Problem 15 - 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. $$y=\sqrt{x}, y=0, \text { and } x=4$$

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(32\pi\) cubic units.

Step 1: Sketch the region

Draw a coordinate plane and plot the given curves: \(y = \sqrt{x}, y = 0\), and \(x = 4\). The region \(R\) is bounded by these curves. Mark it on the coordinate plane.

Step 2: Identify shell elements

We are revolving the region around the x-axis, so the shell's height will be parallel to the y-axis, and its circumference will be at a distance of y from the x-axis. The circumference is given by \(C = 2\pi y\), and the height is the difference between the rightmost curve and the leftmost curve, which is \(h = 4 - (\sqrt{x})\).

Step 3: Set up the integral for the volume

Now, we will find the volume of each shell generated in the region by multiplying the shell's height, circumference, and thickness. The thickness of a shell is given by \(dx\). Hence, the volume is as follows: $$V = \int_a^b 2\pi y h dx$$ where \(a\) and \(b\) are the starting and ending values of \(x\) along the region.

Step 4: Substitute values and set up limits for the integral

For our region, \(a = 0\) and \(b = 4\). The height \(h = 4 - (\sqrt{x})\) and the circumference \(C = 2\pi y\). However, we need to rewrite \(y\) in terms of \(x\). Since \(y = \sqrt{x}\), we get, $$V = \int_0^4 2\pi (\sqrt{x})(4 - (\sqrt{x})) dx$$

Step 5: Evaluate the integral

To evaluate the integral, we first need to simplify the integrand: $$V = \int_0^4 2\pi (4\sqrt{x} - x) dx$$ Now we can evaluate it: $$V = 2\pi \left[ 2\left(\frac{2}{3} x^{\frac{3}{2}}\right) - \frac{1}{2}x^2 \right]_0^4$$

Step 6: Calculate the volume

Substitute the limits into our expression and simplify: $$V = 2\pi \left[ 2\left(\frac{2}{3} (4)^{\frac{3}{2}}\right) - \frac{1}{2}(4)^2 \right] - 2\pi \left[ 2\left(\frac{2}{3} (0)^{\frac{3}{2}}\right) - \frac{1}{2}(0)^2 \right]$$ $$V = 2\pi \left[32 - 16 \right]$$ $$V = 16\cdot 2\pi$$ $$V=32\pi$$ So, the volume of the solid generated when region \(R\) is revolved about the x-axis is \(32\pi\) cubic units.

Key Concepts

Volume of Solid of Revolution

The concept of the Volume of Solid of Revolution plays a crucial role in understanding how to compute volumes for objects created by revolving a region around an axis. When you get a 2D region and spin it around an axis, a 3D solid is formed. To find the volume of this solid, we use integral calculus techniques.
For the Shell Method, consider each thin cylindrical "shell" that makes up your solid. Here’s how it’s done:

• Each shell has a small thickness, denoted as \(dx\) or \(dy\) depending on the axis of revolution.
• The volume of a thin cylindrical shell can be calculated by multiplying its circumference, height, and thickness.
• Integrating these infinitesimal volumes over the boundaries of the region will give you the total volume of the solid.

The advantage of the Shell Method is its efficiency in situations where the Disk or Washer Method may require more complex calculations. It works beautifully when the height of the element (or shell) is expressed in terms of the same variable as that of the thickness.

Definite Integral

A Definite Integral allows us to calculate the exact accumulation of quantities, such as area under curves or volumes. In calculus, when you take a function and integrate it over a specific interval \(a, b\), you find what is known as the definite integral.
This integral tells us the net area between the x-axis and the curve represented by the function over that interval:
\[\int_a^b f(x) \, dx\]
In the context of the Shell Method, a definite integral helps sum up all the tiny volumes of shells to get the total volume over the interval specified.

• The limits of integration \(a\) and \(b\) represent the points between which you are integrating.
• Each infinitesimally small shell volume is integrated to consider the entire region.

This matching up between integrating a function over a variable interval and accumulating volumes or areas directly corresponds to real-world measures.

Revolving Around the X-axis

Revolving Around the X-axis creates a scenario where a 2D area continuum transforms into a 3D solid span. Revolving a shape around this horizontal line follows distinct movements notable in the Shell Method.
When using the Shell Method to revolve around the x-axis, here's how the process works:

• The shell’s height becomes parallel to the y-axis, expressed as \(h = f(x)\) in many cases, varying across x-values.
• Typically, the distance from the x-axis determines the radius of each shell, affecting the overall shape of the resultant 3D object.
• The circumference of each shell is influenced by this radius, specifically the current \(y\) value in computation.

This approach not only uncovers new shaping dimensions but efficiently addresses situations where alternate methods would involve cumbersome transitions or calculations. When revolving around the x-axis, positioned shell properties intertwine, leading toward distinct volume comprehension.