In order to find the area of an ellipse, we may make use of the idea of transformations and our knowledge of the area of a circle. For example, consider R, the region bounded by the (x+3)² (y − 2)² 16 ellipse 9 √ [₁₁A= √ √ 8 will make the tranformation, (x, y) d(u, ‚v) S dA region S is bounded by u²+ v² This means, we must let U= V= X = + = = 1. For easy calculation, we leading to a Jacobian of -dudu, where the transformed which should be easily inverted to obtain = = 1, d(x, y) 0(u, v) and and We calculate the area of the ellipse by multiplying the area of the unit circle and the Jacobian, arriving at (give an exact answer)
In order to find the area of an ellipse, we may make use of the idea of transformations and our knowledge of the area of a circle. For example, consider R, the region bounded by the (x+3)² (y − 2)² 16 ellipse 9 √ [₁₁A= √ √ 8 will make the tranformation, (x, y) d(u, ‚v) S dA region S is bounded by u²+ v² This means, we must let U= V= X = + = = 1. For easy calculation, we leading to a Jacobian of -dudu, where the transformed which should be easily inverted to obtain = = 1, d(x, y) 0(u, v) and and We calculate the area of the ellipse by multiplying the area of the unit circle and the Jacobian, arriving at (give an exact answer)
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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3.8 3
![**Finding the Area of an Ellipse Using Transformations**
To find the area of an ellipse, we can utilize the concept of transformations and our understanding of the area of a circle. Consider the region \( R \), bounded by the ellipse:
\[
\frac{(x+3)^2}{9} + \frac{(y-2)^2}{16} = 1
\]
For simplicity in calculations, we will apply the transformation:
\[
\iint_R dA = \iint_S \frac{\partial(x, y)}{\partial(u, v)} dudv
\]
Here, the transformed region \( S \) is bounded by:
\[
u^2 + v^2 = 1
\]
**Transformation Details:**
This implies we must let:
\[
u = \Box \quad \text{and} \quad v = \Box
\]
These should be easily inverted to obtain:
\[
x = \Box \quad \text{and} \quad y = \Box
\]
This leads to a Jacobian:
\[
\frac{\partial(x, y)}{\partial(u, v)} = \Box
\]
**Calculating the Ellipse Area:**
To determine the area of the ellipse, multiply the area of the unit circle by the Jacobian, resulting in:
\[
\text{Area} = \Box
\]
Please fill in the boxes with the appropriate transformation equations and solve for the exact area of the ellipse.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4e39a1d0-143c-41f0-ae59-e967e0535bad%2F6b54ff54-ff3b-49cd-ad75-ef0b5cd506a7%2Faqfjiq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding the Area of an Ellipse Using Transformations**
To find the area of an ellipse, we can utilize the concept of transformations and our understanding of the area of a circle. Consider the region \( R \), bounded by the ellipse:
\[
\frac{(x+3)^2}{9} + \frac{(y-2)^2}{16} = 1
\]
For simplicity in calculations, we will apply the transformation:
\[
\iint_R dA = \iint_S \frac{\partial(x, y)}{\partial(u, v)} dudv
\]
Here, the transformed region \( S \) is bounded by:
\[
u^2 + v^2 = 1
\]
**Transformation Details:**
This implies we must let:
\[
u = \Box \quad \text{and} \quad v = \Box
\]
These should be easily inverted to obtain:
\[
x = \Box \quad \text{and} \quad y = \Box
\]
This leads to a Jacobian:
\[
\frac{\partial(x, y)}{\partial(u, v)} = \Box
\]
**Calculating the Ellipse Area:**
To determine the area of the ellipse, multiply the area of the unit circle by the Jacobian, resulting in:
\[
\text{Area} = \Box
\]
Please fill in the boxes with the appropriate transformation equations and solve for the exact area of the ellipse.
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