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.
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.
To evaluate the double integral ∬R(3x−y)dA using polar coordinates, we need to change the given Cartesian coordinates to polar coordinates. Let's proceed step by step:
Step 1: Sketch the given region R. In this case, R is the first quadrant enclosed by the circle x^2 + y^2 = 4, the x-axis (x = 0), and the line y = x.
Step 2: Determine the limits of integration in polar coordinates. Since the region R is enclosed by the circle x^2 + y^2 = 4, we can express this circle in polar coordinates as r^2 = 4. Taking the positive square root, we have r = 2.
The line y = x can also be expressed in polar coordinates. Since y = x, we have r * sin(theta) = r * cos(theta). Dividing both sides by r, we get sin(theta) = cos(theta), or tan(theta) = 1. In the first quadrant, theta ranges from 0 to pi/4.
Therefore, the limits of integration for r are from 0 to 2, and the limits of integration for theta are from 0 to pi/4.
Step 3: Convert the integrand to polar coordinates. The integrand (3x−y) needs to be changed to polar coordinates. We will use the following relationships:
x = r * cos(theta)
y = r * sin(theta)
Substituting these into the integral, we get:
∬R(3x−y)dA = ∫[theta=0 to pi/4] ∫[r=0 to 2] (3(r*cos(theta)) - (r*sin(theta))) * r dr dtheta
Step 4: Evaluate the integral. Simplify the integrand and evaluate the integral using the limits of integration:
∬R(3x−y)dA = ∫[theta=0 to pi/4] ∫[r=0 to 2] (3r*cos(theta) - r*sin(theta)) * r dr dtheta
Integrating with respect to r:
= ∫[theta=0 to pi/4] [(3/3)*r^3*cos(theta) - (1/2)*r^3*sin(theta)] | [r=0 to 2] dtheta
= ∫[theta=0 to pi/4] (8*cos(theta) - 4*sin(theta) - 0) dtheta
= ∫[theta=0 to pi/4] (8*cos(theta) - 4*sin(theta)) dtheta
Integrating with respect to theta:
= [8*sin(theta) + 4*cos(theta)] | [theta=0 to pi/4]
= (8*sin(pi/4) + 4*cos(pi/4)) - (8*sin(0) + 4*cos(0))
= (8*(sqrt(2)/2) + 4*(sqrt(2)/2)) - (0 + 4)
= 4*sqrt(2) + 4*sqrt(2) - 4
= 8*sqrt(2) - 4
Therefore, the value of 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, the line x=0, and the line y=x, is equal to 8*sqrt(2) - 4.
To evaluate the double integral ∬R(3x−y)dA in the region R enclosed by the circle x^2+y^2=4, the line x=0, and the line y=x, we can change to polar coordinates.
In polar coordinates, we have x = r*cosθ and y = r*sinθ, where r represents the distance from the origin to the point (x, y) and θ represents the angle the line from the origin to the point (x, y) makes with the positive x-axis.
First, let's find the limits of integration.
The region R is enclosed by the circle x^2 + y^2 = 4, which can be written in polar coordinates as r^2 = 4. Solving for r, we have r = 2.
Since the line x = 0 is the y-axis, we know that θ ranges from 0 to π/2.
The line y = x can also be expressed in polar coordinates as r*sinθ = r*cosθ. Simplifying, we have sinθ = cosθ, which implies that θ = π/4.
Therefore, the limits of integration for r are 0 to 2, and for θ, it is π/4 to π/2.
Now, let's find the Jacobian for the transformation to polar coordinates:
Jacobian, J = r
Next, let's express the integral in terms of polar coordinates:
∬R(3x−y)dA = ∫(θ=π/4 to π/2) ∫(r=0 to 2) (3(r*cosθ) - (r*sinθ)) * r dr dθ
Expanding the equation:
= ∫(θ=π/4 to π/2) ∫(r=0 to 2) (3rcosθ - rsinθ) * r dr dθ
Now, let's solve the inner integral with respect to r:
∫(r=0 to 2) (3rcosθ - rsinθ) * r dr
= ∫(r=0 to 2) (3r^2cosθ - r^2sinθ) dr
= [r^3cosθ/3 - r^3sinθ/3] (r=0 to 2)
= [(8cosθ - 8sinθ)/3 - (0cosθ - 0sinθ)/3]
= (8cosθ - 8sinθ)/3
Now, let's solve the outer integral:
∫(θ=π/4 to π/2) (8cosθ - 8sinθ)/3 dθ
= [(8sinθ + 8cosθ)/3] (θ=π/4 to π/2)
= [(8sin(π/2) + 8cos(π/2))/3 - (8sin(π/4) + 8cos(π/4))/3]
= [8/3 - (8sqrt(2)/3 + 8sqrt(2)/3)]
= 8/3 - 16sqrt(2)/3
Therefore, the value of the double integral is 8/3 - 16sqrt(2)/3.