evaluate integral_0^π integral_0^(π/4) integral_0^sec(θ) sin(2 ϕ) dρ dϕ dθ

Question

please help

Answer 1

 

Evaluate the innermost integral first, where [low limit, high limit]:

(evaluate high limit value and subtract the lower limit value)

∫[0,sec(θ)]sin(2ϕ)dρ=[0,sec(θ)](ρ)sin(2ϕ)=sec(θ)sin(2ϕ) (sin(2ϕ) is considered as a constant in this integral).

Now the second integral:

∫[0,π/4]sec(θ)sin(2ϕ)dϕ=

[0,π/4](-½cos(2ϕ))sec(θ)=

(-½cos(π/2)-(-½cos(0)))sec(θ)=sec(θ)/2 (sec(θ) is considered as a constant in this integral).

Final integral:

∫[0,π](sec(θ)/2)dθ.

To integrate ½sec(θ), multiply by (sec(θ)+tan(θ))/(sec(θ)+tan(θ)):

½(sec²(θ)+sec(θ)tan(θ))/(tan(θ)+sec(θ)).

Integrate wrt θ and we get ½ln(tan(θ)+sec(θ)).

Apply the limits [0,π]:

θ=π: ½ln(0-1) is undefined; θ=0: ½ln(0+1)=0.

Therefore, the triple integral is undefined.