Finding $x$, where $\int_{r-x}^r \sqrt{r^2 - t^2} dt = A$ When solving my practical problem, I would need to compute $x$, given $A$ and $r$, where $x$ appears in the integral:
$$\int_{r-x}^r \sqrt{r^2 - t^2} \text{d}t = A$$
It's quite a few years from my last calculus lesson, so I'm asking here.
thanks
 A: Although this has been presented as a calculus problem, it’s really a problem in high-school geometry. The integral represents the area of half a circular segment $S$. This segment is the area of the circle of radius $r$ that’s to the right of the vertical line $t=x$, in the $ty$-plane.
If the chord (here going between the points $\left(x,\pm\sqrt{r^2-x^2}\,\right)\,$) subtends an angle of $\alpha$, then the sector (pie-wedge) of angle $\alpha$ has area $\alpha r^2/2$. From this we subtract the area of the triangle whose vertices are the origin and the two points I mentioned previously. The altitude is $x$ and the (vertical) base is $2\sqrt{r^2-x^2}$ in length, so the area of the whole triangle has area $x\sqrt{r^2-x^2}$. We have to subtract this quantity from $\alpha r^2/2$ and get rid of the $\alpha$, which is $\arccos(x/r)$. That gives you the area of the whole sector, and our desired quantity is half that, you’re done.
Oh, yes, and if you wanted the “indefinite integral” of the function $\sqrt{r^2-t^2}$, this method is how you find it, though the method of “trigonometric substitution” would also do the same, with more work.
A: The result is a transcendental equation; I know of no closed-form solution:
$$\frac{2 A}{r^2} =  y \sqrt{1-y^2} + \arccos{y}$$
where
$$y = 1-\frac{x}{r}$$
