Transform the region and compute the integral

[STEMucator]

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?

 

[Fightfish]

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.

 

[Dick]

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)?

 

[STEMucator]

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.

 

[STEMucator]

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

The Jacobian of polars is just r so yeah.