Answer:
a) 8π
b) 8/3 π
c) 32/5 π
d) 176/15 π
Step-by-step explanation:
Given lines : y = √x, y = 2, x = 0.
a) The x-axis
using the shell method
y = √x = , x = y^2
h = y^2 , p = y
vol = ( 2π ) [tex]\int\limits^2_0 {ph} \, dy[/tex]
= [tex]( 2\pi ) \int\limits^2_0 {y.y^2} \, dy[/tex]
∴ Vol = 8π
b) The line y = 2 ( using the shell method )
p = 2 - y
h = y^2
vol = ( 2π ) [tex]\int\limits^2_0 {ph} \, dy[/tex]
= [tex]( 2\pi ) \int\limits^2_0 {(2-y).y^2} \, dy[/tex]
= ( 2π ) * [ 2/3 * y^3 - y^4 / 4 ] ²₀
∴ Vol = 8/3 π
c) The y-axis ( using shell method )
h = 2-y = h = 2 - √x
p = x
vol = [tex](2\pi ) \int\limits^4_0 {ph} \, dx[/tex]
= [tex](2\pi ) \int\limits^4_0 {x(2-\sqrt{x} ) } \, dx[/tex]
= ( 2π ) [x^2 - 2/5*x^5/2 ]⁴₀
vol = ( 2π ) ( 16/5 ) = 32/5 π
d) The line x = -1 (using shell method )
p = 1 + x
h = 2√x
vol = [tex](2\pi ) \int\limits^4_0 {ph} \, dx[/tex]
Hence vol = 176/15 π
attached below is the graphical representation of P and h
