The discussion revolves around finding the surface area generated by revolving the function y = x^3/3 about the x-axis, specifically for the interval 1 <= x <= sqrt(7).
Some participants have provided guidance on the substitution method and noted the importance of including constants in the integral. There is an ongoing exploration of how to proceed with the integration process, with no explicit consensus reached yet.
There are indications of potential confusion regarding the correct formulation of the integral and the necessary constants. Participants are also reflecting on their understanding of the substitution technique.
TheLegace said:Homework Statement
Find area of surface generated by revolving about x-axis.
y=x^3/3 1<=x<=sqrt(7)
Homework Equations
find f'(x) = x^2
The Attempt at a Solution
A = integral[ (x^3/3) * [(1+(x^2)^2] ^(1/2) ] ]dx
= integral[ (x^3/3) * [(1+x^4) ^(1/2) ]dx
I just don't know where to go from here.
Any help is appreciated,
Thank You.
nickmai123 said:You forgot the 2\pi in your equation, but you probably have that on paper. Here's what the integral should look like.
A=\frac{2\pi}{3}\int^{\sqrt{7}}_{1}{x^{3}\sqrt{1+x^{4}}dy
Make that du = 4x3dxnickmai123 said:Do what Mark44 said to do.
u=1+x^{4}
du=\frac{x^{3}}{4}dx
Mark44 said:Make that du = 4x3dx