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 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.

Hi Ian,

thanks for your thoughts.

This will not solve my problem, because every multiple integer of pi will cause that tan(n pi) =0.

So it doesn’t matter if you integrate from 0 to 2pi, or from 0 to pi, or from pi to 2pi.

Stay safe!

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

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

Good show Ralph! Thanks for the problem - Ian.

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

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