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

#### sinx

##### Junior Member
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)

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
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)