To evaluate the double integral ∬R(3x−y)dA using polar coordinates, we first need to determine the limits of integration in polar form.
In polar coordinates, the region of interest, R, is described by the inequalities:
0 ≤ r ≤ 2 (since the circle x^2 + y^2 = 4 has a radius of 2)
0 ≤ θ ≤ π/4 (since y = x)
The conversion to polar coordinates is given by x = rcosθ and y = rsinθ. Therefore, the double integral becomes:
∬R(3x−y)dA = ∫[0,π/4]∫[0,2] (3rcosθ - rsinθ) r dr dθ
Simplifying the expression inside the double integral:
(3rcosθ - rsinθ) r = 3r^2cosθ - r^2sinθ = r^3(3cosθ - sinθ)
Now we can evaluate the double integral by integrating with respect to r first and then θ:
∫[0,2] r^3(3cosθ - sinθ) dr = (3cosθ - sinθ) ∫[0,2] r^3 dr = (3cosθ - sinθ) [r^4/4]∣∣∣[0,2]
Plugging in the limits of integration:
(3cosθ - sinθ) [(2^4)/4 - 0] = (3cosθ - sinθ) (8)
Finally, we integrate this expression with respect to θ:
∫[0,π/4] (3cosθ - sinθ)(8) dθ = (8)[3sinθ + cosθ]∣∣[0,π/4]
Plugging in the limits of integration:
(8)[3sinπ/4 + cosπ/4 - 3sin0 - cos0] = 8(3(√2/2) + (√2/2) - 3(0) - 1) = 8(3√2/2 + √2/2 - 1) = 8(2√2/2) = 8√2
Therefore, the value of the double integral ∬R(3x−y)dA when changing to polar coordinates is 8√2.
Evaluate the double integral ∬R(3x−y)dA,
where R
is the region in the first quadrant enclosed by the circle x^2+y^2=4
and the lines x=0
and y=x,
by changing to polar coordinates.
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