Given \(\displaystyle \Omega = \left\{(x,y,z) \in \mathbb{R}^3 \ | \ 1 < x^2+y^2 < 2x, 0 < z < \frac{x^2+y^2}{x^2} \right\}\), evaluate \(\displaystyle \iiint_{\Omega} \frac{y^2}{x^2+y^2} dxdydz\).
I used cylindrical coordinates \(\displaystyle x=r \cos t, y= r \sin t, z=z\) with \(\displaystyle 0 \leq t \leq 2\pi\) and \(\displaystyle r \geq 0\): so the integral becomes
\(\displaystyle \iiint_{D} r \sin^2 t drdtdz\)
With \(\displaystyle D=\left\{(r,t,z) \in \mathbb{R}^3 \ | \ 1 < r^2 < 2 r \cos t,0 < z < \frac{1}{\cos^2 t}\right\}\).
It is then \(\displaystyle 1 < r^2 \iff r>1\) (because \(\displaystyle r \geq 0\)) and it is \(\displaystyle r^2 <2r \cos t \iff r < 2 \cos t\) so it is \(\displaystyle 1< r < 2 \cos t\). The condition on \(\displaystyle z\) is already determined, while \(\displaystyle 0 < z < \frac{1}{\cos^2 t}\) implies \(\displaystyle \frac{1}{\cos^2 t} >0\) which is always verified.
I am not sure how to find other conditions on \(\displaystyle t\) from the inequalities \(\displaystyle 1<r^2<2r \cos t\): maybe from \(\displaystyle 1< r < 2 \cos t\) I can deduce that \(\displaystyle \cos t>\frac{1}{2}\) and so \(\displaystyle 0< t < \pi/3 \lor 5\pi/3 < t < 2\pi\)? Could this work?
Another question: the inequalities like \(\displaystyle a<b<c\) mean in logic "\(\displaystyle a<b\) and \(\displaystyle b<c\) and \(\displaystyle a<c\)"? I believe it is something like "\(\displaystyle a<b\) and \(\displaystyle b<c\)" and this implies \(\displaystyle a<c\), because if it was "and \(\displaystyle a<c\)" instead of the implicatuon then I couldn't deduce \(\displaystyle \cos t>\frac{1}{2}\) from \(\displaystyle 1< r < 2 \cos t\).
I used cylindrical coordinates \(\displaystyle x=r \cos t, y= r \sin t, z=z\) with \(\displaystyle 0 \leq t \leq 2\pi\) and \(\displaystyle r \geq 0\): so the integral becomes
\(\displaystyle \iiint_{D} r \sin^2 t drdtdz\)
With \(\displaystyle D=\left\{(r,t,z) \in \mathbb{R}^3 \ | \ 1 < r^2 < 2 r \cos t,0 < z < \frac{1}{\cos^2 t}\right\}\).
It is then \(\displaystyle 1 < r^2 \iff r>1\) (because \(\displaystyle r \geq 0\)) and it is \(\displaystyle r^2 <2r \cos t \iff r < 2 \cos t\) so it is \(\displaystyle 1< r < 2 \cos t\). The condition on \(\displaystyle z\) is already determined, while \(\displaystyle 0 < z < \frac{1}{\cos^2 t}\) implies \(\displaystyle \frac{1}{\cos^2 t} >0\) which is always verified.
I am not sure how to find other conditions on \(\displaystyle t\) from the inequalities \(\displaystyle 1<r^2<2r \cos t\): maybe from \(\displaystyle 1< r < 2 \cos t\) I can deduce that \(\displaystyle \cos t>\frac{1}{2}\) and so \(\displaystyle 0< t < \pi/3 \lor 5\pi/3 < t < 2\pi\)? Could this work?
Another question: the inequalities like \(\displaystyle a<b<c\) mean in logic "\(\displaystyle a<b\) and \(\displaystyle b<c\) and \(\displaystyle a<c\)"? I believe it is something like "\(\displaystyle a<b\) and \(\displaystyle b<c\)" and this implies \(\displaystyle a<c\), because if it was "and \(\displaystyle a<c\)" instead of the implicatuon then I couldn't deduce \(\displaystyle \cos t>\frac{1}{2}\) from \(\displaystyle 1< r < 2 \cos t\).
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