dunkelheit
New member
- Joined
- Sep 7, 2018
- Messages
- 48
I have a doubt about this triple integral:
\(\displaystyle \iiint_A ydxdydz\)
Where \(\displaystyle A=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2-2x<0,0<z<x,x^2+y^2<1,y>0\}\)
After using cylindrical coordinates I end up with
\(\displaystyle \iiint_B \rho^2 \sin \theta d\rho d\theta dz\)
Where \(\displaystyle B=\{(\rho \cos \theta,\rho \sin \theta,z)\in\mathbb{R}^3:\rho^2-2\rho \cos \theta<0,0<z<\rho \cos \theta,\rho^2<1,\rho \sin \theta>0\}\)
I find out that \(\displaystyle 0<z<\rho \cos \theta\) and that \(\displaystyle 0<\rho<\min \{1,2 \cos \theta \}\); until now I don't have any problems, I know that I have to split the integral in a sum of two integrals due to the fact that because of \(\displaystyle \theta\) the values of the upper bound for \(\displaystyle \rho\) can change.
The doubt is about discussing \(\displaystyle 0<z<\rho \cos \theta\), I have done it by letting \(\displaystyle \rho \cos \theta>0\) and obtaining \(\displaystyle \theta \in \left[0,\frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2},2\pi\right]\) and I know this is the correct way, but I'm not completely aware of why is it!
I mean that I'm not completely sure why I don't have to put \(\displaystyle \rho \cos \theta>z\) instead, because by the description of the domain I have the information that \(\displaystyle z>0\), so I have a lower bound on \(\displaystyle \rho \cos \theta\) that is \(\displaystyle \rho \cos \theta>z\) that implies \(\displaystyle \cos \theta>\frac{z}{\rho}\) and not simply \(\displaystyle \rho \cos \theta>0\).
Can someone help me understand why I can be sure to have \(\displaystyle \rho \cos \theta>0\) from \(\displaystyle 0<z<\rho \cos \theta\)?
Thanks for your time.
\(\displaystyle \iiint_A ydxdydz\)
Where \(\displaystyle A=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2-2x<0,0<z<x,x^2+y^2<1,y>0\}\)
After using cylindrical coordinates I end up with
\(\displaystyle \iiint_B \rho^2 \sin \theta d\rho d\theta dz\)
Where \(\displaystyle B=\{(\rho \cos \theta,\rho \sin \theta,z)\in\mathbb{R}^3:\rho^2-2\rho \cos \theta<0,0<z<\rho \cos \theta,\rho^2<1,\rho \sin \theta>0\}\)
I find out that \(\displaystyle 0<z<\rho \cos \theta\) and that \(\displaystyle 0<\rho<\min \{1,2 \cos \theta \}\); until now I don't have any problems, I know that I have to split the integral in a sum of two integrals due to the fact that because of \(\displaystyle \theta\) the values of the upper bound for \(\displaystyle \rho\) can change.
The doubt is about discussing \(\displaystyle 0<z<\rho \cos \theta\), I have done it by letting \(\displaystyle \rho \cos \theta>0\) and obtaining \(\displaystyle \theta \in \left[0,\frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2},2\pi\right]\) and I know this is the correct way, but I'm not completely aware of why is it!
I mean that I'm not completely sure why I don't have to put \(\displaystyle \rho \cos \theta>z\) instead, because by the description of the domain I have the information that \(\displaystyle z>0\), so I have a lower bound on \(\displaystyle \rho \cos \theta\) that is \(\displaystyle \rho \cos \theta>z\) that implies \(\displaystyle \cos \theta>\frac{z}{\rho}\) and not simply \(\displaystyle \rho \cos \theta>0\).
Can someone help me understand why I can be sure to have \(\displaystyle \rho \cos \theta>0\) from \(\displaystyle 0<z<\rho \cos \theta\)?
Thanks for your time.