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# FP3 Surface of revolution Tweet

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1. FP3 Surface of revolution
The arc of the curve y=(1/2)x^2 between the origin and the point (2,2) is rotated around 4 right angles about the y axis. Find the area of the surface generated.

So I did:

x=rt2*y^(1/2)

so dx/dy= rt2/2 *y^-1/2

(dx/dy)^2= 1/2*1/y

Therefore 2pi*
integral between limits 2 and 0: rt2 rty (1= 0.5y^-1)^1/2 dy

Is this right? I don't know how to integrate this.
Last edited by thethinker; 04-06-2012 at 11:35.
2. Re: FP3 Surface of revolution
(Original post by thethinker)
The arc of the curve y=(1/2)x^2 between the origin and the point (2,2) is rotated around 4 right angles about the x axis. Find the area of the surface generated.

So I did:

x=rt2*y^(1/2)

so dx/dy= rt2/2 *y^-1/2

(dx/dy)^2= 1/2*1/y

Therefore 2pi*
integral between limits 2 and 0: rt2 rty (1= 0.5y^-1)^1/2 dy

Is this right? I don't know how to integrate this.
Don't you need dy/dx rather than dx/dy?
3. Re: FP3 Surface of revolution
(Original post by hassi94)
Don't you need dy/dx rather than dx/dy?
I don't know, my book says you can use either dy/dx or dx/dy.

With dy/dx I get the integral to be

rt2*y^0.5*(1+2y)^0.5

Which I can't integrate
4. Re: FP3 Surface of revolution
(Original post by thethinker)
I don't know, my book says you can use either dy/dx or dx/dy.

With dy/dx I get the integral to be

rt2*y^0.5*(1+2y)^0.5

Which I can't integrate
Surely you need to do dy/dx and integrate with respect to x if it's rotating around the x axis.

dy/dx = x pretty clearly

y = 1/2 x^2

so you get

Can you see what to do now?

Spoiler:
Show
Hint: Need a hyperbolic substitution
5. Re: FP3 Surface of revolution
Hassi94 is correct.
6. Re: FP3 Surface of revolution
(Original post by hassi94)
Surely you need to do dy/dx and integrate with respect to x if it's rotating around the x axis.

dy/dx = x pretty clearly

y = 1/2 x^2

so you get

Can you see what to do now?

Spoiler:
Show
Hint: Need a hyperbolic substitution
Oops sorry the question said y axis
7. Re: FP3 Surface of revolution
(Original post by thethinker)
Oops sorry the question said y axis
The surely you need dx/dy
8. Re: FP3 Surface of revolution
(Original post by thethinker)
The arc of the curve y=(1/2)x^2 between the origin and the point (2,2) is rotated around 4 right angles about the y axis. Find the area of the surface generated.

So I did:

x=rt2*y^(1/2)

so dx/dy= rt2/2 *y^-1/2

(dx/dy)^2= 1/2*1/y

Therefore 2pi*
integral between limits 2 and 0: rt2 rty (1= 0.5y^-1)^1/2 dy

Is this right? I don't know how to integrate this.
It is right if you think to

for integration take out ]
then substitute
and integrate by t.
9. Re: FP3 Surface of revolution
Thanks!