Please point out the mistake on the following.
Find the volume of the solid generated by revolving the area between the curve y=x^3 and the x-axis for 0

Question

Please point out the mistake on the following.
Find the volume of the solid generated by revolving the area between the curve y=x^3 and the x-axis for 0<x<1 about y-axis.

Method 1: Using Shell Method
\ ( V = \int\limits_{a}^{b}2\pi xf(x)dx \ )
\ ( V= 2 pi frac{x^5}{5} |_0^1 =frac{2 pi}{5} \ )

Method 2: Use Washer Method - Since it revolve around y
\ ( V=pi int_{d}^{c} (f(x))^2 dx \ )
where d and c are the new limits.
\ ( V = pi int_{0}^{1} (x^3)^2 dx = frac{pi}{7} \ )

They have different answer! Panic!

campbell_st · Answer

well if you are rotating about the y axis ... don't you integrate with respect to y... 

so your function becomes  $$ x = (y)^{\frac{1}{3}}$$

the limits are 0<= y,= 1

so 

$$V = \pi \int\limits_{0}^{1} (y^{\frac{1}{3}})^2 dy$$