Problem 51 - Single Variable Calculus

Question

The function \(f\) is defined by $$ f=\left\\{\begin{array}{ll} \sqrt{x} & \text { if } 0 \leq x \leq 1 \\ x^{2}-2 x+2 & \text { if } 1

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region under the graph of the given function \(f\) on the interval \([0,2]\) about the x-axis is \(V = \pi\left(\frac{1}{2} + \frac{31}{5} + \frac{56}{3} - 9\right)\).

Step 1: Set up the volume integral for the first part of f

For the first part of f, the function is defined as \(f(x) = \sqrt{x}\) on the interval \([0, 1]\). To find the volume generated from revolving this portion of the graph around the x-axis, we can use the washer method. The washer method has the following formula: \[ V = \pi \int_{a}^{b} (R(x)^2 - r(x)^2)dx \] where V is the volume, \(R(x)\) is the outer radius, \(r(x)\) is the inner radius, and \(a\) and \(b\) are the limits of integration. Here, we have \(R(x) = f(x) = \sqrt{x}\) and \(r(x) = 0\) as there is no inner radius in this case. The limits of integration will be from 0 to 1, so the integral will be: \[ V_1 = \pi \int_{0}^{1} (\sqrt{x})^2 dx \]

Step 2: Evaluate the integral for the first part of f

Now, we simply need to evaluate the integral for the first part of f: \[ V_1 = \pi \int_{0}^{1} x dx \] To evaluate the integral, use the power rule: \[ V_1 = \pi\left[\frac{x^2}{2}\right]_{0}^{1} = \pi\left(\frac{1^2}{2} - \frac{0^2}{2}\right) = \pi\left(\frac{1}{2}\right) = \frac{\pi}{2} \]

Step 3: Set up the volume integral for the second part of f

For the second part of f, the function is defined as \(f(x) = x^2 - 2x + 2\) on the interval \([1, 2]\). To find the volume generated from revolving this portion of the graph around the x-axis, we use the washer method again. This time, we have \(R(x) = f(x) = x^2 - 2x + 2\) and \(r(x) = 0\) as there is no inner radius, and the limits of integration will be from 1 to 2: \[ V_2 = \pi \int_{1}^{2} \left((x^2 - 2x + 2)^2\right)dx \]

Step 4: Evaluate the integral for the second part of f

Now, we need to evaluate the integral for the second part of f: \[ V_2 = \pi \int_{1}^{2} \left(x^4 - 4x^3 + 8x^2 - 4x + 4\right)dx \] Using the power rule for each term, we have: \[ V_2 = \pi\left[\frac{x^5}{5} - x^4 + \frac{8x^3}{3} - 2x^2 + 4x\right]_{1}^{2} \] \[ V_2 = \pi\left(\frac{32}{5} - 16 + \frac{64}{3} - 8 + 8 - \frac{1}{5} + 1 - \frac{8}{3} + 2 - 4\right) \] \[ V_2 = \pi\left(\frac{31}{5} + \frac{56}{3} - 9\right) \]

Step 5: Find the total volume

Finally, we sum the volumes \(V_1\) and \(V_2\) to find the total volume generated by revolving the region under the graph of f around the x-axis: \[ V = V_1 + V_2 = \frac{\pi}{2} + \pi\left(\frac{31}{5} + \frac{56}{3} - 9\right) \] \[ V = \pi\left(\frac{1}{2} + \frac{31}{5} + \frac{56}{3} - 9\right) \] Hence, we have found the volume of the solid generated by revolving the region under the graph of the given function f on the interval \([0,2]\) about the x-axis.

Key Concepts

Integration Techniques

When calculating the volume of solids of revolution, integration techniques become pivotal. These techniques allow us to compute the area under a curve and transform that into three-dimensional volumes. Let's discuss a few essential points to consider:

• Integral Calculus: At its core, integral calculus helps us find the summation of infinitely small quantities, such as curves or slices of a solid. With integration, we break down a complex shape into simpler parts, resulting in a precise measurement of things like area and volume.
• Power Rule: This is a basic yet vital technique when performing integration. It involves increasing the power of a variable by one, and then dividing by this new exponent, such as transforming the integral \( \int x^a dx \) into \( \frac{x^{a+1}}{a+1} \)
• Definite Integrals: These allow us to find exact values for an integral within specified limits, providing a precise measurement of volume for our solid of revolution.

By employing these integration techniques, you can tackle a variety of mathematical problems, including the determination of volumes through methods such as the washer method.

Washer Method

The washer method is a powerful approach for calculating the volume of solids produced by revolving regions around an axis. It works by visualizing the solid as a stack of washers, each contributing a tiny section to the overall volume.

Understanding the Basic Steps

• Visualize the Solid: Imagine the region being revolved, and think about the washers or discs forming the solid. This helps in setting up the integral correctly.
• Find the Radii: Typically, the outer radius \( R(x) \) is the edge of the solid farthest from the axis, while the inner radius \( r(x) \) is the closest edge. For the given scenario, you may not always have an inner radius, simplifying calculations to a single radius.
• Establish the Limits: Determine the limits of integration by analyzing where the solid begins and ends. These limits will help calculate the volume by defining the definite integral.

The washer method formula \( V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \, dx \) captures the process by calculating the sum of the areas of every washer along the axis of rotation.

Definite Integrals

Definite integrals are an essential element in the calculation of volumes, particularly for solids of revolution. Unlike indefinite integrals, which provide a general formula with a constant, definite integrals offer a specific value over a designated interval.

Working with Definite Integrals

• Precision: The key advantage of definite integrals is their precision. They enable the exact calculation of values by applying specific limits.
• Application to Volume: When finding the volume, definite integrals are indispensable. They help sum up numerous small slices or discs to define the solid's entire volume accurately.
• Execution: To evaluate a definite integral, substitute the upper and lower bounds and calculate the difference. By using the power rule, as shown in the exercise solution, we can systematically evaluate definite integrals for solid volumes.

Definite integrals provide the necessary accuracy and concreteness required in comprehensive problem-solving, making them fundamental in calculating revolved solid volumes.