Question

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. $$y=8, y=2 x+2, x=0, \text { and } x=2$$

Short answer

Answer: The volume of the solid generated is 36π cubic units.

Step-by-step solution

Find the Intersection Points

To find the intersection points between y=8 and y=2x+2, set the two equations equal to each other and solve for x: $$8 = 2x + 2$$ $$6 = 2x$$ $$x = 3$$ So there is an error in the problem statement as x=2 is not the intersection point of the two curves and hence, we will use x=3 as the intersection point. The region R is defined by the following curves: $$y = 8, y = 2x+2, x = 0, \text { and } x = 3$$

Find the Height and Radius of the Shell

We are going to use the shell method to find the volume of the solid generated when R is revolved about the x-axis. The height of the shell (h) is the difference between the two curves, and the radius of the shell (r) is equal to y. So, for any given x: $$h(x) = 8 - (2x + 2)$$ $$r(x) = y = 2x + 2$$

Set up the Integral for the Volume Using the Shell Method

The shell method formula for the volume of the solid generated by revolving the region R about the x-axis is: $$V = 2 \pi \int_{a}^{b} r(x) h(x) dx$$ Using the bounds x=0 to x=3 (the region R), the height h(x) and radius r(x) found in Step 2, set up the integral for the volume: $$V = 2 \pi \int_{0}^{3} (2x + 2)(8 - (2x + 2)) dx$$

Solve the Integral for the Volume

To solve the integral, we first need to simplify the integrand by distributing and combining like terms: $$V = 2 \pi \int_{0}^{3} (16x - 2x^2) dx$$ Now, integrate with respect to x: $$V = 2 \pi \left[8x^2 - \frac{2}{3}x^3\right]_{0}^{3}$$ Evaluate the definite integral using the bounds x=0 and x=3: $$V = 2 \pi \left[\left(8(3)^2 - \frac{2}{3}(3)^3\right) - \left(0\right)\right]$$ $$V = 2 \pi \left[72 - 54\right]$$ $$V = 2 \pi (18)$$

Calculate the Volume

Now, multiply the resulting value by 2π to get the final volume: $$V = 36 \pi$$ The volume of the solid generated when the region R is revolved about the x-axis is 36π cubic units.