find the volume of the solid formed by revolving the region bounded by the graphs of y=x^3 y=1 x=2 about the y-axis
when y = 1, x = 1
when y = 8, x = 2so we seem to be talking about the sort of triangle between (1,1) (2,1) and (2,8)
could be vertical cylinders of wall thickness dx and height (x^3-1) and radius x from x = 1 to x = 2
v = integral [2 pi x(x^3-1)]dx from x = 1 to x = 2
2 pi [ x^5/5 - x^2/3 ]
2 pi [ 32/5 - 4/3 ] - 2 pi [1/5-1/3]
2 pi [ 31/5 - 1 ]
check my arithmetic !
using shells of thickness dx, we have
v = ∫[1,2] 2πrh dx
where r=x and h=y-1
v = ∫[1,2] 2πx(x^3-1) dx = 47π/5
using discs (washers) of thickness dy,
v = ∫[1,8] π(R^2-r^2) dy
where R=2 and r=x
v = ∫[1,8] π(4-y^(2/3)) dy = 47π/5
Thank you!!!!
To find the volume of the solid formed by revolving the region bounded by the graphs of y = x^3, y = 1, and x = 2 about the y-axis, we can use the method of cylindrical shells.
First, let's graph the region to visualize it.
The graph of y = x^3 is a cubic function that passes through the origin and has a point of inflection at (0,0). The graph of y = 1 is a horizontal line at y = 1, and the line x = 2 is a vertical line passing through the point (2,0).
To find the volume, we need to integrate the area of the cylindrical shells formed by revolving the region bounded by the curves about the y-axis.
The volume of each cylindrical shell can be approximated by multiplying the circumference of the shell by its height and thickness. The height of each shell is given by the difference between the two curves at a particular y-value, and the thickness is an infinitesimally small change in y.
To set up the integral, we need to express the radius and height of the cylindrical shell in terms of y.
The radius of the shell is the distance from the y-axis to the curve x = 2, which is always 2.
The height of the shell is the difference between the values of y = x^3 and y = 1 at a given y-value. So, the height can be expressed as (x^3 - 1). Since we are revolving the region about the y-axis, we need to express x in terms of y. We can rearrange the equation y = x^3 to get x = y^(1/3).
Now, we can set up the integral to find the volume:
V = ∫[a,b] 2πrh dy
where r is the radius of the shell (2), h is the height of the shell (y^(1/3) - 1), and [a,b] represents the interval over which we need to integrate (the y-values where the curves intersect).
To find the limits of integration, we need to determine the y-values at which the curves intersect. The curve y = x^3 intersects with y = 1 when x = 1 (since 1^3 = 1). So, the limits of integration are from y = 0 to y = 1.
V = ∫[0,1] 2π(2)(y^(1/3) - 1) dy
Evaluating this integral will give you the volume of the solid formed by revolving the region bounded by the graphs of y = x^3, y = 1, and x = 2 about the y-axis.