find the volume of the solid by rotating y=1-x^2, y=0 around the xaxis
using discs, of thickness dx
v = ∫[-1,1] πr^2 dx
where r = y = 1-x^2
So,
v = π∫[-1,1] (1-x^2)^2 dx = 16/15 π
Using shells, of thickness dy, we have to account for the two branches of the parabola, so it's easier just to use symmetry and get
v = 2∫[0,1] 2πrh dy
where r = y and h = x = √(1-y)
v = 4π∫[0,1] y√(1-y) dy = 16/15 π
pi y^2 dx from -1 to +1
same as 2 pi y^2 dx from 0 to 1
2 pi (1-2 x^2 + x^4) dx from 0 to 1
2 pi (1-0) - 4 pi (1^3/3-0) + 2 pi (1^5/5-0)
2 pi - 4 pi/3 + 2 pi/5
(pi/15) ( 30 - 20 + 6)
(16/15) pi
Check my arithmetic !
To find the volume of the solid obtained by rotating the region bounded by the curves y = 1 - x^2 and y = 0 around the x-axis, we can use the method of cylindrical shells.
The key idea behind this method is to approximate the solid volume by summing up the volumes of small, hollow cylindrical shells that make up the solid.
Step 1: Determine the limits of integration
To find the limits of integration, we need to determine the x-values at which the two curves intersect. Setting them equal to each other:
1 - x^2 = 0
Solving for x, we get x = ±1. So the limits of integration will be -1 to 1.
Step 2: Set up the integral
The volume of a cylindrical shell can be calculated using the formula:
V = ∫[a to b] 2πx * h(x) dx,
where a and b are the limits of integration, x is the distance from the axis of rotation, and h(x) is the height of the shell at that particular x-value.
In this case, the radius of the cylindrical shell is x, and the height is given by the difference between the upper and lower functions, which is (1 - x^2) - 0 = 1 - x^2.
Step 3: Evaluate the integral
Substituting the values into the volume formula, we have:
V = ∫[-1 to 1] 2πx * (1 - x^2) dx.
Simplifying the integral:
V = 2π ∫[-1 to 1] (x - x^3) dx.
Using the power rule of integration, and integrating term by term:
V = 2π [(1/2)x^2 - (1/4)x^4] evaluated from -1 to 1.
Evaluating at the limits:
V = 2π [(1/2)(1)^2 - (1/4)(1)^4] - [(1/2)(-1)^2 - (1/4)(-1)^4]
V = 2π [(1/2) - (1/4)] - [(1/2) - (1/4)]
V = 2π [1/4] - [1/4]
V = 2π [1/4]
V = π/2
Therefore, the volume of the solid obtained by rotating the region between y = 1 - x^2 and y = 0 around the x-axis is π/2 cubic units.