Problem 16 - Calculus Early Transcendentals

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. $$y=8, y=2 x+2, x=0, \text { and } x=2$$

Step-by-Step Solution

Answer

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

Step 1: 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: $$y = 8, y = 2x+2, x = 0, \text { and } x = 3$$

Step 2: 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: $$h(x) = 8 - (2x + 2)$$ $$r(x) = y = 2x + 2$$

Step 3: 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: $$V = 2 \pi \int_{0}^{3} (2x + 2)(8 - (2x + 2)) dx$$

Step 4: 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: $$V = 2 \pi \left[\left(8(3)^2 - \frac{2}{3}(3)^3\right) - \left(0\right)\right]$$ $$V = 2 \pi \left[72 - 54\right]$$ $$V = 2 \pi (18)$$

Step 5: 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.