Final answer:
The region enclosed by the curves 2xy=2, 8xy=8, and the line x=0 forms a rectangular solid when revolved around the y-axis. The volume of this solid is 4π units cubed.
Step-by-step explanation:
To illustrate the region enclosed by the given curves, we first need to graph them on a coordinate plane. The curve 2xy=2 can be rewritten as y=1/x, which is a hyperbola with asymptotes x=0 and y=0. The curve 8xy=8 can be rewritten as y=1/(4x), which is also a hyperbola with the same asymptotes. The line x=0 is simply the y-axis. By plotting these curves on a graph, we can see that they intersect at the point (1,1) and form a rectangle with sides x=1 and y=1.
To find the volume of the solid formed by revolving this region around the y-axis, we can use the formula for solid revolution, V=π∫(ƒ(x))^2dx, where ƒ(x) is the function that represents the shape of the cross-section of the solid. In this case, the cross-section is a circle with radius ƒ(x)=1/x, since the region is being revolved around the y-axis.
Thus, the volume can be calculated as V=π∫(1/x)^2dx=π∫(1/x^2)dx=π(-1/x)|_1^1=4π units cubed.
One notable feature of this solid is that the radius of the cross-section changes as x increases, resulting in a non-uniform shape. This can also be seen in the graph, where the distance between the curves decreases as x increases. Additionally, the solid is bounded by the curves, which implies that it has a finite volume. Overall, the solid formed by rotating this region is a unique shape with interesting geometric properties.