Expert: Paul Klarreich - 2/29/2008

Question
I DO NOT EXPECT YOU TO DO THIS PROBLEM.

This time I will give
as much information as possible.

I am asked to find the area of the region
between the given curve and its asymptotes.

y^2=(4-x)/x

This is an Improper Integral.
It was solved by Misc substitution z^2=4-x
and then by trig substitution.

I arrive at
                                              4
lim t>0+ [sqrt(4x-x^2) - Arcsin(sqrt(4-x)/2)] |
                                              0


What I have Found:

Graphing the function Using MathCad
I find that the curves lies in the Quandrant 1
It reaches zero at x = 4 and has a vertical asymptote
at x=0.

After a lot of work I find the area to be 2pi.
MathCad agrees with 2pi
My book claims 4pi

My question is, Am I suppose to double the area
because of the y^2  or sqrt of the function.

Is it possible that the curve also lies in the
4th Quadrant.

I am pretty sure the limits of my Integral are correct.  

Answer
Questioner:   Mark
Category:  Calculus
Private:  No
 
Subject:  Any Suggestions for this Integral
Question:  I DO NOT EXPECT YOU TO DO THIS PROBLEM.

>> Oh ye of little faith!


This time I will give as much information as possible.

I am asked to find the area of the region between the given curve and its asymptotes.

y^2=(4-x)/x

This is an Improper Integral.
It was solved by Misc substitution z^2=4-x
and then by trig substitution.

I arrive at
                                             4
lim t>0+ [sqrt(4x-x^2) - Arcsin(sqrt(4-x)/2)] |
                                             0


What I have Found:

Graphing the function Using MathCad
I find that the curves lies in the Quandrant 1

>> ONE BRANCH OF IT DOES.

It reaches zero at x = 4 and has a vertical asymptote  at x=0.

After a lot of work I find the area to be 2pi.
MathCad agrees with 2pi
My book claims 4pi

My question is, Am I suppose to double the area
because of the y^2  or sqrt of the function.

Is it possible that the curve also lies in the 4th Quadrant.

>> YES!

I am pretty sure the limits of my Integral are correct.
....................................................
Hi, Mark,

I checked your integral [work is below.] and yes, the result of integrating:

{4
| sqrt((4-x)/x) dx
}0

is, indeed, 2pi.  But you are right -- there is a branch of the curve in quadrant 4, and this integral only goes down to the x-axis.  What you want is one of those 'area between two functions' integrals.  

The upper function is   y = + sqrt((4-x)/x).

The lower function is   y = - sqrt((4-x)/x).

And the difference becomes  + sqrt((4-x)/x) - (- + sqrt((4-x)/x)),  

which is   2(sqrt((4-x)/x)

THERE IS YOUR MISSING FACTOR OF 2.
.......................................

My integration: [which matches yours and needs the factor of 2 at the end.]

..........
y = sqrt(4-x)/sqrt(x)

Let z = sqrt(4 - x), Limits of  x = 0 to 4, become  z = 2 to 0

z^2 = 4 - x

x = 4 - z^2

2z dz = - dx

{ sqrt(4-x)
| --------- dx
}  sqrt(x)


{ - z 2z dz  
| ---------
} sqrt(4 - z^2)

{ - 2z^2 dz  
| ---------
} sqrt(4 - z^2)

Set z = 2 sin t, Limits of z = 2 to 0 become   t = pi to 0

dz = 2 cos t dt

sqrt(4 - z^2) = 2 cos t

{0
|   - 4 sin^2 t dt  
}pi

{     1 - cos 2t
| - 4 ------------ dt  
}        2

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

-2t + sin 2t  [ pi to 0]

-2(0 - pi) = 2 pi