Question

VolumesFind the volume of the solid generated by revolving the “triangular” region bounded by the curve y = 4/x³  and the lines x = 1 and y = 1/2 aboutc. the line x = 2

Accepted Answer

First, identify the region bounded by the curve $$y = \frac{4}{x^3$}$$, the vertical line $$x = 1$$, and the horizontal line $$y = \frac{1}{2}. $$Determine the other boundary for $$x by $$solving $$\frac{4}{x^3$} = \frac{1}{2} to $$find the $$x-$$value where the curve meets $$y = \frac{1}{2}. $$Since the solid is generated by revolving the region about the vertical line $$x = 2$$, use the method of cylindrical shells. The formula for the volume using shells is $$V = 2\pi \int_a^b (	ext{radius})(	ext{height}) \, dx. $$Express the radius of a typical shell as the distance from the shell at position $$x to $$the axis of rotation $$x = 2. $$This radius is $$r = 2 - x$$ because the shells are vertical slices between $$x=1$$ and the $$x-$$value found in step 1. The height of each shell is the vertical distance between the curve and the line $$y = \frac{1}{2}. $$Since the region is bounded above by $$y = \frac{4}{x^3$}$$ and below by $$y = \frac{1}{2}$$, the height is $$h = \frac{4}{x^3$} - \frac{1}{2}. $$Set up the integral for the volume as $$V = 2\pi \int_{1}^{x_0$} (2 - x) \left( \frac{4}{x^3$} - \frac{1}{2} ight) dx$$, where $$x_0 is $$the $$x-$$value found in step 1. This integral represents the volume of the solid generated by revolving the region about $$x=2.$$

Volumes
Find the volume of the solid generated by revolving the “triangular” region bounded by the curve y = 4/x³ and the lines x = 1 and y = 1/2 about
c. the line x = 2

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Key Concepts

Volume of Solids of Revolution

Shell Method for Volumes

Setting up Limits and Functions for Integration