Needs help with a volume question (1 Viewer)

[hon1hon2hon3]

continus from above . . . now that for X2 - X1 part, we know y= sin x, thus inverse sin = x, and from how sin behave . . . that all station to central thing . . if u know what i mean . . . we know that X1 = inverse sin y . where X2 = pie - inverse sin y . . . since sin 30 = sin 150 . . same to all . . . sine 15 = sin 165 etc . . . thus now it becomes 3 pie^2 ( 2 inverse sin y - pie )

 

[conics2008]

hmm here is the volume using teh same method as u ( i dont know what is gonig through my mind this looks insane now lol)

R = out radius r = inner radius right

hence let the end point of the typical slice be x1 and x2 respectivly..

there fore outer radius = 2pi - x1 and inner radius = 2pi-x2

hence total area of the base shaper is

d V = PI { 2pi -x1 + 2pi -x2 } { 2pi -x1 -2pi + x2 }

dont forget ot expand and then whats next ?????????????????????

 

[conics2008]

continus from above . . . now that for X2 - X1 part, we know y= sin x, thus inverse sin = x, and from how sin behave . . . that all station to central thing . . if u know what i mean . . . we know that X1 = inverse sin y . where X2 = pie - inverse sin y . . . since sin 30 = sin 150 . . same to all . . . sine 15 = sin 165 etc . . . thus now it becomes 3 pie^2 ( 2 inverse sin y - pie )

i didnt understand all the high light bits..?

do we need to know how sin works when working out volumes..

which paper is this from btw... to be honest ive never came across with a voulme which requires more than 2-3 times integrals ??

 

[conics2008]

one more thing how can u assume that x1 and x2 0 and pi respectivly ??

you cant get a constant with that it must be an approx of the roots ?? read example in the voulme chapter.. u will know what im talking about ??

or maybe im just confused.

 

[hon1hon2hon3]

and as you continues and solve it, u will find out the limit will change , because now is in terms of y , soo that using that y=sin x . u will find the new limit as 0 to 0 . . and thus we need to change it , since y = sin x is symmertrical to pie/2
and so i times it by 2 . . and make it from 0 to pie/2 . . which now becomes 0 to 1. . . and as u continue to solve it . . . now it looks like . . . 6 pie^2 S 2 inverse sin y - pie . . .

 

[hon1hon2hon3]

oh this is not out of any papers , my tutor gave it to me . but the thing is , i forgot to ask him for the solution . . . lol, also i think theres is a easier way in doing this , maybe i am just complicating stuff up . . . is any pros out there can help ="=

 

[hon1hon2hon3]

and now i solved it again , and i find a silly mistake i did, i got 3 pie cube . . . ="= lalala i gave up

 

[conics2008]

you know what im just gonna have a laugh...

=]

stuff this.. stick to shells xD

good luck mate =] stuff all this limits changing business just integrate xD lol.. cya man im back to english now xD

 

[hon1hon2hon3]

thx for your help any way . . . now i feel like not studying for 4 unit . . it stuffs u up = = better go back to 3 unit, then chem then physics and 4 unit last = = thx any way ="=

 

[tom101007]

did anyone actually finish this problem?
i got 6pi^2 + 4pi units cubed

i did it by shells cause when i tried by slices i really didnt want to work out the 2 different radii of the anulus that forms the slice

 

[tom101007]

actually, i made a mistake, and i also got 6pi^2

 

[lost1]

This is how i worked out the two roots (or the two radii) and thus use the slice method:

Let x1 be the end point of larger radius and x2 be end point of smaller radius

If you draw the graph and draw the vertical line x=pi/2 you can see that pi/2 - x1 = x2 - pi/2 --------------(1)
now since the inverse function x=sine-1 y has range -pi/2 to pi/2 it can be seen that x1 is in this range and x2 is out of this range...thus from (1) you can have x2 in terms of x1

now put the integration in terms of x1 and then in terms of y and integrate...and yeh i got 6pi^2 as well

hope Im making some kind of sense...