# Transform the region and compute the integral

• STEMucator
I'm confused.In summary, the conversation discusses using a transformation to map a region defined by a quadratic equation onto a new plane, and then evaluating a double integral using polar coordinates. The key step is to correctly incorporate the Jacobian of the transformation, which was initially overlooked.
STEMucator
Homework Helper

## Homework Statement

This is an exercise from Taylor & Mann page 468 Q5 :

Use the transformation x=au and y=bv to map the region R defined by $\frac{x^2}{a^2} + \frac{y^2}{b^2} ≤ 1$ onto the uv plane.

Evaluate : $\int \int_R \frac{x^2}{a^2} + \frac{y^2}{b^2} dxdy$

with the aid of this transofrm and polar coordinates.

## Homework Equations

$\int \int_R F(x,y) dxdy = \int \int_{R'} G(u,v)|J| dudv$

Where |J| is the Jacobian.

## The Attempt at a Solution

So if R is defined to be $\frac{x^2}{a^2} + \frac{y^2}{b^2} ≤ 1$, then using the transformation x=au and y=bv we define a new region R' by $u^2 + v^2 ≤ 1$

Now I can easily set up a Cartesian integral in terms of u and v, but the point is to use polars to simplify things.

So let u = rcosθ and v = rsinθ and hence R' becomes $r ≤ 1$ since r>0 for $0 ≤ θ ≤ 2π$

The Jacobian of polars is just r, so J = r.

Using all this information, our integral becomes :

$\int_{0}^{1} \int_{0}^{2π} r^3 dθdr = \pi/2$

I'm getting the feeling I'm missing something here as the answer at the back of the book is πab/2 which sadly my integral almost evaluates to, but not quite.

Any pointers?

Just a slight careless mistake when you went from xy-space to uv-space:
dx = a du
dy = b dv
This returns you the factor of ab that you are missing.

Zondrina said:

## Homework Statement

This is an exercise from Taylor & Mann page 468 Q5 :

Use the transformation x=au and y=bv to map the region R defined by $\frac{x^2}{a^2} + \frac{y^2}{b^2} ≤ 1$ onto the uv plane.

Evaluate : $\int \int_R \frac{x^2}{a^2} + \frac{y^2}{b^2} dxdy$

with the aid of this transofrm and polar coordinates.

## Homework Equations

$\int \int_R F(x,y) dxdy = \int \int_{R'} G(u,v)|J| dudv$

Where |J| is the Jacobian.

## The Attempt at a Solution

So if R is defined to be $\frac{x^2}{a^2} + \frac{y^2}{b^2} ≤ 1$, then using the transformation x=au and y=bv we define a new region R' by $u^2 + v^2 ≤ 1$

Now I can easily set up a Cartesian integral in terms of u and v, but the point is to use polars to simplify things.

So let u = rcosθ and v = rsinθ and hence R' becomes $r ≤ 1$ since r>0 for $0 ≤ θ ≤ 2π$

The Jacobian of polars is just r, so J = r.

Using all this information, our integral becomes :

$\int_{0}^{1} \int_{0}^{2π} r^3 dθdr = \pi/2$

I'm getting the feeling I'm missing something here as the answer at the back of the book is πab/2 which sadly my integral almost evaluates to, but not quite.

Any pointers?

What happened to the Jacobian of the transformation (x,y)->(u,v)?

Fightfish said:
Just a slight careless mistake when you went from xy-space to uv-space:
dx = a du
dy = b dv
This returns you the factor of ab that you are missing.

Ahhhhh thank you good sir :) I got the answer I wanted now.

Dick said:
What happened to the Jacobian of the transformation (x,y)->(u,v)?

The Jacobian of polars is just r so yeah.

## 1. What is the meaning of "transforming the region" when computing an integral?

Transforming the region refers to changing the limits of integration in an integral to simplify its evaluation. This can involve changing the variables or using a change of coordinates to convert the integral into a more manageable form.

## 2. Can any region be transformed when computing an integral?

Yes, any region can be transformed as long as the resulting transformed region is still a valid region for the integral to be evaluated over. This means that the transformed region must still be bounded and have a finite area or volume.

## 3. What is the purpose of transforming the region when computing an integral?

The purpose of transforming the region is to make the integral easier to evaluate. By changing the limits of integration, the integral may become simpler or more familiar, making it easier to solve or integrate.

## 4. Are there any specific techniques for transforming regions in integrals?

Yes, there are several techniques for transforming regions in integrals. Some common techniques include substitution, integration by parts, and partial fraction decomposition. These techniques can be used to simplify the integral and transform the region into a more manageable form.

## 5. Can transforming the region in an integral change the value of the integral?

Yes, transforming the region in an integral can change the value of the integral. This is because the limits of integration are changed, and the new limits may result in a different area or volume being calculated. However, the final answer should still be equivalent to the original integral, just in a different form.

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