Multi - Step Integral Problem

[auntjamima]

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: \(\displaystyle \pi*\int_{0}^{1} (1-x^{4})^{\frac{2}{6}},dx\)

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 \(\displaystyle 2*\pi\int_{0}^{1}y(1-y^{6})^{\frac{1}{6}}dx\)

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:
\(\displaystyle -\int_{0}^{1}(2{\pi}xf(x))dfdx/dx\)

Using the substitution I got everything but the negative sign on the outside.

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 don't know what to do for this part
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

 

[tkhunny]

It may make more sense after fixing Part B. Fractional exponent should be 1/4.