My int function is returning the function back instead of an integral

I have this function

 f = (sin(2*r^2)*(16*r^4 + 1)^(1/2))/(r^2 + z^2)^(3/2)

but when I try to integrate it

int(f,r,[0 10])

it returns the function back:

ans =

int((sin(2*r^2)*(16*r^4 + 1)^(1/2))/(r^2 + z^2)^(3/2), r, 0, 10)

How do I find the integral?

[Walter Roberson]

Could you confirm that you have sin(2*r^2) where r is in the range 0 to 10 and r is in radians ? So at the end of the range you would be calculating sin(200 radians) ?

[Marc Adams]

The r and z are from cylindrical coord system, so r is not in radians

[Walter Roberson]

What is the units of r then? If it is degrees then you need to convert degrees to radians for use in the sin() call. (Actually, it would have to be in square root of degrees, since you are using r^2)

Your function does not appear to be linear if r were rescaled to be from 0 to 10*pi/180 or sqrt(10*pi/180)

[Marc Adams]

r is a unit of distance, not angle.

Answer 1

[Walter Roberson]

That function has no obvious closed form integral. You will need to switch to numeric integration, such as with integral() or vpaintegral(). That will require that you have a numeric value for z.

Note: sometimes I would suggest using a taylor series, but if you taylor at r = 0 or r = 5 then the results will be rather off.

[Walter Roberson]

Why are you taking sin() of (distance squared) ? For example at the end of the range are you taking sin(200 in^2) ?

Ask yourself whether your formula is immune to scaling. If you were to replace r with R*c where c is some scaling factor representing a different unit, and were to change the upper bound appropriately, then would the same formula fall out?

The answer is No. For one thing, the +1 in the (16*r^4+1) is not going to scale with the change in units.

We can make an ad-hoc guess that perhaps the 1 needs to be replaced with c^4 where c is the same scaling factor. If we then restrict everything to nonnegative, the formula

sin(2*(R*c)^2)*(16*(R*c)^4+c^4)^(1/2)/((R*c)^2+(Z*c)^2)^(3/2)

simplifies to

sin(2*R^2*c^2)*(16*R^4+1)^(1/2)/(c*(R^2+Z^2)^(3/2))

this has sin() of c^2 times larger and a divisor of c.

If you choose c = 254/100 (inches to mm), and choose z = 5 for the original formula and Z = 5*100/254 for the second formula, (because Z*c is used so 5*100/254*254/100 -> 5 would give the same result as z = 5), and you plot the first formula over r = 0 to 10, and plot the second over R = 0 to 10*100/254: then at first look the graphs are much the same. But if you look carefully, the height of the first peak is different.

If you then proceed to test with r = 7 for the first formula and R = 7 * 100/254 for the second formula, then you get different results -- the formulas are not equal under a distance transform, even if you use c^4 instead of 1 so that the constant can factor out of the sqrt() part.