Find the volume of the solid obtained by revolving the region bounded by
y=22*x−22*(x^2) and the x-axis
around the x-axis.
using discs of thickness dx,
v = ∫[0,1] πr^2 dx
where r=y=22(x-x^2)
v = ∫[0,1] 22π(x-x^2)^2 dx = 11π/15
I misplaced the 22 in the integral. It should be
∫[0,1] π*(22(x-x^2))^2 dx = 242π/15
check: using shells of thickness dy, we have
x = 1/2 ± 1/(2√11) √(11-2y)
The vertex is at (1/2,11/2)
using symmetry, we can just double the volume under one side of the axis of symmetry, so
v = ∫[0,11/2] 2πrh dy
where r=y and h=1/2-x = 1/2-(1/2 - 1/(2√11) √(11-2y)) = √(11-2y)/(2√11)
v = 2∫[0,11/2] 2πy√(11-2y)/(2√11) dy
= 2π/√11 ∫[0,11/2] y√(11-2y) dy = 242π/15
To find the volume of the solid obtained by revolving the region bounded by the curve y = 22x - 22(x^2) and the x-axis around the x-axis, we can use the method of cylindrical shells.
The formula for finding the volume using cylindrical shells is:
V = 2π ∫[a,b] x * f(x) dx
where [a, b] is the interval over which the curve is defined, and f(x) is the function describing the curve.
In this case, since the region is bounded by y = 22x - 22(x^2) and the x-axis, the interval [a, b] can be determined by finding the x-values where the curve intersects the x-axis, which are the roots of the equation 22x - 22(x^2) = 0.
To find the roots, we set y = 0 and solve for x:
0 = 22x - 22(x^2)
0 = 22x(1 - x)
x = 0 or x = 1
So the interval [a, b] is [0, 1].
Now, we can calculate the volume using the formula:
V = 2π ∫[0,1] x * (22x - 22(x^2)) dx
First, let's simplify the integral:
V = 2π ∫[0,1] (22x^2 - 22(x^3)) dx
Now, we can integrate term by term:
V = 2π * [(22/3)x^3 - (22/4)x^4] evaluated from 0 to 1
V = 2π * [(22/3)(1^3) - (22/4)(1^4)] - [(22/3)(0^3) - (22/4)(0^4)]
V = 2π * [(22/3) - (22/4)] - [0 - 0]
V = 2π * [(22/3) - (22/4)]
Now, simplify further:
V = 2π * [(22/3) - (11/2)]
V = 2π * [(44/6) - (33/6)]
V = 2π * (11/6)
Finally, we can calculate the volume:
V ≈ 11π/3
So, the volume of the solid obtained by revolving the region bounded by y = 22x - 22(x^2) and the x-axis around the x-axis is approximately 11π/3 cubic units.