Evaluate ∬R 1/sqrt(x^2+y^2) dx dy, where R is the region bounded by y≥0 and x^2−x+y^2≤0.
The region is the upper half of a circle with radius 1/2 centered at (1/2,0):
(x-1/2)^2 + y^2 = 1/4
So,
∫[0,1]∫[0,√1/4 - (x-1/2)^2) 1/√(x^2+y^2) dy dx
recall that √1/(x^2+y^2) dy = arcsinh(y/x) = log(x + √(x^2+y^2))
oops that's log(y + √(x^2+y^2))
To evaluate the given double integral ∬R 1/sqrt(x^2+y^2) dx dy, we need to find the limits of integration for x and y.
First, let's analyze the region R bounded by y ≥ 0 and x^2 - x + y^2 ≤ 0.
To determine the limits of integration for y, we need to find the upper and lower bounds for y. Since the equation y ≥ 0 represents the region above or on the x-axis, the lower limit for y is 0.
Next, we'll consider the inequality x^2 - x + y^2 ≤ 0. This equation represents a circle centered at (1/2, 0) with a radius of 1/2 (by completing the square). We want to find the limits for x that satisfy this inequality.
To evaluate these limits, we can rewrite the inequality as follows: (x - 1/2)^2 + y^2 ≤ 1/4.
Now, let's solve for x: x - 1/2 ≤ sqrt(1/4 - y^2).
Taking the square root of both sides introduces the positive and negative solution:
x ≤ 1/2 + sqrt(1/4 - y^2) and x ≥ 1/2 - sqrt(1/4 - y^2).
Since y≥0, we can ignore the negative solution. Therefore, the upper limit for x is x ≤ 1/2 + sqrt(1/4 - y^2).
Now, we can set up the double integral:
∬R 1/sqrt(x^2+y^2) dx dy = ∫[0 to y] ∫[1/2 - sqrt(1/4 - y^2) to 1/2 + sqrt(1/4 - y^2)] 1/sqrt(x^2+y^2) dx dy
Now, we need to evaluate this double integral by integrating with respect to x first. Let's consider the inner integral:
∫[1/2 - sqrt(1/4 - y^2) to 1/2 + sqrt(1/4 - y^2)] 1/sqrt(x^2+y^2) dx
Using a trigonometric substitution, let x = (1/2)sinθ:
dx = (1/2)cosθ dθ
x^2 = (1/4)sin^2θ
(x^2 + y^2) = (1/4)sin^2θ + y^2 = (1/4)(sin^2θ + 4y^2)
Plugging in these values, the integral becomes:
(1/2)∫[arcsin(2y)...arcsin(0)] 1/sqrt[(1/4)(sin^2θ + 4y^2)](1/2)cosθ dθ
Simplifying:
(1/4)∫[arcsin(2y)...arcsin(0)] 1/sqrt[(1/4)(sin^2θ + 4y^2)]cosθ dθ
To continue evaluating this integral, we will need to use numerical methods or approximation techniques, as the antiderivative of the integrand does not have a simple closed-form expression.
Finally, the double integral ∬R 1/sqrt(x^2+y^2) dx dy is given by:
∬R 1/sqrt(x^2+y^2) dx dy = (1/4)∫[arcsin(2y)...arcsin(0)] 1/sqrt[(1/4)(sin^2θ + 4y^2)]cosθ dθ
To evaluate the given double integral, we need to find the limits of integration for x and y.
First, let's draw a sketch of the region R bounded by y≥0 and x^2−x+y^2≤0.
Since y≥0, the region R lies in the upper half plane. The inequality x^2−x+y^2≤0 represents a circle centered at (1/2, 0) with a radius of 1/2.
Now, let's find the limits of integration for x and y.
For y, since y≥0, the lower limit of integration for y is 0. To find the upper limit of integration for y, we need to find the equation of the upper boundary of region R.
The equation of the upper boundary is given by x^2−x+y^2=0.
Completing the square for x, we get (x−1/2)^2 + y^2=1/4. This is the equation of a circle centered at (1/2, 0) with a radius of 1/2.
Since the region R lies on or below this circle, the upper limit of integration for y is the equation of the upper boundary of region R, which is y=√(1/4−(x−1/2)^2).
For x, we need to find the limits of integration based on the projection of the region R onto the x-axis.
The projection of the region R onto the x-axis can be represented by the interval [a, b], where a and b are the x-values where the circle intersects the x-axis.
To find these values, we set y=0 in the equation of the circle:
(x−1/2)^2 + 0^2=1/4
(x−1/2)^2=1/4
x−1/2=±1/2
x=1 and x=0.
Therefore, the limits of integration for x are from 0 to 1.
Now, we can write the double integral:
∬R (1/√(x^2+y^2)) dx dy
= ∫[0,1] ∫[0,√(1/4−(x−1/2)^2)] (1/√(x^2+y^2)) dy dx
However, this integral does not have a simple closed-form solution. It would typically be evaluated numerically, using methods such as numerical integration techniques or software like MATLAB or Wolfram Alpha.