Question

Let \(f(x)=\left\\{\begin{array}{cl}x & \text { if } 0 \leq x \leq 2 \\ 2 x-2 & \text { if } 2

Short answer

Answer: The total volume of the solid formed is \(192π\) cubic units.

Step-by-step solution

Set up the Disk Method integrals for each function

For the Disk Method, the volume of each slice of the solid is \(π[f(x)]^2dx\) where we're integrating with respect to \(x\). We'll set up an integral for each function on their respective intervals: 1. For \(f(x) = x\) on \([0, 2]\): \(\int_0^2 π(x)^2 dx\) 2. For \(f(x) = 2x-2\) on \((2,5]\): \(\int_2^5 π(2x-2)^2 dx\) 3. For \(f(x) = -2x+18\) on \((5,6]\): \(\int_5^6 π(-2x+18)^2 dx\) Next, evaluate each integral.

Evaluate the integrals

1. \(π\int_0^2 x^2 dx\): Apply the power rule: \(\int x^2 dx = \frac{1}{3}x^3\). Evaluate from 0 to 2: \(π(\frac{1}{3}(2)^3 - \frac{1}{3}(0)^3)=π(\frac{8}{3})\). 2. \(π\int_2^5 (2x-2)^2 dx\): Expanding the square: \((2x-2)^2 = 4x^2 - 8x + 4\). Evaluate the integral: \(π\int_2^5 (4x^2 - 8x + 4) dx\). Apply the power rule: \(\int (4x^2 - 8x + 4) dx = \frac{4}{3}x^3 - 4x^2 + 4x\). Evaluate from 2 to 5: \(π(\frac{4}{3}(5)^3 - 4(5)^2 + 4(5) - (\frac{4}{3}(2)^3 - 4(2)^2 + 4(2))) =π(180 - \frac{32}{3})\). 3. \(π\int_5^6 (-2x+18)^2 dx\): Expanding the square: \((-2x+18)^2 = 4x^2 - 72x + 324\). Evaluate the integral: \(π\int_5^6 (4x^2 - 72x + 324) dx\). Apply the power rule: \(\int (4x^2 - 72x + 324) dx = \frac{4}{3}x^3 - 36x^2 + 324x\). Evaluate from 5 to 6: \(π(\frac{4}{3}(6)^3 - 36(6)^2 + 324(6) - (\frac{4}{3}(5)^3 - 36(5)^2 + 324(5))) =π(324 - 300)\).

Add the volumes of the three intervals

Now, sum up the volumes we found in Step 2: $$π(\frac{8}{3} + 180 - \frac{32}{3} + 324 - 300)=π(\frac{8}{3} -\frac{32}{3} + 204)$$ $$π(-\frac{24}{3} + 204)=π(192)$$ The total volume of the solid formed when the region is revolved around the \(x\)-axis is \(192π\) cubic units.

Key concepts

Volume of Revolution

The volume of revolution is a fascinating concept that allows us to find the volume of a 3D solid formed by rotating a 2D region around an axis. To tackle this problem, we often use the disk method.

The disk method involves slicing the solid into thin disks perpendicular to the axis of revolution. The formula to find the volume of each disk is \(π[f(x)]^2dx\), where \(f(x)\) represents the radius of the disk. By integrating this formula over the desired interval, we calculate the total volume.

In the original problem, the function \(f(x)\) is divided into three intervals, making it necessary to compute separate integrals for each piece of the piecewise function. Calculating these integrals and adding up the results gives us the total volume of the solid formed.

Piecewise Functions

Piecewise functions are defined by different expressions for different intervals of their domain. They allow for flexibility in modeling real-world situations where a single function wouldn't be sufficient.

In this exercise, we see a piecewise function \(f(x)\) given by three distinct expressions across different ranges of \(x\):

• \(f(x) = x\) for \(0 \leq x \leq 2\)
• \(f(x) = 2x-2\) for \(2 < x \leq 5\)
• \(f(x) = -2x+18\) for \(5 < x \leq 6\)

This variety requires careful consideration when using methods like the disk method, as each segment of the function needs its integral over its respective interval. Understanding how to work with piecewise functions is crucial for tackling complex problems like this one.

Definite Integrals

Definite integrals are a key tool in calculus, allowing us to calculate the area under a curve between two points. When applied to the disk method, they help determine the volume of a solid of revolution.

In this task, we evaluated three definite integrals for different parts of the piecewise function. Each integral represents the volume contribution from a specific segment of the graph.

By integrating the square of the function over its interval and applying the fundamental theorem of calculus, we find volumes:

• \(\int_0^2 x^2 dx\)
• \(\int_2^5 (2x-2)^2 dx\)
• \(\int_5^6 (-2x+18)^2 dx\)

The sum of these integrals, multiplied by \π\, gives the total volume of our solid. Practicing definite integrals strengthens your ability to solve problems involving areas and volumes.