find the volume of the solid of revolution formed by rotating the region bounded by y = x - 2, y = -x + 2…

Explanation:

Step1: Identify the bounds and height for the shell method.

The region is bounded by $x=0$ and the intersection of $y=x-2$ and $y=-x+2$. Setting $x-2 = -x+2$ gives $2x=4$, so $x=2$. The bounds of integration are $[0, 2]$. The height of a cylindrical shell at $x$ is the difference between the upper curve ($y=-x+2$) and the lower curve ($y=x-2$), which is $h(x) = (-x+2) - (x-2) = -2x+4$.

Step2: Set up the volume integral using the shell method.

The volume $V$ using the shell method for rotation about the y-axis is given by $V = \int_{a}^{b} 2\pi x h(x) , dx$. Substituting the bounds and height: $$V = \int_{0}^{2} 2\pi x (-2x+4) , dx$$

Step3: Evaluate the integral.

Factor out $2\pi$ and distribute $x$: $$V = 2\pi \int_{0}^{2} (-2x^2+4x) , dx$$ Find the antiderivative: $$V = 2\pi \left[ -2\frac{x^3}{3} + 4\frac{x^2}{2} \right]{0}^{2}$$ $$V = 2\pi \left[ -\frac{2}{3}x^3 + 2x^2 \right]{0}^{2}$$ Evaluate the antiderivative at the bounds: $$V = 2\pi \left( \left( -\frac{2}{3}(2)^3 + 2(2)^2 \right) - \left( -\frac{2}{3}(0)^3 + 2(0)^2 \right) \right)$$ $$V = 2\pi \left( \left( -\frac{2}{3}(8) + 2(4) \right) - 0 \right)$$ $$V = 2\pi \left( -\frac{16}{3} + 8 \right)$$ $$V = 2\pi \left( -\frac{16}{3} + \frac{24}{3} \right)$$ $$V = 2\pi \left( \frac{8}{3} \right)$$ $$V = \frac{16\pi}{3}$$

Step4: Calculate the numerical value and round.

Calculate the approximate value of $V$: $$V = \frac{16\pi}{3} \approx 16.75516...$$ Round the result to the nearest thousandth: $$V \approx 16.755$$

Answer:

$16.755$