I know it's a bit hard to see on the graph but it's the best I can do for now. This is a graph of:

y = 4sqrt(3) * arctan[ (2tan(x)+1)/sqrt(3) ]

You are trying to evaluate this function in the interval 0 --->2pi. So the problem is that there are 2 discontinuities in this interval for tan(x); pi/2 and 3pi/2. That means we have to break the interval up. However, the graph is symmetrical and I am going to use some very non-rigorous ways to deal with that. So for all you out there who are going to object, go for it. I don't have the gumption to go for rigour today.

I am going to evaluate from pi/2 ---> 3pi/2 which is continuous on the open interval. As this represents one-half of the area of the rotated ellipse, I will then double the value.

I'm not quite sure what you mean by "bone". This is an ellipse which has been rotated out of standard position by an angle of -pi/4. In standard position the equation is

x^2/24 + y^2/8 = 1 making the

semi-major axis = a = sqrt(24) and the semi-minor axis = b = sqrt(8)

The area of an ellipse is pi*a*b = pi*sqrt(24)*sqrt(8) = 8*pi*sqrt(3)

So we agree on that. I think what might be happening is the you are getting positive and negative areas cancelling out. Try evaluating the integral from 0 to pi instead (i.e. area above the x-axis) and then doubling your answer. That should work.

I really don't know what you are doing wrong. But I am sure that

thesis paper writingwriters may help. This is a proper movement to success.Good show Ralph! Thanks for the problem - Ian.

Hi Ian,

You was on the right track.

I checked the results now in two different ways and as you mentioned the discontinuities are the problem.

I came up with the same results by adding up all the individual pieces or take the integral from pi/2 to 3/2pi, see attached.

Thanks for the hint!

Stay safe

Regards

Ralph

I know it's a bit hard to see on the graph but it's the best I can do for now. This is a graph of:

y = 4sqrt(3) * arctan[ (2tan(x)+1)/sqrt(3) ]

You are trying to evaluate this function in the interval 0 --->2pi. So the problem is that there are 2 discontinuities in this interval for tan(x); pi/2 and 3pi/2. That means we have to break the interval up. However, the graph is symmetrical and I am going to use some very non-rigorous ways to deal with that. So for all you out there who are going to object, go for it. I don't have the gumption to go for rigour today.

I am going to evaluate from pi/2 ---> 3pi/2 which is continuous on the open interval. As this represents one-half of the area of the rotated ellipse, I will then double the value.

4*sqrt(3)*arctan(+ inf) - 4*sqrt(3)*arctan(- inf) where arctan(- inf) = - arctan(+ inf)

So we now have:

4*sqrt(3)*arctan(+ inf) + 4*sqrt(3)*arctan(+ inf)

= 4*sqrt(3)*pi/2 + 4*sqrt(3)*pi/2

= 4*sqrt(3)*pi

Doubling gives us 8*sqrt(3)*pi

I'm not quite sure what you mean by "bone". This is an ellipse which has been rotated out of standard position by an angle of -pi/4. In standard position the equation is

x^2/24 + y^2/8 = 1 making the

semi-major axis = a = sqrt(24) and the semi-minor axis = b = sqrt(8)

The area of an ellipse is pi*a*b = pi*sqrt(24)*sqrt(8) = 8*pi*sqrt(3)

So we agree on that. I think what might be happening is the you are getting positive and negative areas cancelling out. Try evaluating the integral from 0 to pi instead (i.e. area above the x-axis) and then doubling your answer. That should work.