Calculus Volume Question (use shell method, disc method, or washer method)?

Question

Consider the area enclosed by y= square root of (5-x) , x=1, and y=0, Find the volume of the solid generated when the region is rotated about:
a) the y axis
b) the line x = -2
the answers should be a) 87.127 and b) 154.147 (these answers may possibly be wrong)

Answer 1

Let V = volume.

a) Rotate about the y axis.
Use the shell method. Integrate from x = 1 to 5.

V = ∫2πxydx = 2π∫x√(5 - x)dx
Let
u = 5 - x and du = -dx
x = 5 - u
when x = 1 then u = 4
when x = 5 then u = 0

= 2π∫x√(5 - x)dx = 2π∫[(5 - u)√u](-du) = -2π∫[5√u - u^(3/2)]du

= -2π[(10/3)u^(3/2) - (2/5)u^(5/2)] | [Evaluated from 4 to 0]

= 0 + 2π[(10/3)4^(3/2) - (2/5)4^(5/2)]

= 2π[(10/3)*8 - (2/5)*32] = 2π[80/3 - 64/5] ≈ 87.127
__________________

a) Rotate about the line x = -2.
Use the shell method. Integrate from x = 1 to 5.

V = ∫2π(x + 2)ydx = 2π∫(x + 2)√(5 - x)dx
Let
u = 5 - x and du = -dx
x = 5 - u
when x = 1 then u = 4
when x = 5 then u = 0

= 2π∫[(x + 2)√(5 - x)]dx = 2π∫[(7 - u)√u](-du)

= -2π∫[7√u - u^(3/2]du

= -2π[(14/3)u^(3/2) - (2/5)u^(5/2)] | [Evaluated from 4 to 0]

= 0 + 2π[(14/3)4^(3/2) - (2/5)4^(5/2)]

= 2π[(14/3)*8 - (2/5)*32] = 2π[112/3 - 64/5] ≈ 154.147