Problem 8 - 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 \(y\) -axis. $$y=6-x, y=0, x=2, \text { and } x=4$$

Step-by-Step Solution

Answer

Question: Find the volume of the solid generated when the region R bounded by the curves \(y=6-x, y=0, x=2,\) and \(x=4\) is revolved about the \(y\)-axis using the shell method. Answer: The volume of the solid generated is \(\frac{112\pi}{3}\).

Step 1: Sketch the curves and the region R

Sketch the given curves \(y=6-x, y=0, x=2,\) and \(x=4\) and identify the region R, which is the region bounded by these curves. This will help us visualize the problem and determine the width and height of the shells.

Step 2: Determine the limits of integration, \(a\) and \(b\)

In this exercise, the limits of integration represent the \(x\) values, at which the shells start and end. Since the region R is bounded by the vertical lines \(x=2\) and \(x=4\), we can set our limits of integration as \(a=2\) and \(b=4\).

Step 3: Find the height function \(h(x)\)

For the shell method, we need to determine the height function, \(h(x)\), which represents the height of the shells at any given \(x\) value between \(2\) and \(4\). In this case, the height of the shell is simply the distance between the curve \(y=6-x\) and the \(x\)-axis. So, we have $$h(x) = (6-x) - 0 = 6-x$$.

Step 4: Apply the shell method formula

Now that we have our limits of integration, \(a=2\) and \(b=4\), and our height function, \(h(x) = 6-x\), we can apply the shell method formula to find the volume of the solid generated when region R is revolved about the \(y\)-axis: $$V = 2\pi\int_a^b x\cdot h(x)\,dx = 2\pi\int_2^4 x(6-x)\,dx$$

Step 5: Evaluate the integral

Evaluate the integral to find the volume of the solid generated: $$V = 2\pi\int_2^4 x(6-x)\,dx = 2\pi\int_2^4 (6x - x^2)\,dx = 2\pi\left[\frac{6x^2}{2} - \frac{x^3}{3}\right]_2^4$$ $$V = 2\pi\left[\left(12 - \frac{64}{3}\right) - \left(6 - \frac{8}{3}\right)\right] = 2\pi\left(\frac{56}{3}\right)$$

Step 6: Calculate the volume

Calculate the volume of the solid generated when region R is revolved about the \(y\)-axis: $$V = 2\pi\left(\frac{56}{3}\right) = \boxed{\frac{112\pi}{3}}$$