Deriving the formula - Area of Ellipse (1 Viewer)

nofate · Jun 20, 2012

Could anyone help where I am going wrong:

nightweaver066 · Jun 20, 2012

What you found was the area bound between the TOP half of the ellipse and the x-axis.

To get the whole area, as the ellipse is symmetrical about the x-axis, the area of the other side would also be (pi)ab/2 so the total area is (pi)ab.

jnney · Jun 21, 2012

A much faster way of doing this is by finding a linear transformation/mapping that maps $f: E \longrightarrow C$ where E is the image of the circle C. We define our linear mapping to be g(u,v) ~ (au,bv) which is equivalent to the matrix multiplication:

$\begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix}\begin{bmatrix} u \\ v\end{bmatrix}$

We make this choice because we notice that if we sub those values into the equation of the ellipse, the a and b cancels out, leaving the equation of the unit circle but in terms of the variables u and v.

We now use the Jacobian Determinant to compute the volume between a domain D and a function f:

$\int_{g(D)} f(x) dx = \int_D f(g(y)) |\det(J_g (y))|dy$

And since we want to find an area (1 Dimensional), it is the equivalent of finding the volume of the solid with an elliptic cross section and a thickness of 1 unit. Very similarly to how you would make all the right entries of the 3x3 matrix become 1 to find the area of a triangle defined by 3 coordinates. But the mapping we have is an affine map, so we have the Jacobian determinant being a constant value, which we will let be equal to say K for some constant K.

Also setting f(y)=1 as explained above and the fact that the affine map has a constant Jacobian, the integral simplifies to become:

$\\ \begin{align*} A_E &= \int \int_{G(E)} dx dy \\\\ &= \det \begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix} \int \int_D du dv \\\\ &= ab \int \int_D du dv \\\\ &= ab \int_{0}^{R} \int_{0}^{2 \pi} r dr d \phi \\\\ &= ab \frac{1}{2} \int_{0}^{2 \pi} R^2 d \phi \\\\ &= ab \pi R^2 \end{align*}$

Where we used Fubini's Theorem towards the end.

So setting R=1, we immediately acquire the area of E, which is $ab \pi$

nightweaver066 · Jun 21, 2012

Hi Carrot.

Parvee · Jun 21, 2012

jnney said:
A much faster way of doing this is by finding a linear transformation/mapping that maps $f: E \longrightarrow C$ where E is the image of the circle C. We define our linear mapping to be g(u,v) ~ (au,bv) which is equivalent to the matrix multiplication:

$\begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix}\begin{bmatrix} u \\ v\end{bmatrix}$

We make this choice because we notice that if we sub those values into the equation of the ellipse, the a and b cancels out, leaving the equation of the unit circle but in terms of the variables u and v.

We now use the Jacobian Determinant to compute the volume between a domain D and a function f:

$\int_{g(D)} f(x) dx = \int_D f(g(y)) |\det(J_g (y))|dy$

And since we want to find an area (1 Dimensional), it is the equivalent of finding the volume of the solid with an elliptic cross section and a thickness of 1 unit. Very similarly to how you would make all the right entries of the 3x3 matrix become 1 to find the area of a triangle defined by 3 coordinates. But the mapping we have is an affine map, so we have the Jacobian determinant being a constant value, which we will let be equal to say K for some constant K.

Also setting f(y)=1 as explained above and the fact that the affine map has a constant Jacobian, the integral simplifies to become:

$\\ \begin{align*} A_E &= \int \int_{G(E)} dx dy \\\\ &= \det \begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix} \int \int_D du dv \\\\ &= ab \int \int_D du dv \\\\ &= ab \int_{0}^{R} \int_{0}^{2 \pi} r dr d \phi \\\\ &= ab \frac{1}{2} \int_{0}^{2 \pi} R^2 d \phi \\\\ &= ab \pi R^2 \end{align*}$

Where we used Fubini's Theorem towards the end.

So setting R=1, we immediately acquire the area of E, which is $ab \pi$

I just had the biggest mind fuck with the accounts -.-

Carrotsticks · Jun 21, 2012

nightweaver066 said:
Hi Carrot.

Wasn't me =)

And jnney you said area is 1 Dimensional. I think you meant 2 Dimensional (unless we exist in the 3rd dimension and view the ellipse from the side!)

seanieg89 · Jun 21, 2012

Green's theorem provides another fast way of finding this area.

Carrotsticks · Jun 21, 2012

seanieg89 said:
Green's theorem provides another fast way of finding this area.

$A = \frac{1}{2} \oint_{\partial D} xdy - ydx= \frac{1}{2} \int_{0}^{2 \pi} (ab \sin^2 t + ab \cos^2 t) dt = \frac{ab}{2} \int_{0}^{2 \pi} dt = ab \pi$

nightweaver066 · Jun 21, 2012

Carrotsticks said:
Wasn't me =)

And jnney you said area is 1 Dimensional. I think you meant 2 Dimensional (unless we exist in the 3rd dimension and view the ellipse from the side!)

I am so confused.

Carrotsticks · Jun 21, 2012

nightweaver066 said:
I am so confused.

A sheet of paper is 2 dimension (actually 3D but with tiny thickness but pretend its 2D).

If I look at it from the side, I don't see a sheet of paper. All I see is a straight line, so it is 1 Dimensional from that POV.

nightweaver066 · Jun 21, 2012

Carrotsticks said:
A sheet of paper is 2 dimension (actually 3D but with tiny thickness but pretend its 2D).

If I look at it from the side, I don't see a sheet of paper. All I see is a straight line, so it is 1 Dimensional from that POV.

lol sorry i didn't mean confused as in with the maths, but with jnney.

Fus Ro Dah · Jun 21, 2012

seanieg89 said:
Green's theorem provides another fast way of finding this area.

Do you know any faster ways? Green's Theorem is the fastest that I know so far.

Carrotsticks said:
$A = \frac{1}{2} \oint_{\partial D} xdy - ydx= \frac{1}{2} \int_{0}^{2 \pi} (ab \sin^2 t + ab \cos^2 t) dt = \frac{ab}{2} \int_{0}^{2 \pi} dt = ab \pi$

For those confused, he used the parametric definition of the ellipse x=acos@ and y=bsin@.

Alkenes · Jun 21, 2012

http://math.ucsd.edu/~wgarner/math10b/area_ellipse.htm
Might help

seanieg89 · Jun 21, 2012

Not really, I cannot imagine something being much faster. I suppose if you take as assumed the formula for the area of a circle, your linear transformation method is just as quick. We don't even need to do any integration as the Jacobian determinant is obviously constant. Just multiply the area of a circle of radius 1 by the determinant of diag(a,b).

Fus Ro Dah · Jun 22, 2012

To be honest nofate's way is still pretty fast I guess because we can just use the area of a circle formula in place of the trig substitution ^^

COLDBOY · Jul 2, 2012

nofate said:
Could anyone help where I am going wrong:

View attachment 25670

Equation of Ellipse: x^2/a^2 + y^2/b^2 = 1

y = sqrt(a^2 - x^2)*(b/a) ( by rearranging making y the subject)

now area underneath ellipse is the integral from -a to a ( where a and -a are the endpoints of the ellipse, which lie on the x-axis)

Area of Ellipse = 2*Int(ydx) = 2*Int(sqrt(a^2 - x^2)*(b/a))dx from -a to a ( i.e 2*Int() because you have the top half of the ellipse and bottom half)

= 2b/a * Int((a^2-x^2)dx) since b/a is a constant we can take this out

Now Int((a^2 -x^2)*dx) from -a to a = pi*a^2/2 (reason is below)

consider the equation of a circle with radius a : x^2 + y^2 = a^2

y = sqrt(a^2 - x^2) (rearrange for y)

now to find the area above the y-axis = Int(sqrt(a^2 - x^2)), i.e. area for 1/2 a circle (semi-circle) = 1/2 * pi * radius^2 = 1/2*pi*a^2

therefore back to the ellipse

Area of Ellipse

= 2b/a * Int((a^2-x^2)dx)

= 2b/a *(1/2)*pi*a^2

= b/a*pi*a^2

= pi*a*b (Q.E.D)

just remember int(sqrt(a^2 - x^2)), is just the area of a semi-circle 0.5*pi*r^2.

You could also find this integral using substitution let x = asin(theta) && dx = acos(theta)d(theta).

But once again, its your choice when i did my 4 unit exam, if i got a question to integrate something which had sqrt(a^2-x^2) from -a to a or 0 to a. I just thought of the area of a semi-circle or a quarter of a circle respectively. Its a pretty quick way to get around things. Especially in the Volumes topic, you have say an ellipse rotating about a vertical axis, (x = p, p is a real number), you get a donut shape. But the point is during the tedious working you would probably approach this integral and instead of having to work it out, you will be like ahhhh, its just 1/2 * pi*r^2 in your mind. Markers probably prefer this method, shows that you understand what you are doing and how to draw relationships from various topics in mathematics.

AbsoluteValue · Jul 2, 2012

Can you say the ellipse is a circle with longer/shorter radii so area is Pi(a)(b) if it was part of a volume question, i.e. the question doesn't explicitly ask you to prove it or do you have to do through the integration all the time?

Carrotsticks · Jul 2, 2012

PORNOGRAPHY said:
Can you say the ellipse is a circle with longer/shorter radii so area is Pi(a)(b) if it was part of a volume question, i.e. the question doesn't explicitly ask you to prove it or do you have to do through the integration all the time?

No, you can't do that.

AbsoluteValue · Jul 2, 2012

Carrotsticks said:
No, you can't do that.

Carrotsticks, try the harder 3 unit question I posted recently please, I'm interested to see what u can come up with

httton · Jul 6, 2012

lol the first solution was alright

you just forgot the square root had a positive and negative case

if you took the negative case as well you'd have teh same intergrand but in negative which in terms of area is exactly the same.

Deriving the formula - Area of Ellipse (1 Viewer)

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