Problem

f(x) = $\displaystyle (1-x^4)^\frac{1}{6}\$

R is the area under the curve from 0 to 1.

Part A: Write down an integral that represents the volume of this solid using the slice method, rotating around the x-axis.

For this part I got an answer that I am pretty sure is right: $\displaystyle pi*\int^1_0 (1-x^4)^\frac{2}{6}\\,dx$

Part B: Write down the integral that represents the volume of this solid using the shell method, rotating around the x-axis.

For this part I put f(x) in terms of y and I got $\displaystyle 2*pi*\int^1_0( y*(1-y^6)^\frac{1}{4}\\,dx$

Part C: The integral you obtain in part (b) should be an integral in y. Now making the substitution y = f(x) show that the integral in part (b) is equal to:Using the substitution I got everything but the negative sign on the outside.

$\displaystyle -\int^1_0(2*pi*x*f(x))\ dfdx/dx$

Part C(2): Hence, show that the integral in part (b) is equal to $\displaystyle -\int^1_0(pi*x)\(d(f^2)/dx)dx$

I did use using the chain rule, d(f^2) = 2f(x) and then the two integrals are the same.

Part D: Using part (c), show that the integral in part (b) is equal to the integral in part (a).No idea what to do

Thanks so much for the help