# Triple Integral with Spherical Coordinates

#### Idyll

Calculate the Integral bellow, where T is the region inside the cylinder xÂ² + yÂ² = 1 and xÂ² + yÂ² + zÂ² = 4.

âˆ« âˆ« âˆ« (xÂ² + yÂ²)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 - yÂ²)
-1 < y < 1
sqrt(xÂ² + yÂ²) < z < sqrt(4 - xÂ² - yÂ²)

xÂ² + yÂ² + zÂ² = ÏÂ²;
ÏÂ² = 4;
Ï = 2

0 < Ï < 2

Then, by analysing y boundries:
-Ï€/2 < Î¸ < Ï€/2

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

2zÂ² = 4;
z = sqrt(2);

And I knew that z = Ïcos(Ï†):

Ïcos(Ï†) = sqrt(2);
2cos(Ï†) = sqrt(2);
cos(Ï†) = sqrt(2)/2 = cos(45Âº) = Ï€/4

Then:
-Ï€/4 < Ï† < Ï€/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):

âˆ« âˆ« âˆ« ÏÂ²sen(Ï†)(ÏÂ²senÂ²(Ï†)cosÂ²(Î¸) + ÏÂ²senÂ²(Î¸)senÂ²(Ï†))dÏdÎ¸Ï†
T

Is that right???

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