Let R be the region defined by the graphs of: y = √(x-1), y = 0, and x = 5 Determine AREA of the region R by using integration with respect to x as well as by integrating with respect to y.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Transcription and Explanation for Educational Context**

---

**Introduction:**
Let \( R \) be the region defined by the graphs of \( y = \sqrt{(x-1)} \), \( y = 0 \), and \( x = 5 \).

**Objective:**
Determine the area of the region \( R \) by using integration with respect to \( x \) as well as by integrating with respect to \( y \).

---

**Area Calculation**

1. **Area by Integrating with Respect to \( x \):**
   \[
   \text{Area} = \int_1^5 \sqrt{x-1} \, dx
   \]
   - Integration steps:
     \[
     = \left[ \frac{(x-1)^{3/2}}{\frac{3}{2}} \right]_1^5
     \]
     \[
     = \left[ \frac{2}{3}(x-1)^{3/2} \right]_1^5
     \]
     \[
     = \frac{2}{3} \left[ (4)^{3/2} - 0 \right]
     \]
     \[
     = \frac{16}{3} \, \text{units}^2
     \]

2. **Area by Integrating with Respect to \( y \):**
   \[
   \text{Area} = \int_0^2 (y^2 - 1) \, dy
   \]
   - Integration steps:
     \[
     = \left[ \frac{y^3}{3} - y \right]_0^2
     \]
     \[
     = \left( \frac{8}{3} - 2 \right) - (0)
     \]
     \[
     A = \frac{2}{3} \, \text{units}^2
     \]

---

**Solid of Revolution:**

Now consider the above region \( R \) is revolved about the line \( x = 6 \).

1. **Determine Volume Using the Shell Method:**

   - Formula and diagram not provided. Proceed with shell method calculations expected in this context.

2. **Determine Volume Using the Disk/Washer Method:**

   \[
   \int_c^d \left[ R(x)^2 - r(y)^
Transcribed Image Text:**Transcription and Explanation for Educational Context** --- **Introduction:** Let \( R \) be the region defined by the graphs of \( y = \sqrt{(x-1)} \), \( y = 0 \), and \( x = 5 \). **Objective:** Determine the area of the region \( R \) by using integration with respect to \( x \) as well as by integrating with respect to \( y \). --- **Area Calculation** 1. **Area by Integrating with Respect to \( x \):** \[ \text{Area} = \int_1^5 \sqrt{x-1} \, dx \] - Integration steps: \[ = \left[ \frac{(x-1)^{3/2}}{\frac{3}{2}} \right]_1^5 \] \[ = \left[ \frac{2}{3}(x-1)^{3/2} \right]_1^5 \] \[ = \frac{2}{3} \left[ (4)^{3/2} - 0 \right] \] \[ = \frac{16}{3} \, \text{units}^2 \] 2. **Area by Integrating with Respect to \( y \):** \[ \text{Area} = \int_0^2 (y^2 - 1) \, dy \] - Integration steps: \[ = \left[ \frac{y^3}{3} - y \right]_0^2 \] \[ = \left( \frac{8}{3} - 2 \right) - (0) \] \[ A = \frac{2}{3} \, \text{units}^2 \] --- **Solid of Revolution:** Now consider the above region \( R \) is revolved about the line \( x = 6 \). 1. **Determine Volume Using the Shell Method:** - Formula and diagram not provided. Proceed with shell method calculations expected in this context. 2. **Determine Volume Using the Disk/Washer Method:** \[ \int_c^d \left[ R(x)^2 - r(y)^
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