Find the volume of the solid generated by revolving the following region about the given axis
The region in the first quadrant bounded above by the curve y=x^2, below by the x-axis and on the right by the line x=1, about the line x=-4
To find the volume of the solid, we will be using the method of cylindrical shells.
First, let's set up the integral to calculate the volume.
The height of each shell will be given by the difference between the top and bottom boundaries of the region, which is y = x^2 and the x-axis.
The radius of each shell will be given by the distance between the line x = -4 and the x-axis, which is 4.
The differential element of volume, dv, is given by dv = 2πrh dx, where r is the radius and h is the height.
Integrating from x = 0 to x = 1, we can calculate the volume as:
V = ∫[0 to 1] (2πrh) dx
= ∫[0 to 1] (2π(4)(x^2)) dx
= 8π ∫[0 to 1] (x^2) dx
Now, we can integrate:
V = 8π * (x^3/3)[0 to 1]
= 8π * (1/3 - 0)
= 8π/3
Therefore, the volume of the solid generated by revolving the region about the line x = -4 is 8π/3 cubic units.
To find the volume of the solid generated by revolving the region about the given axis, we can use the method of cylindrical shells.
First, let's draw a graph of the region to better understand it.
The region is bounded above by the curve y = x^2, below by the x-axis, and on the right by the line x = 1. In the first quadrant, this creates a triangular-shaped region.
To set up the integral for the volume using cylindrical shells, we'll need to consider that the axis of rotation is x = -4. A cylindrical shell is formed by taking a thin strip of the region, with thickness dx, and rotating it about the axis of rotation.
Let's consider a small vertical strip located at x. The height of this strip is given by the function f(x) = x^2, and the length of the strip is dx.
The circumference of the cylindrical shell is the distance traveled by the strip when it's rotated about the axis. In this case, the distance is given by the equation C = 2π(radius), where the radius is x - (-4) = x + 4.
Therefore, the volume of the cylindrical shell is given by dV = 2π(x + 4) * f(x) * dx.
To find the total volume, we integrate this expression over the range of x values that define the region.
∫[from 0 to 1] (2π(x + 4) * x^2) dx
Simplifying the integral:
∫[from 0 to 1] (2πx^3 + 8πx^2) dx
Using the power rule of integration, integrate each term separately:
= (2π/4)x^4 + (8π/3)x^3 evaluated from x = 0 to x = 1
= (π/2)x^4 + (8π/3)x^3 evaluated from x = 0 to x = 1
Evaluating the definite integral:
= (π/2)(1)^4 + (8π/3)(1)^3 - [(π/2)(0)^4 + (8π/3)(0)^3]
= (π/2) + (8π/3) - 0
= (π/2) + (8π/3)
= (π/6)(3 + 16)
= (π/6)(19)
So, the volume of the solid generated by revolving the given region about the line x = -4 is (π/6)(19).
I'd suggest using shells for this one:
v = ∫[0,1] 2πrh dx
where r = x+4 and h = y = x^2
v = 2π∫[0,1](x+4)*x^2 dx
= 2π∫[0,1] x^3 + 4x^2 dx
= 2π (1/4 x^4 + 4/3 x^3) [0,1]
= 2π (1/4 + 4/3)
= 19π/6
It can be done with discs, but you have to make them washers:
v = ∫[0,1] π(R^2-r^2) dy
where R = 5, r=4+x = 4+√y
v = π∫[0,1] (25 - (4+√y)^2) dy
= π (9y - 16/3 y^3/2 - 1/2 y^2) [0,1]
= π (9 - 16/3 - 1/2)
= 19π/6