In the beginning of this answer, I will be fussy (sorry, I'm a mathematician...).
Every line on a plane is parallel to a diameter of a circle lying on the same plane. In fact, a circle doesn't have the diameter, but it has infinitely many diameters. So the object "a parallel to the diameter" is every line of the plane.
Another fussiness about this question is the word "between", because the two shapes (line and circle) do not necessarily surround a closed area: if the line doesn't cross the circle, you have to specify how to enclose this area.
However, the words "cup shape" made me think that you want to find out how to calculate the area of a so-called circular segment:
The area of the "cup shape" is given by the area of the circular sector s=OAPBO minus the area of the triangle Delta=Delta(AOB). They both depend on the angle alpha=hat{AOB}. Given the radius r=AO=BO=PO of the circle, the area of the triangle Delta(AOB) is given by the triangle area formula:
A_{Delta}= 1/2 bar{AO} * bar{BO} sin alpha=1/2 r^2 sin alpha
The area A_s of the circular sector is directly proportional to the angle alpha, so we can use the proportion A_s / alpha = {pi r^2} / {2 pi} to get:
A_s=1/2 r^2 alpha
So the area we're searching for is A=A_{s}-A_{Delta}=1/2 r^2 (alpha - sin alpha).
We can also use the height h of our "cup" (i.e. distance between the point P and the segment bar{AB}) to express the angle alpha. If bar{OH} is the height of the triangle Delta(AOB), then bar{OH}=r-h. The height bar{OH} is also the bisector of the triangle Delta(AOB) (it's an isosceles triangle). That is: hat{AOH}=alpha /2.
Now let's consider the triangle Delta (AOH): it's a right triangle, so by definition of cosine we get bar{OH}=bar{AO} cos(hat{AOH})=r cos(alpha/2). So cos(alpha/2)=bar{OH}/r and this means that cos(alpha/2) = {r-h}/r=1-h/r and alpha=2 arccos(1-h/r).
We finally get the following (awful) formula:
A=1/2 r^2 (2 arccos(1-h/r)-sin(2 arccos(1-h/r)))=1/2 r^2 (2 arccos(1-h/r)-2sin(arccos(1-h/r)) cos(arccos(1-h/r)))=r^2 (arccos(1-h/r)-sqrt(1-(1-h/r)^2) (1-h/r))=r^2 (arccos(1-h/r)-sqrt(h/r(2-h/r)) (1-h/r)).
Note:
We used the following two formulas
sin(2x)=2 sin x cos x
arccos(sin(x))=sqrt{1-x^2}
