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

-1

sqrt(xÂ² + yÂ²)

xÂ² + yÂ² + zÂ² = ÏÂ²;

ÏÂ² = 4;

Ï = 2

0

Then, by analysing y boundries:

-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 = Ïcos(Ï†):

Ïcos(Ï†) = sqrt(2);

2cos(Ï†) = sqrt(2);

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

Then:

-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« ÏÂ²sen(Ï†)(ÏÂ²senÂ²(Ï†)cosÂ²(Î¸) + ÏÂ²senÂ²(Î¸)senÂ²(Ï†))dÏdÎ¸Ï†

T

Is that right???

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__<__1sqrt(xÂ² + yÂ²)

__<__z__<__sqrt(4 - xÂ² - yÂ²)xÂ² + yÂ² + zÂ² = ÏÂ²;

ÏÂ² = 4;

Ï = 2

0

__<__Ï__<__2Then, by analysing y boundries:

-Ï€/2

__<__Î¸__<__Ï€/2And 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 = Ïcos(Ï†):

Ïcos(Ï†) = sqrt(2);

2cos(Ï†) = sqrt(2);

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

Then:

-Ï€/4

__<__Ï†__<__Ï€/4Now, 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« ÏÂ²sen(Ï†)(ÏÂ²senÂ²(Ï†)cosÂ²(Î¸) + ÏÂ²senÂ²(Î¸)senÂ²(Ï†))dÏdÎ¸Ï†

T

Is that right???

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