To solve this problem, first identify the region R in the polar coordinates. Then use these bounds in polar coordinates to set up and evaluate the double integral.
Let's map R to polar coordinates. We know that the polar coordinate system uses (r, θ) where r is the distance from the origin and θ is the angular displacement from the x-axis. R is enclosed by the circle x²+y²=16; this translates to r=4 in polar coordinates. The lines x=0 and y=x translate to θ=0 and θ=π/4 respectively.
Now, express the given function f(x, y) = 3x - y in polar coordinates. In polar coordinates, x = rcos(θ) and y = rsin(θ), so f(r, θ) = 3r cos(θ) - r sin(θ).
Having identified the function and boundaries in polar coordinates, set up the double integral, remembering to include rdθdr. The differential dA in polar coordinates changes into r drdθ
So R becomes:
∫ (from 0 to π/4) ∫ (from 0 to 4) r*(3*r*cos(θ)-r*sin(θ)) drdθ
The inner integral is taken over r, from 0 to 4, and the outer integral is taken over θ, from 0 to π/4.
Evaluate the inner integral first with respect to r,
and then evaluate the outer integral with respect to θ.
When you evaluate these integrals, the result is -64 + 128√2.
So, ∬R(3x − y) dA = -64 + 128√2.