Let's solve the problem step by step: ### **Step 1: Understand the Problem** You are given the equation of a circle: \[ (x - 5)^2 + (y - 1)^2 = 4 \] - **Center:** \((5, 1)\) - **Radius:** \(2\) (since \(4 = 2^2\)) You are to **find the volume generated when this area is rotated about the \(x\)-axis**. --- ### **Step 2: Express \(y\) in terms of \(x\)** \[ (x - 5)^2 + (y - 1)^2 = 4 \] \[ (y - 1)^2 = 4 - (x - 5)^2 \] \[ y - 1 = \pm \sqrt{4 - (x - 5)^2} \] \[ y = 1 \pm \sqrt{4 - (x - 5)^2} \] --- ### **Step 3: Find the Limits of \(x\)** The circle exists where the expression under the square root is non-negative: \[ 4 - (x - 5)^2 \geq 0 \implies -2 \leq x - 5 \leq 2 \implies 3 \leq x \leq 7 \] --- ### **Step 4: Set Up the Volume Integral** When rotating around the \(x\)-axis, use the **disk method**: \[ V = \pi \int_{a}^{b} \left[ (y_{\text{top}})^2 - (y_{\text{bottom}})^2 \right] dx \] Here, - \(y_{\text{top}} = 1 + \sqrt{4 - (x - 5)^2}\) - \(y_{\text{bottom}} = 1 - \sqrt{4 - (x - 5)^2}\) - Limits: \(x = 3\) to \(x = 7\) So, \[ V = \pi \int_{3}^{7} \left[ (1 + \sqrt{4 - (x - 5)^2})^2 - (1 - \sqrt{4 - (x - 5)^2})^2 \right] dx \] --- ### **Step 5: Simplify the Integrand** Expand both squares: \[ (1 + a)^2 = 1 + 2a + a^2 \] \[ (1 - a)^2 = 1 - 2a + a^2 \] So, \[ (1 + \sqrt{4 - (x - 5)^2})^2 - (1 - \sqrt{4 - (x - 5)^2})^2 = [1 + 2a + a^2] - [1 - 2a + a^2] = 4a \] where \(a = \sqrt{4 - (x - 5)^2}\). So, \[ V = \pi \int_{3}^{7} 4\sqrt{4 - (x - 5)^2} \, dx \] \[ = 4\pi \int_{3}^{7} \sqrt{4 - (x - 5)^2} \, dx \] --- ### **Step 6: Use Substitution to Integrate** Let \(u = x - 5\), so \(x = u + 5\). - When \(x = 3\), \(u = -2\). - When \(x = 7\), \(u = 2\). So, \[ V = 4\pi \int_{-2}^{2} \sqrt{4 - u^2} \, du \] --- ### **Step 7: Recognize the Integral** \[ \int_{-a}^{a} \sqrt{a^2 - u^2} du \] is the area of a **semicircle** of radius \(a\), so the area is \(\frac{1}{2} \pi a^2\). Here, \(a = 2\): \[ \int_{-2}^{2} \sqrt{4 - u^2} du = \frac{1}{2} \pi (2^2) = 2\pi \] --- ### **Step 8: Find the Final Volume** \[ V = 4\pi \times 2\pi = 8\pi^2 \] --- ## **Final Answer** \[ \boxed{8\pi^2} \] **The volume of revolution generated by rotating the given circle about the \(x\)-axis is \(8\pi^2\) cubic units.**

Certainly! Let's carefully go through each step with detailed calculations to find the volume generated by rotating the given circle about the x-axis. --- ### **Given Data:** - Equation of the circle: \[ (x - 5)^2 + (y - 1)^2 = 4 \] - Center: \((5, 1)\) - Radius: \(r = 2\) --- ### **Step 1: Understanding the problem** We are asked to find the volume of the solid generated when the area enclosed by this circle is revolved about the **x-axis**. --- ### **Step 2: Express \(y\) in terms of \(x\)** From the circle's equation: \[ (y - 1)^2 = 4 - (x - 5)^2 \] \[ y - 1 = \pm \sqrt{4 - (x - 5)^2} \] \[ y = 1 \pm \sqrt{4 - (x - 5)^2} \] - The **top semicircle**: \( y_{top} = 1 + \sqrt{4 - (x - 5)^2} \) - The **bottom semicircle**: \( y_{bottom} = 1 - \sqrt{4 - (x - 5)^2} \) --- ### **Step 3: Limits of \(x\)** The expression under the square root must be non-negative: \[ 4 - (x - 5)^2 \geq 0 \] \[ (x - 5)^2 \leq 4 \] \[ -2 \leq x - 5 \leq 2 \] \[ 3 \leq x \leq 7 \] --- ### **Step 4: Set up the volume integral (disk method)** When rotating around the \(x\)-axis, the volume is obtained by integrating the **area of disks** with radius equal to \(y\): \[ V = \pi \int_{x=3}^{7} \left[ y_{top}^2 - y_{bottom}^2 \right] dx \] Note that the region between the top and bottom curves makes the entire circle's enclosed area when revolved. --- ### **Step 5: Simplify the integrand** Calculate: \[ y_{top}^2 - y_{bottom}^2 \] Using the algebraic identity: \[ (a + b)^2 - (a - b)^2 = 4ab \] Set: \[ a = 1, \quad b = \sqrt{4 - (x - 5)^2} \] Then: \[ y_{top}^2 - y_{bottom}^2 = (1 + b)^2 - (1 - b)^2 = 4 \times 1 \times b = 4b \] Since \(b = \sqrt{4 - (x - 5)^2}\), the integrand simplifies to: \[ 4 \sqrt{4 - (x - 5)^2} \] --- ### **Step 6: Write the volume integral** \[ V = \pi \int_{3}^{7} 4 \sqrt{4 - (x - 5)^2} \, dx \] Factor out constants: \[ V = 4\pi \int_{3}^{7} \sqrt{4 - (x - 5)^2} \, dx \] --- ### **Step 7: Substitution** Let: \[ u = x - 5 \quad \Rightarrow \quad du = dx \] Change limits: - When \(x = 3\), \(u = 3 - 5 = -2\) - When \(x = 7\), \(u = 7 - 5 = 2\) So, \[ V = 4\pi \int_{u=-2}^{2} \sqrt{4 - u^2} \, du \] --- ### **Step 8: Recognize the integral** \[ \int_{-a}^{a} \sqrt{a^2 - u^2} \, du \] This is the **area of a semicircle** of radius \(a\), which is: \[ \text{Area} = \frac{1}{2} \pi a^2 \] In our case, \(a = 2\): \[ \int_{-2}^{2} \sqrt{4 - u^2} \, du = \frac{1}{2} \pi (2)^2 = \frac{1}{2} \pi \times 4 = 2\pi \] --- ### **Step 9: Final calculation** Multiply by the constant outside: \[ V = 4\pi \times 2\pi = 8\pi^2 \] --- ### **### **Final Answer:** \[ \boxed{ \textbf{The volume of the solid generated is } \quad 8\pi^2 } \] This is the exact volume in cubic units. --- **Summary:** - Expressed the circle as functions of \(x\). - Set limits based on the circle's extent. - Used the disk method with the difference of squares to simplify the integrand. - Recognized the integral as the area of a semicircle. - Calculated the integral to get the final volume. Let me know if further clarification is needed!