evaluate the triple integral

∫(-1 to 1) ∫(0 to 1-x²) ∫(0 to √y) x²y²z² dzdydz

Don't just want the answer cuz everyone can use wolframalpha. I need step by step work.

Been stuck on this problem for a while now, so Willing to pay up to 300fg with clear steps

have you tried changing the order of your integration?

Let I = ∫(-1 to 1) ∫(0 to 1-x²) ∫(0 to √y) x²y²z² dzdydz

Step 1 : assume that x,y are constant, and integrate in regards to z :
I = ∫(-1 to 1) ∫(0 to 1-x²) x²y² ∫(0 to √y) z²dzdydz

(z^3) / 3 is an anti-derivative of z², hence :
∫(0 to √y) z²dz = [ (z^3/3 ] = (√y)^3 / 3

Giving you :
I = ∫(-1 to 1) ∫(0 to 1-x²) x²y² (√y)^3 / 3 dydz

I =(1/3) ∫(-1 to 1) ∫(0 to 1-x²) x²y^(7/2) dydx
since y² (√y)^3 = y^(7/2)

Step 2, assume that x is a constant :

I =(1/3) ∫(-1 to 1) x²∫(0 to 1-x²) y^(7/2) dydx

and (2/9) y^(9/2) is an anti-derivative of y^(7/2) :

∫(0 to 1-x²) y^(7/2) dy = (2/9) [ y^(9/2) ] = (2/9).(1-x²)^(9/2)

Leaving you with :

I =(2/27) ∫(-1 to 1) x²(1-x²)^(9/2)dx
I =(4/27) ∫(0 to 1) x²(1-x²)^(9/2)dx
because the function is even.

Step 3 : Integrate by parts : let F = x, G' = x(1-x²)^(9/2)

I =(4/27) { [ FG ] - ∫ F' G }
I = (4/27) { 0 +(1/11) ∫ (0 to 1) (1-x²)^(11/2) dx }
I = (4/297) ∫ (0 to 1) (1-x²)^(11/2) dx

Step 4 : Now let
x = sin u
dx / du = cos u
dx = cos u du

(1-x²) = 1 - sin² u = cos² u

I =(4/297) ∫(0 to pi/2) (cos² u)^(11/2) cos u du
I =(4/297) ∫(0 to pi/2) (cos u)^12 du

Step 5 : Now it's time for Pascal's triangle : we want to express cos u^12 in terms of cos(12u), cos(10u), cos(8u) ...

cos u ^ 12 = [ ( exp iu + exp -iu ) / 2 ]^12
cos u ^ 12 = (1/2^12).{ exp 12iu + 12.exp 10iu + 66.exp 8iu + 220. exp 6iu + 395.exp 4iu + 792.exp 2iu + 924 + 792.exp -2iu + 395.exp -4iu + 220.exp -6iu + 66.exp -8iu + 12.exp -10iu + exp -12iu }
cos u ^ 12 = (1/2^13).{ cos 12u + 12.cos 10u + 66.cos 8u + 220.cos 6u + 395.cos 4u + 792.cos 2u + 1848 }

∫ cos u ^ 12 du = (1/2^13).{ -(sin 12u)/12 - (12/10).sin 10u - (66/8).sin 8u - (220/6).sin 6u - (395/4).sin 4u + (792/2).sin 2u + 1848.u }
Taking the value between 0 and pi/2 leaves us with (1/2^13)*1848*pi/2 = 1848.pi/16384 = 231.pi/2048

Notice that every (sin pu), when p is even, gives 0 between 0 and pi/2, so in practise, you don't even need to know the coefficients. Only the constant term 1848 is useful.

And in the end, I = (4/297)*(231.pi/2048) = 7.pi / 4608

I'm not sure there's any faster way... but I could be wrong !

thanks. could you explain the pascals triangle part a bit more?

This post was edited by 2wo1ne on Nov 19 2014 08:24pm

Pascal's triangle is used to expand expressions of the form : (a+b)^n, in an abelian (commutative) ring :

(a+b)^1 = 1*a + 1*b
(a+b)^2 = 1*a² + 2*ab + 1*b²
(a+b)^3 = 1*a^3 + 3*a²b + 3*ab² + 1*b^3
(a+b)^4 = 1*a^4 + 4*a^3b + 6*a²b² + 4*ab^3 + 1*b^4

and so on...

The coefficients involved in the process are the binomial coefficients.

Let's rewrite them :

1 1
1 2 1
1 3 3 1
1 4 6 4 1
...

This is the Pascal's triangle. You can find any coefficient by adding up 2 coefficients of the previous raw.

For a 12th power, you need to go on until the 12th raw :
...
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 395 792 924 792 395 220 66 12 1

The bolded coefficient is the only useful one for your purpose.

integrating first with respect to z then y are fairly basic.

I = ∫(-1 to 1) ∫(0 to 1-x²) ∫(0 to √y) x²y²z² dzdydz

(2/27)∫(-1 to 1)∫(0 to 1-x²) x²y² y^(3/2) dydx

(2/27) ∫(-1 to 1) x²(1-x²)^(9/2)dx

this last integral is the trickiest,
let x = sinθ
dx = cosθdθ

I = (2/27) ∫(-π/2 to π/2) sin²θ(1-sin²θ)^(9/2)cosθ dθ

=(2/27) ∫(-π/2 to π/2) sin²θ(cos¹⁰θ) dθ

= (2/27) [∫(-π/2 to π/2) cos¹⁰θ dθ - ∫(-π/2 to π/2) cos¹²θ dθ]

we can solve this using the relation

I_k = ∫(0 to π/2) cos^k(θ) dθ = ∫(0 to π/2) sin^k(θ) dθ =

{ π/(2^(2m+1) * 2mCm if k = 2m
{ (2^(2m))/[(2m+1)(2mCm)] if k = 2m+1

which is known as wallis' integral and I can prove it for you but you can probably find a proof somewhere on the internet. (mostly uses integration by parts and some algebra)
we know cos is an even function from -π/2 to π/2 therefore we just multiply our result by 2
since both our integrals have k = 2m we get

(2/27) [ (π/2¹⁰) * 10C5 - (π/2¹²) * 12C6] = 7π/4608
Resulting in the same value as feanur