Find area between surface $z^2=x^2+y^2$ and $x^2+y^2=2x$.
So, after using polar coordinates $x=r\cos \phi,y=r\sin \phi$ i get $0\le r \le 2\cos\phi$ and $0\le \phi \le \frac{\pi}{2}$. Than i plugged that into formula $\iint \sqrt{1+p^2+q^2}\,dx\,dy$ where $p=\frac{\partial z}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$ and $q=\frac{\partial z}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}$ and got $\frac{\sqrt{2}\pi}{2}$ but correct answer is $2\sqrt{2}\pi$. I probably need to multiply area with $4$ but honestly I'm not sure when to do that, on what it depends etc. I sketched this in plane(paraboloid and cylinder) and I can only see two symmetric areas (instead of four). If someone could clear this for me i would be very thankful.
Best Answer
It's a basic algebra exercise to see that the horizontal cross-section of your cyclinder is a circle centered at $(1,0,0)$ on the $xy$-plane.
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code$\lvert$$\bigodot$It's clear that in your cylindrical coordinates $\begin{cases}x = r\cos \phi& \\ y = r\sin \phi,& \end{cases}$, $\phi$ represents the angle between the segment from $(0,0)$ to $(x,y)$ and the $x$-axis. By setting $0\le \phi \le \dfrac{\pi}{2}$, you've only swept the first quadrant. However, by observing the symmetry along the $xz$-plane (i.e. $y = 0$),
it sufficesyou have to multiply your calculated value by $2$ to get thecorrect answerarea for one slice.As @MathLover has pointed out, there's another slide due to the symmetry along the $xy$-plane, so you have $$\frac{\sqrt2 \pi}{2} \cdot 4 = 2\sqrt2 \pi,$$ which is your desired answer.