Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.y=|x| and y=2−$$x^2$$; about the x-axis

Accepted Answer

First, identify the region R bounded by the curves $$y = |x|$$ and $$y = 2 - x^2$. Since y$$ = |x|$$ is piecewise, consider y = x$ for x$$ \geq 0$$ and y = -x$ for x$$ \u003c 0$$. The region is symmetric about the y-axis, so you can analyze for x$$ \geq 0$$ and then double the volume if needed. Find the points of intersection between the curves by solving $$|x| = 2 - $$x^2. $$For $$x \geq 0$$, solve $$x = 2 - x^2$, which rearranges to x^2$ + x - 2 = 0. $$Find the roots to determine the limits of integration. Set up the volume integral using the method of washers (disks with holes) since the region is revolved about the x-axis. The volume element is $$\pi (R_{outer}^2 - R_{inner}^2) \, dx$$, where $$R_{outer}$$ and $$R_{inner}$$ are the distances from the x-axis to the outer and inner curves respectively. Determine which curve is farther from the x-axis in the interval between the points of intersection. Since the revolution is about the x-axis, the radius is the y-value of the curve. For each $$x$$, $$R_{outer} = \max(|y|)$$ and $$R_{inner} = \min(|y|)$$ between the two curves. Write the integral for the volume as $$V = \pi \int_{a}^{b} \left[(2 - x^2$)^2$$ - (|x|$$)^2$$\right] \, dx$$, where a$ and b$ are the intersection points found earlier. Evaluate this integral to find the volume.

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=|x| and y=2−x^2; about the x-axis

Verified video answer for a similar problem:

Key Concepts

Understanding the Region Bounded by Curves

Method of Disks and Washers

Handling Absolute Value Functions in Integration

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.y=|x| and y=2−x^2; about the x-axis