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/
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