FP3 Surface of revolution
Maths and statistics discussion, revision, exam and homework help.
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FP3 Surface of revolutionThe 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. -
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Re: FP3 Surface of revolutionDon't you need dy/dx rather than dx/dy?(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. -
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Re: FP3 Surface of revolutionI don't know, my book says you can use either dy/dx or dx/dy.(Original post by hassi94)
Don't you need dy/dx rather than dx/dy?
With dy/dx I get the integral to be
rt2*y^0.5*(1+2y)^0.5
Which I can't integrate -
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Re: FP3 Surface of revolutionSurely you need to do dy/dx and integrate with respect to x if it's rotating around the x axis.(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
dy/dx = x pretty clearly
y = 1/2 x^2
so you get
Can you see what to do now?
Spoiler:ShowHint: Need a hyperbolic substitution -
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Re: FP3 Surface of revolutionOops sorry the question said y axis(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:ShowHint: Need a hyperbolic substitution
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Re: FP3 Surface of revolutionIt is right if you think to(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.
for integration take out![\frac{1}{2y} from the root, this
will cancel the factor [latex]\sqrt{2y}](http://www.thestudentroom.co.uk/latexrender/pictures/ef/ef3abfbd4005bac511989d6ef160c3bc.png "\frac{1}{2y} from the root, this
will cancel the factor [latex]\sqrt{2y}")
]
then substitute
and integrate by t.
