Problem 12 - 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. $$y=\sqrt{x}, y=0, \text { and } x=4$$

Step-by-Step Solution

Answer

Question: Find the volume of the solid generated when the region R bounded by the curves \(y = \sqrt{x}, y = 0, \text{ and } x = 4,\) is revolved about the y-axis using the shell method. Answer: The volume of the solid generated when the region R is revolved about the y-axis is \(V = \frac{64}{5}\pi.\)

Step 1: Sketch the region R

First, sketch the region R formed by the curves \(y=\sqrt{x}, y=0, \text { and } x=4\). This helps to visualize the solid shape and can clarify the shell method setup. The region R is enclosed by the square-root curve, the x-axis, and the vertical line \(x=4\).

Step 2: Identify the height and radius of a typical shell

Consider a vertical strip in the region R parallel to the y-axis. When this strip is revolved about the y-axis, it creates a cylindrical shell with a radius \(r = x\) and height \(h = y\). The thickness of the shell is \(\Delta x\). We'll take a limit as \(\Delta x\) approaches 0 to obtain the volume integral.

Step 3: Express the height of the shell in terms of x

To find the volume of the solid, we need to express the height of the shell \(y\) in terms of \(x\). Since we are given \(y = \sqrt{x}\), we can use this expression directly.

Step 4: Write the Shell Volume formula

The Shell Volume formula is given by $$V = 2\pi \int_a^b{r \cdot h \, dx}.$$ In this case, the radius \(r = x\), the height \(h = y = \sqrt{x}\), and \(a = 0\) and \(b = 4\) (the x values that enclose the region R). Substituting these expressions into the formula, we have: $$V = 2\pi \int_0^4 {x \cdot \sqrt{x} \, dx}.$$

Step 5: Simplify the integral and solve for V

To evaluate the integral, first simplify the expression in the integral: $$x\cdot\sqrt{x} = x^{\frac{1}{2}}\cdot x=x^{\frac{3}{2}}.$$ Now integrate: $$V = 2\pi \int_0^4 {x^{\frac{3}{2}} \, dx} = 2\pi \left[\frac{2}{5}x^{\frac{5}{2}}\right]_0^4 = 2\pi \left( \frac{2}{5}(4)^{\frac{5}{2}} - \frac{2}{5}(0)^{\frac{5}{2}} \right).$$ Finally, evaluate the expression to find the volume of the solid: $$V = 2 \pi \cdot \frac{2}{5} \cdot 32 =\frac{64}{5}\pi.$$

Step 6: Final Answer

The volume of the solid generated when the region R is revolved about the y-axis is $$V = \frac{64}{5}\pi.$$

Key Concepts

Volume of Revolution

When we talk about the **volume of revolution**, we refer to finding the volume of a three-dimensional solid formed by rotating a two-dimensional shape around an axis. In this exercise, the region bounded by the curves is revolved around the y-axis. This transformation allows us to use geometric calculus techniques to measure the "occupied space" of complex shapes.

One efficient method for calculating these types of volumes is the Shell Method. This approach involves taking a thin vertical strip from the region, rotating it around the specified axis, and forming a cylindrical shell. The volume of each shell is calculated and then integrated (summed) over the entire region.

The method is beneficial when dealing with shapes or curves that are easier to describe in terms of x-values as is the case in our exercise. By revoliving around the vertical (y) axis, the Shell Method simplifies the process significantly, as we observed with the equation transformation methods used to solve the integral. Understanding this concept can pave the way for mastering various volume-related problems in calculus.

Integral Calculus

In this exercise, **integral calculus** is used to summarize the volume of infinitely many infinitesimally thin shells. We achieve this by determining the definite integral of the function representing our region of interest.

The integral employed in this solution calculates the combined volume of the individual shells produced by rotating each slice of the region around the y-axis. For the bounded region defined by the square root function, this integral takes the form:

• Integrate from the lower boundary, where the curve starts (\(x = 0\)), to the upper boundary (\(x = 4\)).
• Use the equation setup: \(\int_0^4 {x \cdot \sqrt{x} \, dx\}\), where \(x\) represents the radius and \(\sqrt{x}\) is the vertical height of each cylindrical shell.

Integral calculus is essential for taking the concept of infinite skin-thin elements and creating a real-world volume of an object within a continuous space. By practicing and solving such continuous calculus problems, students not only develop problem-solving skills but also grasp the conceptual beauty of mathematical modeling.

Region Bounded by Curves

A **region bounded by curves** defines the specific area or shape formed between multiple mathematical functions. In this problem, the region is bounded by three curves: \(y = \sqrt{x}\),\(y = 0\) (the x-axis), and the vertical line\(x = 4\). Understanding how these boundaries enclose a specific area is crucial for setting up integrals or geometric calculations.

To identify this region, we need to plot each curve and understand where they intersect. In our problem:

• \(y = \sqrt{x}\) is a curve that starts at the origin (0,0) and rises to the point (4,2).
• \(y = 0\) represents the x-axis itself, serving as a flat boundary.
• \(x = 4\) is a vertical boundary cutting through \(y = \sqrt{x}\).

Visualizing how these curves interact helps in setting up the problem correctly and ensuring accurate solutions. Understanding bounded regions simplifies many types of calculus problems, as it gives us a concrete frame of reference to apply calculus techniques like integration.