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.
Answer:
To evaluate the double integral using polar coordinates, we need to express the differential area element d𝑨 in terms of polar coordinates. The conversion formulas are:
𝑥 = 𝑟cos𝜃
𝑦 = 𝑟sin𝜃
d𝑨 = 𝑟d𝑟d𝜃
The region R in the first quadrant enclosed by the circle x^2 + y^2 = 4, x = 0, and y = x can be expressed in polar coordinates as follows:
𝑧: 𝜃 ∈ [0, π/4]
𝑟: 𝑟 ∈ [0, 2]
Now, let's substitute the expressions for 𝑥, 𝑦, and d𝑨 into the integral:
∬𝑅 (3𝑥 − 𝑦) d𝐴 = ∫[𝜃=0 to 𝜃=π/4] ∫[𝑟=0 to 𝑟=2] (3(𝑟cos𝜃) − (𝑟sin𝜃)) 𝑟 d𝑟 d𝜃
Expanding and simplifying the expression inside the integral:
∫[𝜃=0 to 𝜃=π/4] ∫[𝑟=0 to 𝑟=2] (3𝑟cos𝜃 − 𝑟sin𝜃) 𝑟 d𝑟 d𝜃
= ∫[𝜃=0 to 𝜃=π/4] ∫[𝑟=0 to 𝑟=2] (3𝑟^2cos𝜃 − 𝑟^2sin𝜃) d𝑟 d𝜃
Now, we evaluate the inner integral by integrating with respect to r:
∫[𝑟=0 to 𝑟=2] (3𝑟^2cos𝜃 − 𝑟^2sin𝜃) d𝑟
= (𝑟^3cos𝜃/3 − 𝑟^3sin𝜃/3) evaluated from 𝑟=0 to 𝑟=2
= (8cos𝜃/3 − 8sin𝜃/3) - (0) = 8cos𝜃/3 - 8sin𝜃/3
Now, we evaluate the outer integral by integrating with respect to 𝜃:
∫[𝜃=0 to 𝜃=π/4] (8cos𝜃/3 - 8sin𝜃/3) d𝜃
= (8sin𝜃/3 + 8cos𝜃/3) evaluated from 𝜃=0 to 𝜃=π/4
= (8(sin(π/4))/3 + 8(cos(π/4))/3) - (8(0)/3)
= (8/√2)/3 + (8/√2)/3
= (8√2 + 8√2)/3
= 16√2/3
Therefore, the value of the double integral is 16√2/3.
To evaluate the double integral ∬R(3x−y)dA using polar coordinates, we first need to express the region of integration R in terms of polar coordinates.
In polar coordinates, x = r*cos(theta) and y = r*sin(theta).
The equation of the circle x^2 + y^2 = 4 becomes r^2*cos^2(theta) + r^2*sin^2(theta) = 4.
Simplifying this equation gives r^2 = 4, which means r = 2.
The line x=0 corresponds to theta = π/2, and the line y = x corresponds to theta = π/4.
Therefore, the region of integration R in polar coordinates is defined by 0 ≤ r ≤ 2 and π/4 ≤ θ ≤ π/2.
Now, let's express the integrand (3x−y) in terms of polar coordinates.
Substituting x = r*cos(theta) and y = r*sin(theta) into (3x−y), we get:
(3x−y) = 3(r*cos(theta)) − (r*sin(theta)).
Now, we can evaluate the double integral using polar coordinates:
∬R(3x−y)dA = ∫(π/4 to π/2)∫(0 to 2) (3(r*cos(theta)) − (r*sin(theta))) * r dr dθ.
Let's integrate this expression step-by-step.