Please answer both questions and be detailed, thank you in advance! **Problem 6: Surface of Revolution****Objective:**Find the area of the surface of revolution generated by revolving the graph of \( y = \ln x \), bounded by \( y = 0 \) over the interval \([1, e]\) about the \( y \)-axis.**Instructions:**Set up and simplify the integral. Do not evaluate the integral.**Solution Outline:**1. **Identify the function and bounds:** - Function: \( y = \ln x \) - Bounds: \( x \in [1, e] \) - Revolving around the \( y \)-axis.2. **Set Up the Integral:** - The formula for the surface area \( S \) of a function \( x = g(y) \) revolved around the \( y \)-axis from \( y = c \) to \( y = d \) is: \[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \]3. **Rewriting the function:** - Given \( y = \ln x \), we can express \( x \) as a function of \( y \): \( x = e^y \).4. **Compute the derivative:** - \(\frac{dx}{dy} = \frac{d}{dy}(e^y) = e^y\).5. **Set up the integral with the appropriate bounds:** - Since \( x = e^y \) when \( y \in [0, 1] \) (changing the bounds to match those for \( y \)): \[ S = 2\pi \int_{0}^{1} e^y \sqrt{1 + (e^y)^2} \, dy \]6. **Simplify the integrand:** - The integral becomes: \[ S = 2\pi \int_{0}^{1} e^y \sqrt{1 + e^{2y}} \, dy \]By following these steps, we have set up and simplified the integral for the area of the surface of revolution. The integral is ready for evaluation.**Note:** Since the instruction specifies not to evaluate the integral, we ### Problem 5: Arc Length of a Function**Problem Statement:**Find the arc length of the graph of \( y = \frac{1}{14} x^7 + \frac{1}{10x^5} \) over the interval \([1, 2]\).**Solution Steps:**To find the arc length of a graph \( y = f(x) \) over the interval \([a, b]\), we use the formula:\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]1. **Function and Derivative:** Given the function: \[ y = \frac{1}{14} x^7 + \frac{1}{10x^5} \] First, we need to calculate the derivative \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{14} x^7 + \frac{1}{10x^5} \right) \]2. **Derivative Calculation:** Using the power rule: \[ \frac{dy}{dx} = \frac{1}{14} \cdot 7x^6 - \frac{1}{10} \cdot 5x^{-6} \] \[ \frac{dy}{dx} = \frac{7}{14} x^6 - \frac{5}{10} x^{-6} \] \[ \frac{dy}{dx} = \frac{1}{2} x^6 - \frac{1}{2} x^{-6} \] \[ \frac{dy}{dx} = \frac{1}{2} x^6 - \frac{1}{2x^6} \]3. **Arc Length Integral:** Next, we plug the derivative into the arc length formula: \[ L = \int_{1}^{2} \sqrt{1 + \left( \frac{1}{2} x^6 - \frac{1}{2x^6} \right)^2} \, dx \]4. **Simplified Expression:** We simplify inside the integral: \[ \left( \frac{1}{

Name: Please answer both questions and be detailed, thank you in advance! **
Uploaded: 2024-03-05T11:52:21.000Z
Channel: Bartleby
Description: Please answer both questions and be detailed, thank you in advance! **Problem 6: Surface of Revolution****Objective:**Find the area of the surface of revolution generated by revolving the graph of \( y = \ln x \), bounded by \( y = 0 \) over the inte