Elliptic Area Using Integral

[knowLittle]

Homework Statement

(x+y)^2 + (y-2)^2 =4

2. Homework Equations
y^2 = (2-x)y - (x^2)/2

Equation of an ellipse:
$\left( \dfrac {x-h} {a}\right) ^{2}+\left( \dfrac {y-g} {b}\right) ^{2}=1$
From this, we know that (h,g) is the center of the ellipse.
and the radius along the x and y-axis is a and b.

3. The Attempt at a Solution
I assume that y and x in parametric form is analogous than for a circle, but in this case the radius is not equal everywhere and the center is not the origin.
$y=h+b\sin \theta$
$x=g+a\cos \theta$

dx = -a sin(theta) d(theta)

Area = INT(pi/2 to 0)[-b sin(theta) * a sin(theta) * d(theta)]
= INT(0 to pi/2)[absin^2(theta) * d(theta)]
= INT(0 to pi/2)[(ab/2)(1-cos(2theta) * d(theta)]
= (ab/2)[(theta) - (1/2)sin(2theta)] from 0 to pi/2
= (ab/2)[pi/2 - 0]
= pi*ab/4

The total area of the ellipse will be 4 times this area, so:

Area of ellipse = pi*ab
However, I do not know how to acquire a and b from
the expanded equation of the ellipse:
y^2 = (2-x)y - (x^2)/2

(x^2)/2 +y^2= (2-x)y
Can I say that the radius along the x-axis is sqrt(2) and the radius along the y-axis is 1?
In other words a=sqrt(2) and b=1. Then,
(2-x)y=1
2y-xy=1, but I don't know how to interpret this.

Please help.

 

[SteamKing]

I think if you graph the original equation, you will find the major and minor axes of the ellipse are rotated with respect to the x-y coordinate axes. Hint: expand the original equation and collect terms. There is a non-zero xy term in the expansion.

 

[knowLittle]

I think if you graph the original equation, you will find the major and minor axes of the ellipse are rotated with respect to the x-y coordinate axes. Hint: expand the original equation and collect terms. There is a non-zero xy term in the expansion.

I know that the axis are rotate with respect of the x-y coordinates. I thought that the equation was still valid in these cases.

Yes. I see the xy term and I showed it in the first post. I am trying to compare it with 1, since it fits the equation. However, this equation would be wrong. I realize now.
I have trouble setting
(x+y)^2 + (y-2)^2 =4 in the form
$\left( \dfrac {x-h} {a}\right) ^{2}+\left( \dfrac {y-g} {b}\right) ^{2}=1$

 

[SteamKing]

Try this article on for size:
http://en.wikipedia.org/wiki/Rotation_of_axes

 

[SammyS]

I know that the axis are rotate with respect of the x-y coordinates. I thought that the equation was still valid in these cases.

Yes. I see the xy term and I showed it in the first post. I am trying to compare it with 1, since it fits the equation. However, this equation would be wrong. I realize now.
I have trouble setting
(x+y)^2 + (y-2)^2 =4 in the form
$\left( \dfrac {x-h} {a}\right) ^{2}+\left( \dfrac {y-g} {b}\right) ^{2}=1$

You cannot put the equation for this ellipse directly into standard form. To do that, you would need to rotate the coordinate system , or rotate the ellipse.

If you need to find the area of the ellipse via integration (That's not stated explicitly in the body of the Original Post.), that can be achieved without the rotation.

You could do the integration in polar coordinates. The ellipse passes through the origin and is tangent to the x-axis. The result of expressing r as a function of θ is rather intimidating from the standpoint of taking the resulting integral.

You can solve the equation of the ellipse for x or for y and integrate. Since x only appears once in the equation, it makes sense to solve for x as a function of y. I haven't tried my hand at the resulting integral so I don't know how well this approach will work out.