Use polar coordinates to evaluate the integral [0,2]∫ [0,x√3]∫1/√(𝑥^2+𝑦^2)dydx
No. You have to change the variables of integration to r,θ
Recall that the area element dy dx becomes r dr dθ
Now look at the region. It is a triangle withe vertices at (0,0), (2,0), (2,√3)
In polar coordinates, r = 2 secθ
The point (2,2√3) is at θ = π/3
So the integral becomes
∫[0,π/3] ∫[0,2secθ] 1/r * r dr dθ
= ∫[0,π/3] ∫[0,2secθ] 1 dr dθ
= ∫[0,π/3] 2secθ dθ
= ln|secθ + tanθ| [0,π/3]
= ln|2 + √3|
note that you have ∫∫ 1/r dA
so can you analyze the graph and determine the domain in polar coordinates?
Are these right [0,2]∫ 1/r(y)[y=0,y=x√3]dx=>[0,2]∫(x√3)/rdx=>(√3)/r(x^2)/2[0,2]=(2√3)/r
To evaluate the given double integral using polar coordinates, we need to express the integrand and the limits of integration in terms of polar coordinates.
In polar coordinates, we have x = rcos(θ) and y = rsin(θ), where r represents the radial distance from the origin, and θ represents the angle measured counterclockwise from the positive x-axis.
Let's start by expressing the limits of integration in polar form. The domain of integration is described by the inequalities 0 ≤ x ≤ 2 and 0 ≤ y ≤ x√3. Substituting x and y in terms of polar coordinates, we get:
0 ≤ rcos(θ) ≤ 2 (1)
0 ≤ rsin(θ) ≤ rcos(θ)√3 (2)
From equation (1), we can rewrite it as follows:
0 ≤ r ≤ 2/cos(θ) (3)
From equation (2), we divide both sides by r, assuming r is not zero:
0 ≤ sin(θ) ≤ cos(θ)√3 (4)
Now, we have expressed the limits of integration in polar form. Let's express the integrand in terms of polar coordinates.
The integrand is given as 1/√(𝑥^2+𝑦^2). Substituting x and y in terms of polar coordinates, we get:
1/√(r^2cos^2(θ) + r^2sin^2(θ))
Simplifying, we get:
1/√(r^2(cos^2(θ) + sin^2(θ)))
Since cos^2(θ) + sin^2(θ) = 1, the integrand simplifies to:
1/r
Now that we have the limits of integration and the integrand expressed in polar coordinates, we can rewrite the double integral as:
∫(θ=0 to θ=π/3) ∫(r=0 to r=2/cos(θ)) (1/r) rdrdθ
Evaluating this integral will give us the desired result.