1. Find the surface area formed by revolve region along the x- axis f(x) = -1/square root of 3 *x +1
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

The basic expression for the integrand is 2πyds, where y=f(x) and ds is an infinitesimal segment of the curve. ds is related to dy and dx by ds²=dy²+dx², so (ds/dx)²=(dy/dx)²+1 and ds=√((dy/dx)²+1). The surface area A=2πʃyds=2πʃy√((dy/dx)²+1)dx.

If y=-1/√(3x+1)=-(3x+1)^-½ then dy/dx=3/(2(3x+1)^(3/2)), (dy/dx)²=9/(4(3x+1)³), and √((dy/dx)²+1=√((9+4(3x+1)³)/(4(3x+1)³)).



f(x) has a vertical asymptote at x=-⅓ and the x-axis is a horizontal asymptote. The limits for the integral are therefore -⅓≤x≤∞ if the whole curve is to be rotated around the x-axis. At x→-⅓, the integrand approaches infinity, so the surface area is infinite. Below f(x) is shown as the red curve and the asymptote as a vertical red line.

by Top Rated User (600k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
81,978 questions
86,378 answers
72,192 users