evaluate the triple integral of x√(y^2+4) that lies above the parabolic cylinder y^2 +4Z=16 and the planes x=0?

Bounds for z:

z = 0 to z = 4 - y^2/4.

This projects to y^2 = 16 ==> y = -4 or 4, along with x = 0 to x = 1.

So, the integral equals

∫(x = 0 to 1) ∫(y = -4 to 4) ∫(z = 0 to 4 - y^2/4) x√(y^2 + 4) dz dy dx

= ∫(x = 0 to 1) ∫(y = -4 to 4) (4 - y^2/4) x√(y^2 + 4) dy dx

= [∫(x = 0 to 1) x dx] * [∫(y = -4 to 4) (4 - y^2/4) √(y^2 + 4) dy]

= (1/2) * [2 ∫(y = 0 to 4) (4 - y^2/4) √(y^2 + 4) dy]

= ∫(y = 0 to 4) (1/4) (16 - y^2) √(y^2 + 4) dy.

Now, let y = 2 tan θ, dy = 2 sec^2(θ) dθ.

==> ∫(θ = 0 to arctan(1/2)) (1/4) (16 - 4 tan^2(θ)) * 2 sec θ * (2 sec^2(θ) dθ)

= ∫(θ = 0 to arctan(1/2)) [16 - 4 (sec^2(θ) - 1)] sec^3(θ) dθ

= ∫(θ = 0 to arctan(1/2)) [20 sec^3(θ) - 4 sec^5(θ)] dθ

Using the reduction formula (obtained from integration by parts)

∫ sec^n(x) dx = (1/(n-1)) * [sec^(n-2)(x) tan x + (n - 2) ∫ sec^(n-2)(x) dx]:

∫ sec^3(x) dx = (1/2) [sec x tan x + ∫ sec x dx]

...................= (1/2) [sec x tan x + ln |sec x + tan x|] + C

∫ sec^5(x) dx = (1/4) [sec^3(x) tan x + 3 ∫ sec^3(x) dx]

...................= (1/4) {sec^3(x) tan x + (3/2) [sec x tan x + ln |sec x + tan x|]} + C

...................= (1/8) [2 sec^3(x) tan x + 3 sec x tan x + 3 ln |sec x + tan x|] + C

Therefore, ∫(θ = 0 to arctan(1/2)) [20 sec^3(θ) - 4 sec^5(θ)] dθ

= 10 [sec x tan x + ln |sec x + tan x|] - (1/2) [2 sec^3(x) tan x + 3 sec x tan x + 3 ln |sec x + tan x|]

{for θ = 0 to arctan(1/2)}

= (1/2) [17 sec x tan x + 17 ln |sec x + tan x| - 2 sec^3(x) tan x] {for θ = 0 to arctan(1/2)}

= (1/2) [17(√5/2)(1/2) + 17 ln |√5/2 + 1/2| - 2 (√5/2)^3 (1/2)] - 0

= (1/2) [17√5/4 + 17 ln [(1 + √5)/2] - 5√5/8]

I hope this helps!

wow i feel stupid