The discussion revolves around finding the surface area of the curve defined by the equation x = ln(y) - y^2/8 when rotated about the y-axis, specifically over the interval [1, e]. The participants are examining the setup of the integral for surface area calculation.
The conversation is ongoing, with participants providing insights into their calculations and questioning each other's approaches. There is no explicit consensus yet, but some guidance on the derivative calculation has been shared.
Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There is an emphasis on verifying the correctness of mathematical expressions derived from the original equation.
NWeid1 said:Homework Statement
Find the area of the surface obtained from rotating the curve x = ln(y) - y^2/8 on [1,e] about the y-axis.
Homework Equations
SA = \int 2\pi*(f(x))*\sqrt{1+[f'(x)]^2}dx
The Attempt at a Solution
SA = \int 2\pi*(ln(y) - \frac{y^2}{8}))*\sqrt{1+\frac{9}{16*y^2}}dy from 1 to e
I don't know how to solve this integral, which makes me thing I set it up wron.g
NWeid1 said:That was the equation that my prof. gave me, lol
NWeid1 said:oh ok. I took the derivative of x = lny - \frac{y^2}{8} and got
x = \frac{1}{y} - \frac{1}{4y} and squared it to get
\frac{1}{y^2} - \frac{1}{2y^2} + \frac{1}{16y^2} which is
\frac{16}{16y^2} - \frac{8}{16y^2} + \frac{1}{16y^2} which is
\frac{9}{16y^2}