Revolve the region bounded by y=x and y=x^2 about the y axis. In cubic units, the resulting volume is?
You can use discs, integrating along y:
V = Int(pi (R^2 - r^2) dy)[0,1]
where R = y and r = sqrt(y)
= pi*Int(y - y^2)dy[0,1]
= pi(1/2 y^2 - 1/3 y^3)[0,1]
= pi(1/2 - 1/3)
= pi/6
Or, you can use shells, integrating along x:
V = Int(2pi*r*h dx)[0,1]
where r = x h = x-x^2
= 2pi*Int(x(x-x^2) dx)[0,1]
= 2pi(x^2 - x^3 dx)[0,1]
= 2pi(1/3 x^3 - 1/4 x^4)[0,1]
= 2pi(1/3 - 1/4)
= 2pi(1/12)
= pi/6
Revolve the region bounded by y = 4x and y = x2 about the y-axis. In cubic units, the resulting volume is
To find the resulting volume when rotating the region bounded by the curves y=x and y=x^2 about the y-axis, we need to use the method of cylindrical shells.
First, let's find the points of intersection between the curves y=x and y=x^2:
Set y=x and y=x^2 equal to each other:
x = x^2
Rearrange the equation:
x - x^2 = 0
Factor out the common factor:
x(1 - x) = 0
Set each factor equal to zero and solve for x:
x = 0 or x = 1
So, the region bounded by the curves y=x and y=x^2 lies between x=0 and x=1.
Next, we determine the height of a typical cylindrical shell at any given value of x. The height is the difference between the curves y=x and y=x^2:
height = (x) - (x^2) = x - x^2
The radius of the cylindrical shell is the distance from the y-axis to the curve y=x. Since we are rotating the region about the y-axis, the radius is simply x.
Now, we are ready to compute the volume using the integral:
V = ∫[from x=0 to x=1] 2πr(x)h(x) dx
V = ∫[from x=0 to x=1] 2π(x)(x - x^2) dx
V = 2π ∫[from x=0 to x=1] (x^2 - x^3) dx
Integrating term by term:
V = 2π [ (x^3/3) - (x^4/4) ] [from x=0 to x=1]
V = 2π [ (1/3) - (1/4) ]
V = 2π [ (4/12) - (3/12) ]
V = 2π [1/12]
V = π/6 cubic units
Therefore, the resulting volume is π/6 cubic units.