# P5 Surface of RevolutionWatch

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#1
im stuck on a question.
here it is (number 7) :

Anyone any idea where im going wrong? seems too long winded to be only 8 marks.
And sorry for size. jus took a pic to save having to type all
Thx
0
14 years ago
#2
The area is given by

A = 2 pi Int[0->h] y ds

Now y = 2 rt(a x)

So that dy/dx = 2 rt(a) (1/2) /rt(x) = rt(a/x)

And (ds/dx)^2 = 1 + a/x

Then

A = 2 pi Int[0->h] 2 rt(a x) rt(1+a/x) dx

= 4 pi rt(a) Int[0->h] rt(x+a)

= 4 pi rt(a) [2/3 (x+a)^(3/2) ] [0->h]

= 8/3 pi rt(a) [(a+h)^3/2 - a^(3/2)]
0
#3
(Original post by RichE)

And (ds/dx)^2 = 1 + a/x
Thanx i understand it all except this line. which is crucial.
How did you arrive at that?
Thanx
0
14 years ago
#4
(Original post by Syncman)
Thanx i understand it all except this line. which is crucial.
How did you arrive at that?
Thanx
From the equation

(ds/dx)^2 = 1 + (dy/dx)^2
0
#5
i hate to seem like the idiot here.
But i still get where that comes from.
Can u refernce somewhere where i can see the proof?
0
14 years ago
#6
That's pretty much the definition of arc-length.

It's not really anything other than Pythagoras' Theorem at an infinitesimal level so that

ds^2 = dx^2 + dy^2.

Isn't the defintion of arc-length given somewhere in your P5 manual?
0
#7
yeah. wow thanx dude.
that makes these questions a little easier!
0
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