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Triple integral

laura123

Member
May 2, 2013
28
Hi! I have some problems with the integral
\(\displaystyle \iiint_T\sqrt{x^2+y^2}z^4e^{z^4}dx\ dy\ dz\)
where
\(\displaystyle T=\{(x,y,z)\in\mathbb{R}^3:\sqrt{x^2+y^2}\leq z\leq 1\}\)
I have tried to change it to spherical and cylindrical coordinates but... nothing
Can someone help me?
Thanks
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: triple integral

Hi! I have some problems with the integral
\(\displaystyle \iiint_T\sqrt{x^2+y^2}z^4e^{z^4}dx\ dy\ dz\)
where
\(\displaystyle T=\{(x,y,z)\in\mathbb{R}^3:\sqrt{x^2+y^2}\leq z\leq 1\}\)
I have tried to change it to spherical and cylindrical coordinates but... nothing
Can someone help me?
Thanks
Wellcome on MHB laura!... may be that the best way is to separate the integrations in the following way...

$\displaystyle I = \int_{A} \sqrt{x^{2}+y^{2}}\ dx\ dy\ \int_{\sqrt{x^{2}+y^{2}}}^{1} z^{4}\ e^{z^{4}}\ dz$ (1)

... where A is the unit circle on the xy plane. First You solve the integral in z and then You can change the integral in x and y in polar coordinates...

Kind regards

$\chi$ $\sigma$
 

laura123

Member
May 2, 2013
28
Re: triple integral

Thanks $\chi\sigma$ but the problem is in $\int z^4 e^{z^4}dz$...
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: triple integral

Hi! I have some problems with the integral
\(\displaystyle \iiint_T\sqrt{x^2+y^2}z^4e^{z^4}dx\ dy\ dz\)
where
\(\displaystyle T=\{(x,y,z)\in\mathbb{R}^3:\sqrt{x^2+y^2}\leq z\leq 1\}\)
I have tried to change it to spherical and cylindrical coordinates but... nothing
Can someone help me?
Thanks
Step 1. Convert to cylindrical coordinates: $$\iiint_T\sqrt{x^2+y^2}z^4e^{z^4}dx\, dy\, dz = \int_0^1\int_r^1\int_0^{2\pi} rz^4e^{z^4}\,rd\theta\, dz\, dr = 2\pi\int_0^1\int_r^1r^2z^4e^{z^4}dz\, dr$$ (doing the $\theta$ integral first).

Step 2. Change the order of integration. The integral is over a triangular region in the $(r,z)$-plane. If $z$ goes from $r$ to $1$ (and then $r$ goes from $0$ to $1$), then when you switch the order, you find that $r$ goes from $0$ to $z$ (and then $z$ goes from $0$ to $1$). Thus the integral becomes $$2\pi\int_0^1\int_0^zr^2z^4e^{z^4}dr\,dz = 2\pi\int_0^1\frac{z^3}3z^4e^{z^4}dz,$$ which you can evaluate by making the substitution $u=z^4$.