Find positive numbers a and b so that the change of variables s = ax, t = by transforms the integral S S dx dy into a = for the region R, the elliptical region x2²/25 + y²/16 ≤ 1 and the region T, the circle s² + t² = 1. b What is (x,y) a(s,t) Ə(x,y) a(s,t) = Ə(x, y) ||, ost) in this case? ds dt
Find positive numbers a and b so that the change of variables s = ax, t = by transforms the integral S S dx dy into a = for the region R, the elliptical region x2²/25 + y²/16 ≤ 1 and the region T, the circle s² + t² = 1. b What is (x,y) a(s,t) Ə(x,y) a(s,t) = Ə(x, y) ||, ost) in this case? ds dt
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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![### Changing Variables in Double Integrals: An Example
**Problem Statement:**
Find positive numbers \(a\) and \(b\) so that the change of variables \(s = ax, t = by\) transforms the integral
\[
\iint_{R} dx \, dy
\]
into
\[
\iint_{T} \left|\frac{\partial(x,y)}{\partial(s,t)}\right| ds \, dt
\]
for the regions \(R\) and \(T\):
- The region \(R\) is the elliptical region defined by \(x^2/25 + y^2/16 \leq 1\).
- The region \(T\) is the circular region defined by \(s^2 + t^2 = 1\).
**To Do:**
1. Determine the values of \(a\) and \(b\).
2. Find the value of the Jacobian determinant \(\left|\frac{\partial(x,y)}{\partial(s,t)}\right|\) for this change of variables.
**Solution Steps:**
1. **Finding \(a\) and \(b\):**
First, we need to transform the ellipse \(x^2/25 + y^2/16 = 1\) into a circle \(s^2 + t^2 = 1\).
\[
s = ax \implies a^2x^2
\]
\[
t = by \implies b^2y^2
\]
So, we have:
\[
\frac{a^2 x^2}{25} + \frac{b^2 y^2}{16} = 1
\]
Since \(x=5\) and \(y=4\) are points on the ellipse that correspond to the points on the unit circle when scaled properly, we get:
\[
a = \frac{1}{5}
\]
\[
b = \frac{1}{4}
\]
2. **Finding the Jacobian determinant:**
The Jacobian determinant \(\left|\frac{\partial(x,y)}{\partial(s,t)}\right|\) is calculated as:
\[
\frac{\partial(x,y)}{\partial(s,t)} = \begin{vmatrix}
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbff2935-77bb-4550-bfd1-d595e6271f30%2F524c5f3d-6844-4835-9d7e-693b674d042c%2Fjai3ld_processed.png&w=3840&q=75)
Transcribed Image Text:### Changing Variables in Double Integrals: An Example
**Problem Statement:**
Find positive numbers \(a\) and \(b\) so that the change of variables \(s = ax, t = by\) transforms the integral
\[
\iint_{R} dx \, dy
\]
into
\[
\iint_{T} \left|\frac{\partial(x,y)}{\partial(s,t)}\right| ds \, dt
\]
for the regions \(R\) and \(T\):
- The region \(R\) is the elliptical region defined by \(x^2/25 + y^2/16 \leq 1\).
- The region \(T\) is the circular region defined by \(s^2 + t^2 = 1\).
**To Do:**
1. Determine the values of \(a\) and \(b\).
2. Find the value of the Jacobian determinant \(\left|\frac{\partial(x,y)}{\partial(s,t)}\right|\) for this change of variables.
**Solution Steps:**
1. **Finding \(a\) and \(b\):**
First, we need to transform the ellipse \(x^2/25 + y^2/16 = 1\) into a circle \(s^2 + t^2 = 1\).
\[
s = ax \implies a^2x^2
\]
\[
t = by \implies b^2y^2
\]
So, we have:
\[
\frac{a^2 x^2}{25} + \frac{b^2 y^2}{16} = 1
\]
Since \(x=5\) and \(y=4\) are points on the ellipse that correspond to the points on the unit circle when scaled properly, we get:
\[
a = \frac{1}{5}
\]
\[
b = \frac{1}{4}
\]
2. **Finding the Jacobian determinant:**
The Jacobian determinant \(\left|\frac{\partial(x,y)}{\partial(s,t)}\right|\) is calculated as:
\[
\frac{\partial(x,y)}{\partial(s,t)} = \begin{vmatrix}
\
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