Guidance for Leibniz Rule integration under derivatives

Let f(x.y) be a continuous function in R^2. Set F(t)= the double integral with region x^2+y^2<=t of f(x,y)dA.

Write another expression for F(t) by converting the integral to polar coordinates.
I got F(t)= int((dtheta)int(f(rcostheta,rsintheta)r dr). theta=0..2pi, r=0..sqrt(t)

Apply Leibniz Rule to the integral to find the derivative of F'(t). it's okay to give the answer in polar coordinates.

for this part i'm just confused. do i need to say that the integral becomes 2pi*int(f(rcostheta,rsintheta)rdr from r=0..sqrt(t) then take F'(t)=2pi* int(df/dt(f(rcostheta,rsintheta)rdr))? does that even become anything else?

My best guess:

F(t) = ∫∫ f(x,y)dA
.......=∫ (θ = 0 to 2π)∫(r = 0 to t) f(r,θ) r dr dθ

F '(t) = .......=∫ (θ = 0 to 2π) t * f(t, θ) dθ
:)

Sorry, I just saw your response. Your were so right about √t.
In that case, ∂F/∂t ∫ [r from 0 to √t ] f(r, θ)r dr would be
f(√t, θ)*√t*1/(2√t)
=1/2 * f(√t, θ)
Overall, F '(t) = ∫[θ from 0 to 2π ]1/2 * f(√t, θ) dθ. :)

Report Abuse