note that you have ∫∫ 1/r dA
so can you analyze the graph and determine the domain in polar coordinates?
Use polar coordinates to evaluate the integral [0,2]∫ [0,x√3]∫1/√(𝑥^2+𝑦^2)dydx
3 answers
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
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|
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|