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 May 1st, 2013, 12:05 AM #1 Newbie   Joined: Apr 2013 Posts: 2 Thanks: 0 triple integral Hi! I have some problems with the integral $\iiint_T\sqrt{x^2+y^2}\ z^4e^{z^4} dx\ dy\ dz$ where $T=\{(x,y,z)\in\mathbb{R}^3:\sqrt{x^2+y^2}\leq z\leq 1\}$. I have to change it to spherical coordinates but... nothing.. can someone help me? Thanks
May 1st, 2013, 12:16 PM   #2
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Re: triple integral

Quote:
 Originally Posted by laura123 Hi! I have some problems with the integral $\iiint_T\sqrt{x^2+y^2}\ z^4e^{z^4} dx\ dy\ dz$ where $T=\{(x,y,z)\in\mathbb{R}^3:\sqrt{x^2+y^2}\leq z\leq 1\}$. I have to change it to spherical coordinates but... nothing.. can someone help me? Thanks
I suggest you convert to cylindrical coordinates: (x,y) -> (r,?) and z -> z.

 May 1st, 2013, 03:37 PM #3 Newbie   Joined: Apr 2013 Posts: 2 Thanks: 0 Re: triple integral thank you mathman, but the problem is in the integral $\int z^4 e^{z^4}dz$...it's a nonelementary integral..
 May 2nd, 2013, 03:48 PM #4 Global Moderator   Joined: May 2007 Posts: 6,527 Thanks: 588 Re: triple integral gradshteyn and ryzhik pdf download Google the above. You can get it free. Look for atsol..... If you can't find it there, it probably can't be done analytically.
May 3rd, 2013, 07:19 AM   #5
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Re: triple integral

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 Originally Posted by laura123 $\int z^4 e^{z^4}dz$
The integral is equivalent to $\frac{z e^{z^4}}{4} - \xi_{4} \frac{\Gamma$$\frac{1}{4}, -z^4$$}{16}$ where $\xi_4$ is the 4-th root of unity.

In fact, this can be arbitrarily generalized for any exponent replaced by 4 in the given integral.

May 9th, 2013, 10:11 PM   #6
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Re: triple integral

Quote:
 Originally Posted by laura123 Hi! I have some problems with the integral $\iiint_T\sqrt{x^2+y^2}\ z^4e^{z^4} dx\ dy\ dz$ where $T=\{(x,y,z)\in\mathbb{R}^3:\sqrt{x^2+y^2}\leq z\leq 1\}$. I have to change it to spherical coordinates but... nothing.. can someone help me? Thanks
Assuming I have this right...

$\iiint_T\ \rho^2\sin(\phi)\sqrt{(\rho\sin(\phi)\cos(\theta)) ^2 + (\rho\sin(\phi)\sin(\theta))^2}\ {(\rho\cos(\phi))}^4e^{(\rho\cos(\phi))^4}\ d\theta d\phi d\rho$

And simplify? Or am I completely wrong?

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