The region in the first quadrant bounded by the 𝑦-axis and the curve 𝑥 = 2𝑦2 − 𝑦3 (graph please) is revolved about the 𝑥-axis. Find the volume?
x = y^2(2-y)
So the region in the 1st quadrant is a hump from (0,0) to (0,2)
To revolve that around the x-axis it is best to use shells of radius y, thickness dy and height x. So, the total volume is
∫[0,2] 2πy(2y^2-y^3) dy = 16π/5
To find the volume of the region in the first quadrant bounded by the 𝑦-axis and the curve 𝑥 = 2𝑦^2 − 𝑦^3 when revolved about the 𝑥-axis, we can use the method of cylindrical shells.
Cylindrical shells method uses infinitesimally thin cylindrical shells to approximate the volume. The volume of each shell can be calculated using the formula:
Volume of a shell = 2π(radius)(height)(thickness)
Let's break down the steps to find the volume using this method:
1. Determine the bounds of integration:
Since the region is bounded by the 𝑦-axis and the curve 𝑥 = 2𝑦^2 − 𝑦^3, we need to find the values of 𝑦 where the curve intersects the 𝑦-axis. To do this, we set 𝑥 equal to zero:
0 = 2𝑦^2 − 𝑦^3
Factoring out 𝑦^2, we get:
0 = 𝑦^2(2 − 𝑦)
So, 𝑦 = 0 or 𝑦 = 2. We will integrate from 𝑦 = 0 to 𝑦 = 2.
2. Set up the integral:
The volume of each cylindrical shell is given by:
dV = 2π(radius)(height)(thickness)
The radius of each shell is the distance from the 𝑦-axis to the curve at a particular 𝑦-value, which is simply 𝑥. So, the radius, 𝑟, is equal to 𝑦.
The height of each shell is the difference between the curve at 𝑦 and the 𝑦-axis. So, the height, ℎ, is equal to 𝑥 (which is 𝑥 = 2𝑦^2 − 𝑦^3).
The thickness of each shell is an infinitesimally small change in 𝑦, denoted as Δ𝑦. We will later take the limit as Δ𝑦 approaches zero to find the exact volume.
The integral to find the volume becomes:
V = ∫[0 to 2] 2π(𝑦)(2𝑦^2 − 𝑦^3) d𝑦
3. Evaluate the integral:
We can now evaluate the integral to find the volume. Simplify the integrand and integrate with respect to 𝑦:
V = ∫[0 to 2] 4π(𝑦^3 − 𝑦^4) d𝑦
Integrating term by term, the volume formula becomes:
V = 4π(𝑦^4/4 − 𝑦^5/5) from 0 to 2
Substituting the upper and lower limits of integration:
V = 4π((2^4/4 − 2^5/5) − (0^4/4 − 0^5/5))
V = 4π((16/4 − 32/5) − (0))
V = 4π(4 − 32/5)
V = 4π(20/5 − 32/5)
V = 4π(-12/5)
V = -48π/5
Hence, the volume of the region in the first quadrant bounded by the 𝑦-axis and the curve 𝑥 = 2𝑦^2 − 𝑦^3, when revolved about the 𝑥-axis, is -48π/5 (negative since it is below the 𝑥-axis).