Question

43–55. Volumes of solids Choose the general slicing method, the disk/washer method, or the shell method to answer the following questions.

The region bounded by the graphs of y = 2x,y = 6−x, and y = 0 is revolved about the line y = −2 and the line x = −2. Find the volumes of the resulting solids. Which one is greater?

Accepted Answer

First, identify the region bounded by the curves: $$y = 2x$$, $$y = 6 - x$$, and $$y = 0. $$Sketching or analyzing these will help visualize the area to be revolved. Determine the points of intersection between $$y = 2x$$ and $$y = 6 - x by $$setting $$2x = 6 - x$$ and solving for $$x. $$This gives the limits of integration for the region. For the solid revolved about the line $$y = -2$$, use the disk/washer method because the axis of rotation is horizontal and outside the region. Express the radius of each disk/washer as the distance from the curve to $$y = -2. $$Set up the volume integral for revolution about $$y = -2 as$$ $$V = \pi \int_{a}^{b} \left[ (R_{outer})^2$ - (R_{inner})^2$ \right] \, dx$$, where $$R_{outer}$$ and $$R_{inner}$$ are the distances from the curves to the line $$y = -2. $$For the solid revolved about the line $$x = -2$$, use the shell method because the axis of rotation is vertical and outside the region. Express the height of each shell as the difference between the two curves and the radius as the distance from $$x to$$ $$-2. $$Set up the volume integral as $$V = 2\pi \int_{a}^{b} (radius)(height) \, dx.$$

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Key Concepts

Volume of Solids of Revolution

Disk/Washer and Shell Methods

Setting up the Integral with Correct Bounds and Axis