integral [from x to 0] (r^2 - x^2)^{1/2} dx : different units

sinx

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int(r2-x2)1/2dx, limits x=x, 0 (x is upper limit, 0 is lower limit)
=int r(1-x2/r2)1/2dx

sub x/r=sin&
=r2int cos2&d&, limits &=sin-1(x/r), 0 ??
cos2&=1/2(1+cos2&)
=r2/2 int(1+cos2&)d&

sub u=2&
1/2du=d&
=r2/4 int(1+cosu)du limits u=2sin-1(x/r),0 ??
=r2/4 (u+sinu) evaluated at limits u=2sin-1(x/r),0
=r2/4 ((2sin-1(x/r) + sin 2sin-1(x/r))-(0+sin0))
=r2/4 (2sin-1(x/r) + sin (2sin-1(x/r)))

besides the questions in red about changing limits, here is the main question;
2sin-1(x/r) is an angle; sin (2sin-1(x/r) is a number, the units are not the same.
in fact, the expression
u+sinu gives units are that are not the same (whether or not my limits are correct).
when multiplied by the constant in front (r2/4), you get inches2, or mm2 or whatever r is.
this should be the units of the integral, since it is an area.
I know radians are 'unitless', but do i ignore the fact that i am adding an angle to the sin of an angle?
is sin& 'unitless' also? (of course it is, inches/inches, answered my own question here)
still do you add radians to a dimentionless number?

Is there a way to determine sin
(2sin-1&)?,
that is
sin(sin-1&) is just &, but what about sin(2sin-1&)?, [I'm sure it is not 2&].


 
Last edited:

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
4,152
int(r2-x2)1/2dx, limits x=x, 0 (x is upper limit, 0 is lower limit)
=int r(1-x2/r2)1/2dx

sub x/r=sin&
=r2int cos2&d&, limits &=sin-1(x/r), 0 ??
cos2&=1/2(1+cos2&)
=r2/2 int(1+cos2&)d&

sub u=2&
1/2du=d&
=r2/4 int(1+cosu)du limits u=2sin-1(x/r),0 ??
=r2/4 (u+sinu) evaluated at limits u=2sin-1(x/r),0
=r2/4 ((2sin-1(x/r) + sin 2sin-1(x/r))-(0+sin0))
=r2/4 (2sin-1(x/r) + sin (2sin-1(x/r)))

besides the questions in red about changing limits, here is the main question;
2sin-1(x/r) is an angle; sin (2sin-1(x/r) is a number, the units are not the same.
in fact, the expression
u+sinu gives units are that are not the same (whether or not my limits are correct).
when multiplied by the constant in front (r2/4), you get inches2, or mm2 or whatever r is.
this should be the units of the integral, since it is an area.
I know radians are 'unitless', but do i ignore the fact that i am adding an angle to the sin of an angle?
is sin& 'unitless' also? (of course it is, inches/inches, answered my own question here)
still do you add radians to a dimentionless number?

Is there a way to determine sin
(2sin-1&)?,
that is
sin(sin-1&) is just &, but what about sin(2sin-1&)?, [I'm sure it is not 2&].
First, all your limits are correct (though you should know that limits are read from bottom to top, so it's "from 0 to x"; and it's really a little wrong to use x both as the variable of integration and in the limit). It would have been easier to read if you'd used either an actual theta (θ) or a letter like t, but that's okay.

As far as calculus is concerned, trig functions take mere numbers as argument, and produce mere numbers -- all dimensionless quantities. We don't think in terms of radians as a unit in this context. That is, the sine or inverse sine is just a function from the real numbers to the real numbers. A good presentation of the unit circle model will make this clear; we just wrap a number line around the unit circle, with no units involved.

To evaluate sin(2sin-1u) [you didn't mean theta there, did you?], you can just plug numbers into your calculator. But if you had a reason to want to simplify the expression, you can use the double-angle formula, sin(2θ) = 2sin(θ)cos(θ). So you get sin(2sin-1u) = 2sin(sin-1u)cos(sin-1u) = 2u√(1-u2). [This is valid without worrying about the sign of the cosine because the inverse sine is always in a quadrant where the cosine is positive.]
 
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