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### Area of an Ellipse

```
Date: 08/03/2001 at 23:00:19
From: Tate
Subject: Proof

I have been given a problem to be solved via integration.

I have to show that the area enclosed by the ellipse
x^2/a^2 + y^2/b^2 = 1 is pi*a*b, where a is the longer
horizontal part and b is the shorter vertical part of an ellipse.

Can you offer some suggestions?
```

```
Date: 08/05/2001 at 03:36:51
From: Doctor Jeremiah
Subject: Re: Proof

Hi Tate,

This ellipse is centered on (0,0), so all we have to do is solve the
equation for y to get:

x^2/a^2 + y^2/b^2 = 1
b^2*x^2/a^2 + b^2*y^2/b^2 = b^2*1
b^2*x^2/a^2 + y^2 = b^2
b^2*x^2/a^2 + y^2 - b^2*x^2/a^2 = b^2 - b^2*x^2/a^2
y^2 = b^2 - b^2*x^2/a^2
y^2 = b^2/a^2(a^2 - x^2)
sqrt( y^2 ) = sqrt( b^2/a^2(a^2 - x^2) )
y = +- b/a * sqrt(a^2 - x^2)

This is two equations. One has a plus sign and the other has a minus
sign. If we write them as two equations we get:

y = b/a * sqrt(a^2 - x^2)
y = -b/a * sqrt(a^2 - x^2)

The one without the minus sign is the part above the x-axis, and the
one with the minus sign is the part below the x-axis.

To calculate area between the curve and the x-axis we need to
integrate one equation, but to calculate the total area (above and
below the x-axis) we need to integrate both equations and then add the
areas.

But both equations have the same curve and the same area, so we could
just integrate one equation and multiply the area by two.

If we do that we end up with:

/
A = 2 * | -b/a * sqrt(a^2 - x^2) dx
/

Try solving that, and if you get stuck again please write back.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/05/2001 at 10:04:36
From: Tate
Subject: Re: Proof

To solve
/
A = 2 * | -b/a * sqrt(a^2 - x^2) dx
/

I began by pulling out -b/a, as these are constants. The remaining
sqrt(a^2 - x^2) dx looks like a good candidate for trig substitution,
so I let u = x = a sin theta. Then, my du = a Cos theta.

Here is where the train wreck begins.

/
A = -2(b/a) * | sqrt(a^2 -(a^2 sin^2 theta)) a Cos theta d(theta)
/

Where do I go now? If I change the Sin^2 to (1-Cos^2), I run into a
dead end. Same if I use the double angle identity.

My proof is suppose to end up with pi*a*b. How am I going to end up
with pi in my solution with a definate integral? If limits are
introduced at some point, how do they come to be?
```

```
Date: 08/10/2001 at 08:48:59
From: Doctor Jeremiah
Subject: Re: Proof

Hi Tate,

First let me say that when I picked the equation to make the integral
out of, I should have picked the positive one:

y = b/a * sqrt(a^2 - x^2)

I picked the negative one, which means the area will be negative.
Oops. So I am going to change that now. The equation for the area will
be:

y = b/a * sqrt(a^2 - x^2)  <== one above the x-axis

x=a
/
A = 2 * | b/a * sqrt(a^2 - x^2) dx
/
x=-a

Notice that it's a positive integral now.

Your idea of the trig substitution was exactly the right thing to do,
but you need to substitute so that the entire square root disappears.

I think I would solve this integral like this:

The limits on the integral must be -a to a because in the equation
x^2/a^2 + y^2/b^2 = 1 the maximum and minimum values of the ellipse
occur when it touches the x-axis. When it touches the x-axis, the
value of y is 0. When y = 0, then x = a or x = -a depending on the
square root.

x=a
/
A = 2 * | b/a * sqrt(a^2 - x^2) dx
/
x=-a

let  x = a*sin(t)

When we do this our limits of x = a and x = -a become:

a = a*sin(t)   -a = a*sin(t)
1 = sin(t)     -1 = sin(t)
Pi/2 = t       -Pi/2 = t

But more importantly the contents of our square root become:

let  x = a*sin(t)
then x^2 = a^2*sin^2(t)
and  a^2 - x^2 = a^2 - a^2*sin^2(t) = a^2(1-sin^2(t))

Now you probably remember that sin^2(t) + cos^2(t) = 1. That means
that cos^2(t) = 1 - sin^2(t),

so   a^2 - x^2 = a^2(1-sin^2(t)) = a^2*cos^2(t)
and  sqrt(a^2 - x^2) = sqrt( a^2*cos^2(t) ) = a*cos(t)

Now since x = a*cos(t), then dx must be

dx/dt = d/dt( a*sin(t) ) = a*cos(t)*dt

Plugging these into the integral we get:

t=Pi/2
/
A = 2 * | b/a * sqrt(a^2 - x^2) dx
/
t=-Pi/2

t=Pi/2
/
A = 2 * | b/a * a*cos(t) * a*cos(t)*dt
/
t=-Pi/2

t=Pi/2
/
A = 2ba | cos^2(t) dt
/
t=-Pi/2

Now the easiest way to solve the integral of cos^2(t) is to look it up
in a book, but I don't have my book handy. So the integral of cos^2(t)
is:

By parts:

/                               /
| cos^2(t) dt = cos(t)*sin(t) - | sin(t) * -sin(t) dt
/                               /
u = cos(t)      dv = cos(t)*dt
du = -sin(t)*dt   v = sin(t)

/                               /
| cos^2(t) dt = cos(t)*sin(t) + | sin^2(t) dt
/                               /

And since sin^2(t) + cos^2(t) = 1 then sin^2(t) = 1 - cos^2(t)

/                               /
| cos^2(t) dt = cos(t)*sin(t) + | 1 - cos^2(t) dt
/                               /

/                               /      /
| cos^2(t) dt = cos(t)*sin(t) + | dt - | cos^2(t) dt
/                               /      /

consolidate the cos^2(t) integrals:

/                               /
2 | cos^2(t) dt = cos(t)*sin(t) + | dt
/                               /

/
2 | cos^2(t) dt = cos(t)*sin(t) + t
/

/
| cos^2(t) dt = cos(t)*sin(t)/2 + t/2
/

Cool, huh?

So now we can plug this into the original equation:

t=Pi/2                                           t=Pi/2
/                                               |
A = 2ba | cos^2(t) dt = -2ba ( cos(t)*sin(t)/2 + t/2 )  |
/                                               |
t=-Pi/2                                          t=-Pi/2

t=Pi/2
|
A = 2ba ( cos(t)*sin(t)/2 + t/2 )  |
|
t=-Pi/2

A = [ 2ba ( cos(Pi/2)*sin(Pi/2)/2 + (Pi/2)/2 ) ]
- [ 2ba ( cos(-Pi/2)*sin(-Pi/2)/2 + (-Pi/2)/2 ) ]

A = [ 2ba ( (0)*(1)/2 + Pi/4 ) ]
- [ 2ba ( (0)*(-1)/2 - Pi/4 ) ]

A = [ 2ba*(Pi/4) ] - [ 2ba*(-Pi/4) ]

A = [ ba*Pi/2 ] - [ -ba*Pi/2 ]

A = ba*Pi/2 + ba*Pi/2

A = ba*Pi

And there you are!

Let me know if I skimmed over the details too quickly and I will be
happy to elaborate on what I did.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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