Introduction to Volume of Revolution

The problem of finding the volume of a solid generated by rotating a region around a specified line is a classic problem in calculus. This article will guide you through the process of finding the volume of the region bounded by the curves y 3 - x2, y 2^x, the x-axis, and the y-axis, when this region is rotated around the line y -1. We will cover the mathematical steps and provide detailed solutions for both the positive and negative x regions.

Mechanics of Volume of Revolution

When a two-dimensional planar region is rotated around a line, the resulting solid can be analyzed using the method of disks or washers, depending on whether the area being rotated is completely enclosed or not. Here, the region is bounded by the curves y 3 - x2 and y 2^x, and it intersects the x-axis and the y-axis.

Defining the Regions

First, we need to identify the x-coordinates of the points of intersection of the curves y 3 - x2 and y 2^x. Through analysis, we find that these intersections occur at x 1 and x 2, as well as approximately x -1.64 and y 0.32. The exact values for the latter pair of points are transcendental, meaning they cannot be expressed as a finite combination of numbers and basic operations.

Volume of Revolution for the Positive X Region

To find the volume of the solid of revolution generated by rotating the region bounded by y 3 - x2, y 2^x, and the curves from x 0 to x 1 around the line y -1, we use the following integral:

V ∫ 0 1 π [ 2 x 1 2 - 1 2 ] d x ∫ 1 3 π [ 4 x 1 2 - 1 2 ] d x [ 7 ln 4 4 π ( 3 5 - 47 15 ) ] appox 24.37 cubic units

This solution gives the volume of the solid generated by rotating the region from x 0 to x 1 around the line y -1. The result is a volume of approximately 24.37 cubic units.

Volume of Revolution for the Negative X Region

The volume of the solid generated by rotating the region from x -√3 to x -1.64 (approximately) around the line y -1 can be described by the following integral:

V ∫ - 3 - 1.64 π [ 4 x 1 2 - 1 2 ] d x ∫ - 1.64 0 π [ 2 x 1 2 - 1 2 ] d x

While the exact solution for this integral requires further computation, it provides a method to evaluate the volume for the region from x -√3 to x -1.64, as well as the region from x -1.64 to x 0.