Integrating Elliptical Density: A Simplified Approach Using Cross Products

[mantgx]

Homework Statement

∫∫_D√(9x²+4y²) dx dyD is the region: x²/4+y²/9=1

My understanding is that i have to integrate the function of a density to calculate the mass of plate which is ellipse. Problem is i can't and shouldn't be able to integrate this integral at my level, so am i missing some way of simplification?

 

[Orodruin]

Have you considered a change of variables (or two)?

 

[mantgx]

to polar cordinates?

 

[Orodruin]

Try getting the coordinates to a form where making the change to polar coordinates make more sense first. Hint: An ellipse does not give the simplest boundary condition in polar coordinates.

 

[STEMucator]

Why not let $x = 2u$ and $y = 3v$.

Then switch to polar.

 

[mantgx]

Why not let $x = 2u$ and $y = 3v$.

Then switch to polar.

i do not understand how this will help.

can please someone solve the problem for me I am desperate

 

[Orodruin]

Why don't you try doing it? It will be much more instructive than if we solve it for you.

 

[STEMucator]

i do not understand how this will help.

can please someone solve the problem for me I am desperate

There's a theorem you should be familiar with, it should be along the lines of:

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

Where $|J|$ is the Jacobian matrix of the transformation, $R$ and $R'$ are the regions. You have an integral, which looks like the left side of the above theorem. The theorem tells you that you can change the region and co-ordinates of a double integral, and get the same answer.

So making the substitution I noted in an earlier post, what happens to the region $D$? It gets mapped to some region $D'$ through an invertible map. What is the result of plugging the substitution into the integral on the right?

 

[mantgx]

There's a theorem you should be familiar with, it should be along the lines of:

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

Where $|J|$ is the Jacobian matrix of the transformation, $R$ and $R'$ are the regions. You have an integral, which looks like the left side of the above theorem. The theorem tells you that you can change the region and co-ordinates of a double integral, and get the same answer.

So making the substitution I noted in an earlier post, what happens to the region $D$? It gets mapped to some region $D'$ through an invertible map. What is the result of plugging the substitution into the integral on the right?

∫∫_R'√(36u^2+12v^2) J du dv

?

 

[STEMucator]

∫∫_R'√(36u^2+12v^2) J du dv

?

No, the integral becomes $36 \int \int_{D'} \sqrt{u^2 + v^2} |J| dudv$. Where of course $dx = 2du$ and $dy = 3dv$.

The region $D → D' := u^2 + v^2 = 1$. Notice this is the equation of a circle with radius 1.

The Jacobian is the matrix of partial derivatives of the transformation. Can you compute it?

For reference: http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

 

[Orodruin]

What is $4\times 3^2$ and what is the expression for $J$?

If you are not familiar with jacobians, try the following instead: what is $du$ if $u = x/2$? (It is ok to use this type of reasoning as long as you are making parameter transformations of the type considered here, i.e., u = f(x) and v = g(y))

 

[Orodruin]

No, the integral becomes $36 \int \int_{D'} \sqrt{u^2 + v^2} |J| dudv$.

$\sqrt{36} = 6$ ...

 

[mantgx]

ok, i understand a little now.

J=6 ?

 

[mantgx]

is the final solution 36pi?

 

[Orodruin]

Yes, J=6 (or rather |J|=6) in the first variable change.

What is the area element $du\, dv$ expressed in polar coordinates?

 

[mantgx]

24pi

 

[STEMucator]

$\sqrt{36} = 6$ ...

Don't forget $dx$ and $dy$.

 

[Orodruin]

Don't forget $dx$ and $dy$.

I would have agreed if you did not still have the jacobian in your expression.

 

[Orodruin]

24pi

What are your integration limits in polar coordinates? (The answer is the same as I get by doing it in my head, which I do not trust at 1am, but I just want to make sure you've got it right)

 

[mantgx]

r=(0,1)
fi=(0,2pi)

 

[Orodruin]

:thumbs:

I am assuming you got 36π by forgetting the r in the area element? (du dv = r dr dθ, i.e., |J|=r for the change to polar coordinates)

 

[STEMucator]

I would have agreed if you did not still have the jacobian in your expression.

Whoops, my bad. Didn't notice that while typing.

 

[mantgx]

your assumption is correct

thank you guys 100x times

 

[Orodruin]

Whoops, my bad. Didn't notice that while typing.

Copy and Paste, the source of 98% of all typos in published papers - also 73.4% of all statistics you will read are made up on the spot ;)

 

[ehild]

What is the shape of the region bounded by the line $\frac{x^2}{4}+\frac{y^2}{9}=1$ ?

You can rewrite the square root in the integrand as $\sqrt{9x^2+4y^2}=6\sqrt{x^2/4+y^2/9}=6$, so you have the integral $\int _D{dxdy}$ which is the area of the shape bounded by the line $\frac{x^2}{2^2}+\frac{y^2}{9}=1$.

ehild

 

[Orodruin]

What is the shape of the region bounded by the line $\frac{x^2}{4}+\frac{y^2}{9}=1$ ?

You can rewrite the square root in the integrand as $\sqrt{9x^2+4y^2}=6\sqrt{x^2/4+y^2/9}=6$, so you have the integral $\int _D{dxdy}$ which is the area of the shape bounded by the line $\frac{x^2}{2^2}+\frac{y^2}{9}=1$.

ehild

The integrand is not constant in the area. There is an obvious typo in the formulation where D is defined as being $\frac{x^2}{2^2}+\frac{y^2}{9}=1$. It should read $\frac{x^2}{2^2}+\frac{y^2}{9}\leq 1$. Otherwise D would have measure 0, which would result in 0.

The problem was already solved by OP: you make the substitution
$$
x = 2 r \cos\theta,\quad y=3 r\sin\theta
$$
which brings the entire problem to an integral of $36r$ over the unit disk.

 

[HallsofIvy]

Another way of looking at the Jacobian: In analogy with polar coordinates an "obvious" substitution or parameterization is x= 2r cos(\theta), y= 3r sin(\theta). We can represent the surface by the "position vector" \vec{v}= 2r cos(\theta)\vec{i}+ 3r sin(\theta)\vec{j}. The derivative vectors, \vec{v}_r= 2 cos(\theta)\vec{i}- 3 sin(\theta)\vec{j} and \vec{v}_\theta= -2r sin(\theta)\vec{i}+ 3r cos(\theta)\vec{k} are tangent to the surface and contain metric information in their directions. Their cross product, 6r \vec{k}, is perpendicular to the surface and gives the differential of area: dxdy= 6r drd\theta.

The integrand is \sqrt{9x^2+ 4y^2}= \sqrt{9(4r^2 cos^2(\theta))+ 4(9r^2 sin^2(\theta))}= 6r
So that the integral is \int_{r= 0}^1\int_{\theta= 0}^{2\pi} (6r)(6r drd\theta)= 36\int_{r=0}^1\int_{\theta= 0}^{2\pi} r^2 drd\theta

 

[Orodruin]

I would agree but for the introduction of the 3D cross product to solve a 2D problem, since the method is much more general, applies to arbitrary dimensions, and there is no need to reference a 3D space. Instead I would reference the determinant as the area (or more generally, the n-volume, of the parallelogram spanned by the columns/rows), which for two vectors in 3D just so happens to be the magnitude of the cross product.