Question

Evaluate the integral ∬ᵣ (2x−y)dA, where R is the region in the first quadrant enclosed by the circle x²+y² =4 and the lines x=0, and y=x by changing the polar coordinates.

Answer 1

The double integral ∫∫ṛ (2x−y)dA over the specified region can be evaluated by converting to polar coordinates, setting up the bounds for r and θ, and then performing the double integral step by step.

Step-by-step explanation:

Evaluating the Integral in Polar Coordinates

To evaluate the integral ∫∫ṛ (2x−y)dA, where R is the region in the first quadrant enclosed by the circle x²+y² = 4 and the lines x=0, and y=x, we change the integral into polar coordinates. In polar coordinates (r, θ), the conversion formulas are x=r·cos(θ) and y=r·sin(θ). Thus, the integral becomes ∫∫ṛ (2r·cos(θ) − r·sin(θ))rdrdθ by factoring out the common r. Additionally, dA converts to rdrdθ in polar coordinates.

The region of integration is now described by the bounds of r from 0 to 2 (since the radius of the circle is 2) and θ from 0 to π/4 (since the region is enclosed by y=x and y=0 in the first quadrant, which translates to θ from 0 to π/4 in polar coordinates). The integral becomes:

∫0π/4∫02(2r·cos(θ) − r·sin(θ))rdrdθ = ∫0π/4 ( ∫02(2r²·cos(θ) − r²·sin(θ))dr ) dθ

By computing the definite integral over r first, followed by integration over θ, we find the value of the double integral. Performing these integrations will give us the answer to the problem.