Let R be the region between the graph of y=x3 and the x-axis, and between x=0 and x=1.
Compute the volume of the region obtained by revolving R around the y-axis
with discs (washers),
v = ∫[0,1] π (R^2-r^2) dy
where r=x and R=1
v = π∫[0,1] (1 - y^(2/3)) dy
= π (y - 3/5 y^(5/3)) [0,1]
= 2/5 π
with shells,
v = ∫[0,1] 2π rh dx
where r=x and h=y
v = 2π∫[0,1] x*x^3 dx
= 2π (1/5 x^5) [0,1]
= 2/5 π
First off, thank you answering my question.
I just have one question. So does it matter which method we use?
I am just trying to find a better way of understanding this problem from a visual perspective.
generally it doesn't matter. It depends on whether it is easier to express y as a function of x, or vice-versa. For example if y=e^x + 4(x^2-1) then it's pretty tough to come up with y-1
In this case, it was easy.
To compute the volume of the region obtained by revolving the region R around the y-axis, we can use the method of cylindrical shells.
First, let's visualize the region R. The graph of y=x^3 is a cubic curve that starts at the origin (0,0) and passes through the points (1,1) and (-1,-1). The region R is the area between this curve and the x-axis, bounded by x=0 and x=1.
To find the volume using cylindrical shells, we need to integrate the circumference of each shell multiplied by its height. The circumference of each shell is given by 2πy, and the height is given by the differential dx. So, the volume element dV of a cylindrical shell is given by dV = 2πy dx.
To express y in terms of x, we rearrange the equation y = x^3 to get x = y^(1/3).
Now, we need to determine the limits of integration for x. We are revolving the region R around the y-axis, so the slices of the region will have their heights measured along the y-axis. The y-values range from 0 to 1, as given by the bounds of the region.
Therefore, the integral for the volume V of the region can be set up as follows:
V = ∫[from 0 to 1] (2πy) * dx
V = ∫[from 0 to 1] (2π * y * dx)
V = ∫[from 0 to 1] (2π * y * 1) dx
V = 2π * ∫[from 0 to 1] y dx
Now, substituting x = y^(1/3), we rewrite the integral in terms of x:
V = 2π * ∫[from 0 to 1] (x^3) dx
Evaluating this integral, we get:
V = 2π * [x^4/4] [from 0 to 1]
V = 2π * (1/4 - 0)
V = π/2
Therefore, the volume of the region obtained by revolving R around the y-axis is π/2.