What is the surface area of the solid created by revolving f(x) = x^3 , x in [1,2]f(x)=x3,x∈[1,2] around the x axis?
1 Answer
Explanation:
The surface area of a solid created by a function revolved the
S=2piint_a^bf(x)sqrt(1+(f'(x))^2)dx
Here we see that
S=2piint_1^2x^3sqrt(1+9x^4)dx
We can integrate this. Let
S=(2pi)/36int_1^2 36x^3sqrt(1+9x^4)dx
Before going from
S=pi/18int_10^145sqrtudu
S=pi/18int_10^145u^(1/2)du
S=pi/18[2/3u^(3/2)]_10^145
S=pi/18(2/3(145)^(3/2)-2/3(10)^(3/2))
S=pi/27(145^(3/2)-10^(3/2))approx199.4805