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Integrating with the Arc Length Formula


Date: 10/19/98 at 23:43:11
From: Johnny Dinh
Subject: Arc Areas of specific equations.

My math teacher tells me there are only 5 types of functions that can
be integrated in closed form when using the arc length formula, which
is the integral from a to b of the square root of 1 + y'^2.

I got the first three:

   1)  x^2 + y^2 = R^2
   2)  y = x^(3/2) on [0,1]
   3)  y = x^2 

I can't seem to solve these last two in closed form:

   4)  x^(2/3) + y^(2/3) = 1
   5)  y = e^-x

Please help.


Date: 10/20/98 at 12:20:30
From: Doctor Rob
Subject: Re: Arc Areas of specific equations.

Hello,

Just so we're all on the same level, I'm going to refer you to:

   http://mathforum.org/dr.math/problems/sue7.28.98.html   

for a good reminder of the arc length formula. 

Now for the problems:

4) Differentiate implicitly and solve for y', and then find 
sqrt(1+y'^2) in terms of both x and y. Now use the equation 4) to 
substitute for y^(2/3), thereby eliminating y, and reduce the 
integrand to a function of x alone. This should give you an easy 
function to integrate.

5) This lead to the integral of sqrt(1+e^[-2*x])*dx. Let
u = sqrt[1+e^(-2*x)], so u^2 = 1 + e^(-2*x), and x = -(1/2)*ln(u^2-1), 
so dx = -u*du/(u^2-1). That reduces the problem to integrating a 
rational function, which is easy.

Another approach is to multiply numerator and denominator of the 
integrand by sqrt(1+e^[-2*x]), and split the fraction into two parts.  
One will be an arcsinh if you use u = e^(-x), and the other will be a 
square root once you multiply the numerator and denominator by e^x, 
and let u = e^x.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   


Date: 10/21/98 at 15:27:36
From: Johnny Dinh
Subject: Arc length and surface area of rotation question

I recently sent in a question regarding integrating these functions in
closed form when using the arc length formula.

4) x^(2/3) + y^(2/3) = 1

I'm still having problems with this one. This ended up with 
sqrt(1+x^(-2/3) - x^(-4/3) dx. My classmates and I tried many 
substitutions, all of them ended up nowhere. The best one seemed to be 
x = u^(-3/2) which eliminated fractions as powers. This gave us:

   -3/2 Integral of sqrt(1+ u - u^2) u^(-5/2) du

We tried many things with this also, but still unsure how to solve 
this.

5) y = e^-x  

Your help here was greatly needed and appreciated.

Our class was given an additional 4 problems today concerning the
surface area of rotation. These problems, I believe were to be done
with the formula 2pi * the integral of f(x)* sqrt(1+f(x)'^2)dx

We got these two:

   1) y = x from x = 0 to x = h
   2) y = sqrt(x) from x = 0 to x = h

We need help with:

   3) y = x^(3/2) from x = 0 to x = h
   4) y = sin(x) from x = 0 to x = pi

Could you please help us with problem 4 on the arc length and problem 
3 on the surface area? Also thanks for your help once again. Our class 
has spent 4 hours this week in the library trying to solve these 
problems.


Date: 10/26/98 at 09:08:12
From: Doctor Rob
Subject: Re: arc length and surface area of rotation question

4) Differentiating implicitly, you should get:

   (2/3)*x^(-1/3) + (2/3)*y^(-1/3)*y' = 0

             y' = -(y/x)^(1/3)

       1 + y'^2 = 1 + (y/x)^(2/3)
                = [x^(2/3)+y^(2/3)]/x^(2/3)
                = x^(-2/3)

   sqrt(1+y'^2) = x^(-1/3)

Now you should be able to finish.

3)
   y = sin(x)
   y' = cos(x)
   INT[y*sqrt(1+y'^2)*dx] = INT[sin(x)*sqrt(1+cos^2[x])*dx]

Put cos(x) = u, sin(x)*dx = -du, and get:

   INT[sqrt(1+u^2)*du]

This you should be able to do using an inverse hyperbolic function.  
When done, resubstitute for u its equal cos(x), and evaluate between 
the limits.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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