A (y) = π ( outer radius )^2 - π ( inner radius )^2
A (y) = π ( 1 - 0 )^2 - π ( (y/4)^(1/5) - 0 )^2
A (y) = π ( 1 - (y/4)^(2/5) )
4
∫ π ( 1 - (y/4)^(2/5) ) dy = 8π/7
0
Using shell method:
y = 4x^5
height ===> 4x^5
radius ====> x
1
∫ 2π * x * 4x^5 dx = 8π/7
0
----------
using the Disk Method:
about x = 1
y = 4x^5 ---> x = (y/4)^(1/5)
4
∫ π ( 1 - (y/4)^(1/5) )^2 dy = 4π/21
0
Using shell method:
y = 4x^5
height ===> 4x^5
radius ====> 1 - x
1
∫ 2π * (1 - x) * 4x^5 dx = 4π/21
0
==========
II)
since the diameters are the height here, the area of the circle is:
π * r^2 or π * (d/2)^2
<---- can't tell: there are no limits
∫ π * (d/2)^2 dx
0
<---- can't tell: there are no limits
∫ π * (3x^4/2)^2 dx
0
===========
III)
y = sin x
about x-axis
using the Disk Method:
π. . . . . . . . . . . . .2π
∫ π ( sin(x) )^2 dx + ∫ π ( 0 - sin(x) )^2 dx = π^2
0. . . . . .. . . . . . . .π
Using the Shell Method:
1
∫ 2π * y * sin^-1(y) = π^2/4. . we have to multiply by 4 since we are integrating to the max.
0
http://www.wolframalpha.com/input/?i=y+%…
we have 4 of them.
4 * π^2/4 = π^2
about y-axis
using the Shell Method:
y = sin(x)
π
∫ 2π * x * sin(x) dx = 2π^2 (we have two of them) ----> 2 * 2π^2 = 4π^2
0
========
I will assume that the x-axis is boundary.
y=e^-x+1
x= -3, x = 2 and about line y = -1
using the Disk Method:
y = e^-x + 1
A(x) = π * ( e^-x + 1 - -1 )^2 - π * ( 0 - -1 )^2
A(x) = π * ( ( e^-x + 2 )^2 - 1 )
2
∫ π * ( ( e^-x + 2 )^2 - 1 ) dx ≈ 931.50
-3
==========
Shell Method:
1
∫ 2π * x * ( 0 - ln(x) ) dx ≈ 1.5
1/12
ln(x) bewfore x = 1, it is below the x-axis
==========
Using Disk: