Find the surface area of the part of the plane z = 3+ 5x + 2y that lies above the rectangle [1,4] × [0,3]. X
Find the surface area of the part of the plane z = 3+ 5x + 2y that lies above the rectangle [1,4] × [0,3]. X
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Problem Statement:**
Find the surface area of the part of the plane \( z = 3 + 5x + 2y \) that lies above the rectangle \([1, 4] \times [0, 3]\).
**Solution Approach:**
To find the area, we need to integrate the differential area over the given region. The rectangular region in the xy-plane has vertices at (1, 0), (1, 3), (4, 0), and (4, 3).
The function for the plane is \( z(x, y) = 3 + 5x + 2y \). The general formula for the surface area of a function \( z = f(x, y) \) over a region \( R \) is given by:
\[
A = \iint_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy
\]
**Partial Derivatives:**
1. \(\frac{\partial z}{\partial x} = 5\)
2. \(\frac{\partial z}{\partial y} = 2\)
**Integrating over the rectangle \([1, 4] \times [0, 3]\):**
\[
A = \int_0^3 \int_1^4 \sqrt{1 + 5^2 + 2^2} \, dx \, dy
\]
Simplifying inside the square root:
\[
1 + 25 + 4 = 30
\]
Thus, the integral becomes:
\[
A = \int_0^3 \int_1^4 \sqrt{30} \, dx \, dy
\]
This simplifies to:
\[
A = \sqrt{30} \times \text{(area of the base rectangle)}
\]
The area of the rectangle is:
\[
(4 - 1) \times (3 - 0) = 3 \times 3 = 9
\]
Therefore, the surface area:
\[
A = 9 \sqrt{30}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07dc3289-c262-41f1-9fc5-a8d18500de1b%2Fdf3a88c5-9c94-4750-8bb6-c7a7da94ee7d%2F7uoye8_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the surface area of the part of the plane \( z = 3 + 5x + 2y \) that lies above the rectangle \([1, 4] \times [0, 3]\).
**Solution Approach:**
To find the area, we need to integrate the differential area over the given region. The rectangular region in the xy-plane has vertices at (1, 0), (1, 3), (4, 0), and (4, 3).
The function for the plane is \( z(x, y) = 3 + 5x + 2y \). The general formula for the surface area of a function \( z = f(x, y) \) over a region \( R \) is given by:
\[
A = \iint_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy
\]
**Partial Derivatives:**
1. \(\frac{\partial z}{\partial x} = 5\)
2. \(\frac{\partial z}{\partial y} = 2\)
**Integrating over the rectangle \([1, 4] \times [0, 3]\):**
\[
A = \int_0^3 \int_1^4 \sqrt{1 + 5^2 + 2^2} \, dx \, dy
\]
Simplifying inside the square root:
\[
1 + 25 + 4 = 30
\]
Thus, the integral becomes:
\[
A = \int_0^3 \int_1^4 \sqrt{30} \, dx \, dy
\]
This simplifies to:
\[
A = \sqrt{30} \times \text{(area of the base rectangle)}
\]
The area of the rectangle is:
\[
(4 - 1) \times (3 - 0) = 3 \times 3 = 9
\]
Therefore, the surface area:
\[
A = 9 \sqrt{30}
\]
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