Find the volume V of the solid obtained by rotating the region bounded
by curves y=x and y= √𝒙 about the x-axis
Easy to see that the intersection of y = x and y = √x is (0,0) and (1,1)
In that interval y = √x is above y = x, so we have
Vol = π ∫ (x - x^2) dx from 0 to 1
= π [ x^2/2 - x^3/3 | from 0 to 1
= π( 1/2 - 1/3 - 0)
= π/6 units^3
To find the volume of the solid obtained by rotating the region bounded by curves y = x and y = √x about the x-axis, we can use the method of cylindrical shells.
Here's the step-by-step explanation on how to find the volume using this method:
1. First, let's find the points of intersection between the two curves. Set the equations of the curves equal to each other and solve for x:
x = √x
Square both sides to eliminate the square root:
x^2 = x
Rearrange the equation:
x^2 - x = 0
Factor out x from the equation:
x(x - 1) = 0
Set each factor equal to zero:
x = 0 or x - 1 = 0
Solve for x:
x = 0 or x = 1
So we have two points of intersection: (0, 0) and (1, 1).
2. Now, let's set up the integral to find the volume. The volume of a solid obtained by rotating a region is given by the integral:
V = ∫[a,b] 2πy * h * dx
In this case, the height, h, is given by the difference between the top curve, y = √x, and the bottom curve, y = x.
So h = √x - x.
The limits of integration, a and b, are the x-values where the curves intersect. In this case, a = 0 and b = 1.
Substituting the values into the integral, we get:
V = ∫[0,1] 2πx(√x - x) dx
3. Now, let's evaluate the integral using basic integration rules. Simplify the expression inside the integral:
V = 2π ∫[0,1] (x^(3/2) - x^2) dx
Integrate each term separately:
V = 2π [(2/5)x^(5/2) - (1/3)x^3] between 0 and 1
Evaluate the integral at the upper and lower limits:
V = 2π [((2/5)(1)^(5/2) - (1/3)(1)^3) - ((2/5)(0)^(5/2) - (1/3)(0)^3)]
Simplify the expression:
V = 2π [(2/5) - (1/3)]
Multiply and divide to simplify further:
V = 2π [(6/15) - (5/15)]
V = 2π (1/15)
V = 2π/15
Therefore, the volume of the solid obtained by rotating the region bounded by curves y = x and y = √x about the x-axis is (2π/15) cubic units.
Well they cross at (1,1) so I suppose you mean between x = 0 and x = 1
so the outer radius is ro = x^.5
and the inner radius is ri = x
and we want the integral from x = 0 to x = 1 of
[ pi ro^2 - pi ri^2 ] dx
= pi [ x - x^2 ] dx
= pi [ (1/2) x^2 - (1/3) x^3] at x = 1 - at x = 0
= pi [1/2 - 1/3] = pi / 6
using shells of thickness dy, we have
v = ∫[0,1] 2πrh dy
where r = y and h = y-y^2
v = ∫[0,1] 2πy(y-y^2) dy = π/6