y=x^2 and x=2 intersect at (2,4)
the radius of the rotated region = 2 - x
= 2 - √y
V = π∫ (2-√y)^2 dy from 0 to 4
= π∫ (4 - 4√y + y) dy from 0 to 4
= π[ 4y - (8/3)y^(3/2) + (1/2)y^2 ] from 0 to 4
= π (16 - (8/3)(8) + 8 - 0)
= 8π/3
check my arithmetic and thinking.
The region R is bounded by the x-axis, x = 2, and y = x^2. What is the the volume of the solid formed by revolving R about the line x = 2?
2 answers
y=x^2 intersects x=2 at (2,4)
using shells,
v = ∫2πrh dx [0,2]
where
r = 2-x
h = y
v = 2π∫(2-x)(x^2)dx [0,2]
= 2π∫2x^2 - x^3 dx [0,2]
= 2π(2/3 x^3 - 1/4 x^4)[0,2]
= 2π(16/3 - 4)
= 2π(4/3)
= 8π/3
using discs,
v = ∫πr^2 dy [0,4]
= π∫(2-√y)^2 dy [0,4]
= π∫(4 - 4√y + y)dy [0,4]
= π(4x - 8/3y√y + 1/2 y^2)[0,4]
= π(16 - 64/3 + 8)
= 8π/3
using shells,
v = ∫2πrh dx [0,2]
where
r = 2-x
h = y
v = 2π∫(2-x)(x^2)dx [0,2]
= 2π∫2x^2 - x^3 dx [0,2]
= 2π(2/3 x^3 - 1/4 x^4)[0,2]
= 2π(16/3 - 4)
= 2π(4/3)
= 8π/3
using discs,
v = ∫πr^2 dy [0,4]
= π∫(2-√y)^2 dy [0,4]
= π∫(4 - 4√y + y)dy [0,4]
= π(4x - 8/3y√y + 1/2 y^2)[0,4]
= π(16 - 64/3 + 8)
= 8π/3
2 answers