Expert: Paul Klarreich - 9/25/2007

Question
find the area of the region delta in the xy plane bounded by the graphs of:
delta: y^2=x^2{(1-x)/(1+x)} as x is [0,1]
assume distance in meters

obviously it is discontinuous at x=-1, but that isn't in the restriction anyway.

length of an arc: y= 2(x)^1/2  x is [0,1]
hint advises that you substitute x=tan^2, answer in meters.

for this one, i used the formula of the integral from 0 to 1 of the square root of 1+{f'(x)}^2. in doing this, i first found dy/dx of y=2(x)^1/2 as (x)^-1/2. square this and you have under the radical 1+ 1/x. then i substituted x for tan^2.... then you get 1+1/tan^2, which is 1+cot^2, which is csc^2, take the square root of that and you get the integral of cscudu, which is ln abs(cscu-cotu) from 0 to 1 + c. then: ln abs(csc(1)-cot(1))- ln abs(csc(0)-cot(0)) + c. and this is where i am unsure. but maybe i did it all wrong anyway?

Answer
Questioner:   Amery
Category:  Advanced Math
Private:  No
 
Subject:  Area and length of an arc using an Integral (calculus 2)
Question:  find the area of the region delta in the xy plane bounded by the graphs of:
delta: y^2=x^2{(1-x)/(1+x)} as x is [0,1]
assume distance in meters

obviously it is discontinuous at x=-1, but that isn't in the restriction anyway.
...........

length of an arc: y= 2(x)^1/2  x is [0,1]
hint advises that you substitute x=tan^2, answer in meters.

for this one, i used the formula of the integral from 0 to 1 of the square root of 1+{f'(x)}^2. in doing this, i first found dy/dx of y=2(x)^1/2 as (x)^-1/2. square this and you have under the radical 1+ 1/x. then i substituted x for tan^2.... then you get 1+1/tan^2, which is 1+cot^2, which is csc^2, take the square root of that and you get the integral of csc u du,

>> You forgot something here.  See below.


which is ln abs(cscu-cotu) from 0 to 1 + c. then: ln abs(csc(1)-cot(1))- ln abs(csc(0)-cot(0)) + c. and this is where i am unsure. but maybe i did it all wrong anyway?
..............................................
Hi, Amery,

This looks like two separate questions.

First: Area under  y = x sqrt( (1-x)/(1+x) )  from 0 to 1.

I know that you are supposed to get the area between the two 'branches' of y^2 = ..., but the symmetry lets you just take that area above and double it.

Now I hope you made an error in inputting this expression, because there is no reasonable way to integrate this.  Not impossible, but I am an old man now, and I have a couple of other things I would like to do before I go.

[How do I know it's bad?  Look up:

THE INTEGRATOR

on the web, at http://integrals.wolfram.com/index.jsp

and you will see the mess created by this integral.

..........................................

Second:

Arc length of  y= 2(x)^1/2, where   x is in [0,1]

Yes, the arc length is given by:

{1
| sqrt(1 +  (f')^2 ) dx
}0

and  f'(x) = x^-1/2, so you will have:

=======================================================
WARNING: USE COURIER FONT TO VIEW THIS
===================================================

{1
| sqrt(1 +  (x^-1/2)^2 ) dx =
}0

{1
| sqrt(1 +  1/x) dx =
}0

{1
| sqrt((x +  1)/x) dx =
}0

Now the INTEGRATOR gives this answer:

        1
Sqrt[1 + -] Sqrt[x]
        x

            ArcSinh[Sqrt[x]]
 (Sqrt[x] + ----------------)
              Sqrt[1 + x]

[the two lines are multiplied]

but I don't think you are supposed to be using inverse hyperbolic functions.  So perhaps we find another way.  You got the hint that you should do a trig substitution:  

x = tan^2 t

dx = 2 tan t sec^2(t) dt    << don't forget this part.

and your integral transforms to:  ( T stands for  tan t )

{
| sqrt((T^2 +  1)/T^2) 2 T sec^2(t) dt =
}

{
| 2 sqrt((sec^2(t)/T^2)  T sec^2(t) dt =
}

{
| 2 sec(t)/T * T sec^2(t) dt =
}

{
| 2 sec(t) sec^2(t) dt =
}

Now the Integrator gives this answer, which might give us a hint:

         t        t
(-Log[Cos[-] - Sin[-]] +
         2        2

           t        t
   Log[Cos[-] + Sin[-]] + Sec[t] Tan[t] )
           2        2


Now I suppose I could work this out, given the answer, perhaps along these lines:

{
| sec(t) sec^2(t) dt =
}

{       1
| --------------- dt
} cos t cos^2(t)

{      cos t
| --------------- dt
} cos^2(t) cos^2(t)

{            cos t
| ----------------------------- dt
} (1 - sin^2(t))(1 - sin^2(t))

Let  u = sin t,  du = cos t dt, and you have:

{         1
| ------------------ du
} (1 - u^2)(1 - u^2)

and now the method of Partial Fractions should work.  (But not in the few years I have left, so I will leave that to you.)