12.6- Surface Area 1. Find the area of the part of the surface z = 1+ 3x + 2y²that lies above the triangle with vertices (0,0), (0,1), and (2,1).
12.6- Surface Area 1. Find the area of the part of the surface z = 1+ 3x + 2y²that lies above the triangle with vertices (0,0), (0,1), and (2,1).
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### 12.6 – Surface Area
1. **Problem Statement:**
Find the area of the part of the surface defined by \( z = 1 + 3x + 2y^2 \) that lies above the triangle with vertices \((0,0)\), \((0,1)\), and \((2,1)\).
### Explanation:
To solve this problem, you will need to set up a surface integral to find the area of the surface above the triangular region. The surface is described by the equation \( z = 1 + 3x + 2y^2 \), and you are tasked with finding the surface area above the specified triangular region in the xy-plane.
#### Steps to Approach:
1. **Identify the Region \( D \):**
- The region \( D \) in the xy-plane is defined by the triangle with vertices \((0,0)\), \((0,1)\), and \((2,1)\).
- The boundaries of \( D \) can be described using the lines connecting these points.
2. **Set Up the Surface Integral:**
- Use the formula for the surface area \( A \) of a surface \( z = f(x,y) \):
\[
A = \iint_{D} \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2} \, dA
\]
- Calculate the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
3. **Substitute in the Function:**
- For \( f(x, y) = 1 + 3x + 2y^2 \):
\[
\frac{\partial f}{\partial x} = 3
\]
\[
\frac{\partial f}{\partial y} = 4y
\]
- Therefore, the integral becomes:
\[
A = \iint_{D} \sqrt{1 + 3^2 + (4y)^2} \, dA
\]
\[
= \iint_{D} \sqrt{1 + 9 + 16y^2} \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ed410c2-9d2c-4a3f-9ddd-880f6763e1c0%2F3645e5a1-ad05-4d04-a584-b329024f1b3a%2Fx9gr735_processed.png&w=3840&q=75)
Transcribed Image Text:### 12.6 – Surface Area
1. **Problem Statement:**
Find the area of the part of the surface defined by \( z = 1 + 3x + 2y^2 \) that lies above the triangle with vertices \((0,0)\), \((0,1)\), and \((2,1)\).
### Explanation:
To solve this problem, you will need to set up a surface integral to find the area of the surface above the triangular region. The surface is described by the equation \( z = 1 + 3x + 2y^2 \), and you are tasked with finding the surface area above the specified triangular region in the xy-plane.
#### Steps to Approach:
1. **Identify the Region \( D \):**
- The region \( D \) in the xy-plane is defined by the triangle with vertices \((0,0)\), \((0,1)\), and \((2,1)\).
- The boundaries of \( D \) can be described using the lines connecting these points.
2. **Set Up the Surface Integral:**
- Use the formula for the surface area \( A \) of a surface \( z = f(x,y) \):
\[
A = \iint_{D} \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2} \, dA
\]
- Calculate the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
3. **Substitute in the Function:**
- For \( f(x, y) = 1 + 3x + 2y^2 \):
\[
\frac{\partial f}{\partial x} = 3
\]
\[
\frac{\partial f}{\partial y} = 4y
\]
- Therefore, the integral becomes:
\[
A = \iint_{D} \sqrt{1 + 3^2 + (4y)^2} \, dA
\]
\[
= \iint_{D} \sqrt{1 + 9 + 16y^2} \,
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