integration to find in terms of q?

Question

2 * INT (0 to q) [sqrt(r^2 - x^2) - h] dx

i know i can use a trigonometric substitiution in order to solve the intergral in terms of q .......but what substitution and how???

and to make things worse, how do i find q as a function of h to substitute back into the previous equation to get the area as a function of h

Answer 1

2∫ [√(r² - x² ) - h ]dx

Let x = rsinθ then dx = rcosθ dθ

2∫ [√(r² - r²sin²θ) - h] rcosθ dθ

= 2∫ [√(r²cos²θ) - h] rcosθ dθ

= 2∫ [(rcosθ) - h] rcosθ dθ

= 2∫ [(r²cos²θ) - hrcosθ] dθ

= 2∫ [(r²(1 + cos2θ)/2 - hrcosθ] dθ

= ∫ [(r²(1 + cos2θ) - 2hrcosθ] dθ

= r²(θ + ½sin2θ) - 2hrsinθ + c

= r²(arcsin(x/r) + sinθcosθ) - 2hr(x/r) + c

= r²(arcsin(x/r) + (x/r)(√(1 - (x/r)²)) - 2hx + c

now you can substitute your limits for x

Answer 2

let x=rsin t

t=arc sin (x/r)

r^-x^2=r^2cos^2 t

separate integrals.
integral rcost dx-integralhdx

do not forget 2

Answer: 2*{(qrcos(arcsin(q/r))-qh}