Volumes help needed (1 Viewer)

[blackops23]

Hi guys, here's a question which I was trying to do using cylindrical shells (slicing involves integrating inverse trig)

Q. Find the volumes of the solid of revolution when the arc of the curve y=sinx from x=0 to x=pi/2 is rotated about the y-axis.

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My working out:

A(y) = 2pi(r)(h) --> r is radius, h is height of cylinder
A(y) = 2pi(x)(sinx)
Therefore volume of shell with width (DELTA y) = 2pi(x)(sinx).(delta y)

Therefore V= (2pi)*INT xsinx dx, limits of integral: 0(<)x(<)pi/2
V=2pi(1) = 2pi u^3

But the answer says V= [(pi^3)/4 - 2pi] u^3

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Can someone please point out what I did wrong? And tell me how to fix my mistake?

Thankyou, appreciate the help

 

[blackops23]

no i hav got it. what you did wrong was u were rotating area under the curve round the y-acis while what they want is u to rotate arc f the curve around the y-axis, so here u wuld use slicing rather than shells.

v = pi x^2 delta y
v = pi (sin inverse y)^2 delta y and solve between 0 to y to 1

umm... sorry i don't get it, what does it mean by "rotating the arc"? isn't an arc just a 1 dimensional line? What exactly are you meant to do?

Thanks

 

[Hermes1]

completely forget what i said b4. okay so what they want you to do is rotate that curve of y = sinx around the y-axis.
so do this:
V =
V =

now solve that integral between 0 and pi/2 and you will get that answer they had.

 

[Hermes1]

what you were doing was rotating area between curve and the x-axis round the y-axis. they want to get the area between curve and y-axis and rotate round y-axis. thats what they mean by rotating arc round the ACIS

 

[blackops23]

what you were doing was rotating area between curve and the x-axis round the y-axis. they want to get the area between curve and y-axis and rotate round y-axis. thats what they mean by rotating arc round the ACIS

Ahh thank you now I know what the question was asking for.

But there's another question I'm having trouble with:

Q. The region inside the circle x^2 + y^2 = a^2 is rotated about the line x=2a to form a solid called a torus. Find it's volume by the means of cylindrical shells.

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My working out:

A(x) = (2pi)(r)(h) --> radius and height of the cylinder of thickness : delta x
A(x) = (2pi)(2y)(2a - x)

Now y^2 = a^2 - x^2
so y=sqrt(a^2 - x^2)

therefore:

A(x) = (4pi)(2a - x)[sqrt(a^2 - x^2)]

skipping a few steps....

V = 4pi* INT (2a-x)[sqrt(a^2 - x^2)] dx .... integral limits: -a(<)x(<)a

Question is, how do i integrate (2a-x)[sqrt(a^2 - x^2)]??

Or is there a more efficient method to do it?

Thank you very much, appreciate the help

 

[Hermes1]

well to integrate that between a to -a u use the fact that

sqrt(a^2-x^2) is a semicircle therefore you can just use

therefore for the 2a(a^2-x^2) part you just get 2a multiplied by pia^2/2

notice the a is the radius,

and for the x(a^2-x^2)^0.5 you ll notice that it is an odd function, therefore between limits of same magnitude but opposite sign the integral of an odd function is zero. therefore this second part of the integral is just zero.

therefore youll get 4pi times by pia^2/2 as ur answer

that ll give you 4pi^2a^3 as ur overall answer

 

[Drongoski]

bravo arji! Not allowed to rep u since I haven't spread enough.

 

[Hermes1]

this is the most efficient way of solving the question and u hav to learn it bcuz ur goin to be usin it everytime.
now you wont always be rotating a circle with centre the origin.

therefore you could sometimes get an integral like (2R-x)(R^2-(x-2R)^2)^0.5 between limits 3R and R

in this case ull first have to make a substitution like for this example you would say let u = x-2R and then after geting the integral interms of u u use the method i said above in previous post.

these type of questions are very common, typically around question 6. TBH u want to ensure you get all the marks in volumes section bcuz it is just integration once u get an expression.

 

[Hermes1]

bravo arji! Not allowed to rep u since I haven't spread enough.

lol how did everyone work out i changed my username.

 

[blackops23]

well to integrate that between a to -a u use the fact that

sqrt(a^2-x^2) is a semicircle therefore you can just use

therefore for the 2a(a^2-x^2) part you just get 2a multiplied by pia^2/2

notice the a is the radius,

and for the x(a^2-x^2)^0.5 you ll notice that it is an odd function, therefore between limits of same magnitude but opposite sign the integral of an odd function is zero. therefore this second part of the integral is just zero.

therefore youll get 4pi times by pia^2/2 as ur answer

that ll give you 4pi^2a^3 as ur overall answer

this is the most efficient way of solving the question and u hav to learn it bcuz ur goin to be usin it everytime.
now you wont always be rotating a circle with centre the origin.

therefore you could sometimes get an integral like (2R-x)(R^2-(x-2R)^2)^0.5 between limits 3R and R

in this case ull first have to make a substitution like for this example you would say let u = x-2R and then after geting the integral interms of u u use the method i said above in previous post.

these type of questions are very common, typically around question 6. TBH u want to ensure you get all the marks in volumes section bcuz it is just integration once u get an expression.

^^repped