The region in the first quadrant bounded by the š¦-axis and the curve š„ = 2š¦2 ā š¦3 (graph please) is revolved about the š„-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
ā«[0,2] 2Ļy(2y^2-y^3) dy = 16Ļ/5
To find the volume of the region in the first quadrant bounded by the š¦-axis and the curve š„ = 2š¦^2 ā š¦^3 when revolved about the š„-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 = 2Ļ(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 š¦-axis and the curve š„ = 2š¦^2 ā š¦^3, we need to find the values of š¦ where the curve intersects the š¦-axis. To do this, we set š„ equal to zero:
0 = 2š¦^2 ā š¦^3
Factoring out š¦^2, we get:
0 = š¦^2(2 ā š¦)
So, š¦ = 0 or š¦ = 2. We will integrate from š¦ = 0 to š¦ = 2.
2. Set up the integral:
The volume of each cylindrical shell is given by:
dV = 2Ļ(radius)(height)(thickness)
The radius of each shell is the distance from the š¦-axis to the curve at a particular š¦-value, which is simply š„. So, the radius, š, is equal to š¦.
The height of each shell is the difference between the curve at š¦ and the š¦-axis. So, the height, ā, is equal to š„ (which is š„ = 2š¦^2 ā š¦^3).
The thickness of each shell is an infinitesimally small change in š¦, denoted as Īš¦. We will later take the limit as Īš¦ approaches zero to find the exact volume.
The integral to find the volume becomes:
V = ā«[0 to 2] 2Ļ(š¦)(2š¦^2 ā š¦^3) dš¦
3. Evaluate the integral:
We can now evaluate the integral to find the volume. Simplify the integrand and integrate with respect to š¦:
V = ā«[0 to 2] 4Ļ(š¦^3 ā š¦^4) dš¦
Integrating term by term, the volume formula becomes:
V = 4Ļ(š¦^4/4 ā š¦^5/5) from 0 to 2
Substituting the upper and lower limits of integration:
V = 4Ļ((2^4/4 ā 2^5/5) ā (0^4/4 ā 0^5/5))
V = 4Ļ((16/4 ā 32/5) ā (0))
V = 4Ļ(4 ā 32/5)
V = 4Ļ(20/5 ā 32/5)
V = 4Ļ(-12/5)
V = -48Ļ/5
Hence, the volume of the region in the first quadrant bounded by the š¦-axis and the curve š„ = 2š¦^2 ā š¦^3, when revolved about the š„-axis, is -48Ļ/5 (negative since it is below the š„-axis).