Triple integral, spherical coordinates

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Discussion Overview

The discussion revolves around the evaluation of a triple integral using spherical coordinates, specifically addressing the limits of integration for the angle theta. Participants are exploring the transformation of the given integral and the geometric interpretation of the region of integration.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about how the limits for theta were derived, specifically questioning the range $$\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$$.
  • Another participant provides a hint involving the equations $$z=\sqrt{x^2+y^2}$$ and $$x^2+y^2+z^2=1$$, leading to the conclusion that $$z=\frac{\sqrt{2}}{2}$$.
  • A participant requests the original problem statement due to an image not displaying correctly.
  • One participant reiterates the hint about the equations and expresses uncertainty regarding the variable P in the context of the integral limits.
  • Another participant states that using spherical coordinates leads to the conclusion that $$\theta=\pi/4$$, and provides a proposed set of limits for the region D.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the limits of integration for theta, with multiple interpretations and hints being offered. The discussion remains unresolved regarding the correct limits and the role of variable P.

Contextual Notes

There are limitations in the clarity of the original problem due to an image not being displayed, which may affect participants' understanding and contributions. Additionally, the derivation of limits involves assumptions about the geometric interpretation of the region D.

Petrus
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Hello MHB,
1y4d39.jpg

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

$$\int\int\int_D(x^2y^2z)dxdydz$$
where D is $$D={(x,y,z);0\leq z \leq \sqrt{x^2+y^2}, x^2+y^2+z^2 \leq 1}$$
 
Last edited:
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Re: Triple integral, spherical cordinates

Hint:
$$\left \{ \begin{matrix} z=\sqrt{x^2+y^2}\\x^2+y^2+z^2=1\end{matrix}\right.\Rightarrow \left \{ \begin{matrix} z^2=x^2+y^2\\x^2+y^2+z^2=1\end{matrix}\right. \Rightarrow 2z^2=1\Rightarrow z=\frac{\sqrt{2}}{2}$$
 
Re: Triple integral, spherical cordinates

Could you please re-post the original problem? The tinypic image is not showing up for me.
 
Re: Triple integral, spherical cordinates

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

Regards,
$$|\pi\rangle$$
 
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:
$$D\equiv \left \{ \begin{matrix} 0\leq \varphi \leq 2\pi& \\\pi/4\leq \theta\leq \pi/2\\0\leq r\leq 1\end{matrix}\right.$$
 

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