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Triple integral, spherical coordinates

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,

So when I change to space polar I Dont understand how facit got \(\displaystyle \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}\)
Regards,
\(\displaystyle |\pi\rangle\)

\(\displaystyle \int\int\int_D(x^2y^2z)dxdydz\)
where D is \(\displaystyle D={(x,y,z);0\leq z \leq \sqrt{x^2+y^2}, x^2+y^2+z^2 \leq 1}\)
 
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Re: Triple integral, spherical cordinates

Hint:
$$\left \{ \begin{matrix} z=\sqrt{x^2+y^2}\\x^2+y^2+z^2=1\end{matrix}\right.\Rightarrow \left \{ \begin{matrix} z^2=x^2+y^2\\x^2+y^2+z^2=1\end{matrix}\right. \Rightarrow 2z^2=1\Rightarrow z=\frac{\sqrt{2}}{2}$$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: Triple integral, spherical cordinates

Could you please re-post the original problem? The tinypic image is not showing up for me.
 

Petrus

Well-known member
Feb 21, 2013
739
Re: Triple integral, spherical cordinates

Hint:
$$\left \{ \begin{matrix} z=\sqrt{x^2+y^2}\\x^2+y^2+z^2=1\end{matrix}\right.\Rightarrow \left \{ \begin{matrix} z^2=x^2+y^2\\x^2+y^2+z^2=1\end{matrix}\right. \Rightarrow 2z^2=1\Rightarrow z=\frac{\sqrt{2}}{2}$$
\(\displaystyle z=\cos(\theta)P\)
what happend to P? I am somehow unsure with geting the integral limit

Regards,
\(\displaystyle |\pi\rangle\)
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Re: Triple integral, spherical cordinates

Using the spherical coordinates:
$$P(r,\theta,\varphi)\qquad (0\leq r <+\infty,\;0\leq \varphi \leq 2\pi,\;0\leq \theta \leq \pi),$$
the equality $\sqrt{2}/2=1\cos \theta$ implies $\theta=\pi/4,$ so:
$$D\equiv \left \{ \begin{matrix} 0\leq \varphi \leq 2\pi& \\\pi/4\leq \theta\leq \pi/2\\0\leq r\leq 1\end{matrix}\right.$$