Triple Integral with Spherical Coordinates

Aug 2017
2
0
Brazil
Calculate the Integral bellow, where T is the region inside the cylinder x² + y² = 1 and x² + y² + z² = 4.

∫ ∫ ∫ (x² + y²)dV
T

I'm having trouble to convert the coordinates...
I did it like this (feel free to correct me, after all, probrably I'm wrong)

0 < x <sqrt(1 - y²)
-1 < y < 1
sqrt(x² + y²) < z < sqrt(4 - x² - y²)

x² + y² + z² = ρ²;
ρ² = 4;
ρ = 2

0 < ρ < 2

Then, by analysing y boundries:
-π/2 < θ < π/2

And then, to find φ, I merged the two equations above I used to find the boundries:

2z² = 4;
z = sqrt(2);

And I knew that z = ρcos(φ):

ρcos(φ) = sqrt(2);
2cos(φ) = sqrt(2);
cos(φ) = sqrt(2)/2 = cos(45º) = π/4

Then:
-π/4 < φ < π/4

Now, with the boundries found, I set the Integral ready like this (since it's very hard to place the boundries, imagine that I did it bellow):

∫ ∫ ∫ ρ²sen(φ)(ρ²sen²(φ)cos²(θ) + ρ²sen²(θ)sen²(φ))dρdθφ
T

Is that right???
 
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