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
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**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:

\[
A = 9 \sqrt{30}
\]
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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