Volume of a solid #2!?

Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=x^2, y=1, about y=2! Please Help me!

Answer 1

Using washer method pi S[a,b] (R^2 -r^2) dx
where R is the bigger radius
and r is smaller radius
S is the integral sign
[a,b] is the limit of integration

R = (x^2) - 2
r = 1 - 2

pi S[-1,1] (x^2 - 2)^2 - (-1)^2 dx

Since it is symmetric around the y axis we can write it as

2pi S [ 0,1] (x^2 - 2)^2 - (-1)^2 dx

When you do the integration you should get 56pi/15

Answer 2

You have a washer. The outer radius goes from the axis rotation (y=2) down to the parabola, so R = 2-x². The inner radius is contant at 1 (the difference between y = 1 and y = 2. y=1 intersects the parabola at x=1, -1. But since this region is symmetrical over the y-axis, you can calculate the volume from 0 to 1 and then double it: So it's the integral from 0 to 1

2 * π ∫(2-x²)² - 1 dx
=2 * π ∫( x^4 - 4x² + 3 dx

= 2π [(1/5)x^5 - (4/3)x^3 + 3x] from 0 to 1 = 56π/15

Answer 3

Pi multiplied by the integral from 0 to 1 of [ (2-x)^2-(2-1)^2 ] dx

Answer is 5.864306287

Answer 4

Use the cylindrical shell method.

y = x²
x = √y

Integrate from y = 1 to y = 2.

2∫2π(2 - y)x dy = 4π∫(2 - y)√y dy = 4π∫[2√y - y^(3/2)]dy

= 4π[(4/3)y^(3/2) - (2/5)y^(5/2)] | [Evaluated from y = 1 to 2]

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

= 4π[(8/3)√2 - (8/5)√2] - 4π[14/15]

= 32π√2[1/3 - 1/5] - 56π/15 = 32π√2[2/15] - 56π/15

= 64π√2/15 - 56π/15 = 8π/15[8√2 - 7] ≈ 7.227688