case u are waiting for it . . i will keep tpying . .

 

[hon1hon2hon3]

then from pie ( 4 pie - X1 - X2 ) ( X2 - X1 ) . . . can be simplified to pie ( 4 - ( X1 + X2)) (X2 - X1) . . . . and we know X1 + X2 = pie . . . dont ask why , its how sin behave. . . thus the equations now looks like pie ( 3 pie ) (X2 - X1)

 

[conics2008]

yeh im at teh same spot now how da fuck do u find x2 -x 1

see what i mean those are the roots im talking about lol...... you can just assume sin^1(x1) and sin^-1(x2)

see its too complicated and undoable ?