Trigonometry (MindTap Course List)
10th Edition
ISBN:9781337278461
Author:Ron Larson
Publisher:Ron Larson
Chapter6: Topics In Analytic Geometry
Section6.6: Parametric Equations
Problem 5ECP: Write parametric equations for a cycloid traced by a point P on a circle of radius a as the circle...
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![**Problem: Calculating the Surface Area from a Rotated Curve**
Find the area of the surface obtained by rotating the curve
\[ y = \sqrt{2x} \]
from \( x = 0 \) to \( x = 5 \) about the x-axis.
**Solution**
The area is \[\boxed{\phantom{0000}} \] square units.
**Diagrams and Graphs Explanation:**
In this problem, we calculate the area of the surface generated by rotating a given curve about the x-axis. The given curve is \( y = \sqrt{2x} \), and the interval for x is from 0 to 5.
To solve this:
1. **Formula for Surface Area of Revolution:**
The formula to find the surface area \( S \) when rotating a function \( y = f(x) \) about the x-axis from \( x = a \) to \( x = b \) is:
\[ S = 2\pi \int_{a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
2. **Compute \( \frac{dy}{dx} \):**
First, we need to find the derivative of \( y = \sqrt{2x} \).
\[ y = \sqrt{2x} \implies \frac{dy}{dx} = \frac{d}{dx} (\sqrt{2x}) = \frac{1}{\sqrt{2x}} \cdot 2 = \frac{1}{\sqrt{2x}} \]
3. **Substitute into the Surface Area Formula:**
Substitute \( y \) and \( \frac{dy}{dx} \) into the surface area formula:
\[ S = 2\pi \int_{0}^{5} (\sqrt{2x}) \sqrt{1 + \left( \frac{1}{\sqrt{2x}} \right)^2} \, dx \]
Simplify inside the integral:
\[ \left( \frac{1}{\sqrt{2x}} \right)^2 = \frac{1}{2x} \]
\[ S = 2\pi \int_{0}^{5} (\sqrt{2x}) \sqrt{1 +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f3d9715-5083-4150-a4a6-02c5c7b14656%2F8264bfb4-6b07-4493-a172-ee085e91f771%2F5bximn_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Calculating the Surface Area from a Rotated Curve**
Find the area of the surface obtained by rotating the curve
\[ y = \sqrt{2x} \]
from \( x = 0 \) to \( x = 5 \) about the x-axis.
**Solution**
The area is \[\boxed{\phantom{0000}} \] square units.
**Diagrams and Graphs Explanation:**
In this problem, we calculate the area of the surface generated by rotating a given curve about the x-axis. The given curve is \( y = \sqrt{2x} \), and the interval for x is from 0 to 5.
To solve this:
1. **Formula for Surface Area of Revolution:**
The formula to find the surface area \( S \) when rotating a function \( y = f(x) \) about the x-axis from \( x = a \) to \( x = b \) is:
\[ S = 2\pi \int_{a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
2. **Compute \( \frac{dy}{dx} \):**
First, we need to find the derivative of \( y = \sqrt{2x} \).
\[ y = \sqrt{2x} \implies \frac{dy}{dx} = \frac{d}{dx} (\sqrt{2x}) = \frac{1}{\sqrt{2x}} \cdot 2 = \frac{1}{\sqrt{2x}} \]
3. **Substitute into the Surface Area Formula:**
Substitute \( y \) and \( \frac{dy}{dx} \) into the surface area formula:
\[ S = 2\pi \int_{0}^{5} (\sqrt{2x}) \sqrt{1 + \left( \frac{1}{\sqrt{2x}} \right)^2} \, dx \]
Simplify inside the integral:
\[ \left( \frac{1}{\sqrt{2x}} \right)^2 = \frac{1}{2x} \]
\[ S = 2\pi \int_{0}^{5} (\sqrt{2x}) \sqrt{1 +
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