Help with calculus problems

I. Consider region bounded by curve of y=4x^5, x=1, and the x axis. Find area, and determine volume of solid when it's revolved around:
x-axis, line y=4, y axis, line x=1.

II. Find volume of solid with bounded region as base and cross sections perpendicular to x axis are circles whose diameters touch curve y=3x^4 and x axis.

III. Set up, but don't evaluate integral to represent volume of solid generated by revolving bounded region around give line.
1. y=sin x'
a) 0 b) about x axis
c) about y axis

2. y=e^ -x+1
a) x= -3
b) x=2
c) about line y= -1

3. y= lnx
a) 1/12 b) about y axis

4. y=tan x
a) x= 3pi/4
b) x= 5pi/4
c) about x axis
d) about line x=pi

5. y=x^4
a) y=x^3-4x^2+4x
b) about x axis
c) about line y=2
d) about y axis
e) about line x = -2

Thanks in advance. :)

I)
using the Disk Method:
about x-axis
y = 4x^5

1
∫ π ( 4x^5 )^2 dx = 16π/11
0

Using shell method:

y = 4x^5 ----> (y/4)^(1/5) = x
height ===> 1 - (y/4)^(1/5)
radius ====> y

limits:
when x = 0 ----> y = 0
when x = 1 ----> y = 4

4
∫ 2π * y * ( 1 - (y/4)^(1/5) ) dy = 16π/11
0

------------

using the Washer Method:
about y = 4
y = 4x^5

A (x) = π ( outer radius )^2 - π ( inner radius )^2
A (x) = π ( 4 - 0 )^2 - π ( 4 - 4x^5 )^2
A (x) = π ( 16 - (16 - 2 * 4 * 4x^5 + 16x^10 )
A (x) = π ( 16 - (16 - 32x^5 + 16x^10 ) )
A (x) = π ( 16 - 16 + 32x^5 - 16x^10 )
A (x) = π ( 32x^5 - 16x^10 )

1
∫ π ( 32x^5 - 16x^10 ) dx = 128π/33
0

Using shell method:

y = 4x^5 ----> (y/4)^(1/5) = x
height ===> 1 - (y/4)^(1/5)
radius ====> 4 - y

4
∫ 2π * (4 - y) * ( 1 - (y/4)^(1/5) ) dy = 128π/33
0

------------
using the Washer Method:
about y-axis
y = 4x^5 ---> x = (y/4)^(1/5)