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I have to find the parametric equation of the surface of the sphere inside the cylinder and above the $z=0$ plane, as shown in this picture.

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$$ \text{Sphere: }x^2 + y^2 + z^2 = 1\\ \text{Cylinder: }x^2 + y^2 = x $$

I came up with this:

$$ \vec r(x,y)=\left(x, y, \sqrt{1-(x^2 + y^2)}\right), \qquad (x,y) \in D $$

With $D$ being the projection of the cylinder on the $XY$ plane. However, it doesn't feel right, because the cylinder isn't on $(0,0)$.

Could it be this?

$$ \vec r(r,\theta)=\left(r \cos(\theta) + r, r \sin(\theta), \sqrt{1-r^2}\right), \qquad r\in[0,0.5],\ \theta\in [0,2\Pi] $$

Update: Seeing as the cylinder is shifted on the x axis (its center is on $(0.5, 0)$ with $r=0.5$), I thought about shifting the coordinate system so that $(0,0)$ is on the center of the cylinder. Here's the new equation and the new parametric surface:

$$ \text{Sphere: }(x+0.5)^2 + y^2 + z^2 = 1\\ \text{Cylinder: }x^2 + y^2 = 0.5^2 $$

$$ \vec r(r,\theta)= \begin{cases} r \cos(\theta) \\ r \sin(\theta) \\ \sqrt{1-(r \cos{\theta} + 0.5)^2 - r^2 \sin^2{\theta}} = \sqrt{0.75 -r^2 - r \cos{\theta}} \end{cases}, \ \ r\in[0,0.5],\ \theta\in [0,2\Pi] $$

Is this better?

Then I have to calculate $\vec r_r$ and $\vec r_\theta$.

$$ \text{A} = \int_0^{2\Pi} \int_0^{0.5} \left\lVert \vec r_r \times \vec r_\theta \right\rVert r\operatorname d\!r \operatorname d\!\theta\, $$

However, I haven't taken into consideration that it's only the surface above the $z=0$ plane. I could always divide the answer by 2, but the $z$ part of $\vec r(r,\theta)$ has a square root. Does that mean that it's only considering the positive part of the $z$ axis (because it has to be positive)?

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  • $\begingroup$ Where does the $+r$ come from? In polar coordinates $x = r\cos\theta$ $\endgroup$
    – Varun Iyer
    Aug 6, 2014 at 0:11
  • $\begingroup$ The cylinder is shifted to the right on the x axis (take a look at the cylinder's equation). Its center is actually on (r,0). That's why I added $+r$ to $\vec{x}(r,\theta)$. $\endgroup$ Aug 6, 2014 at 0:22
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    $\begingroup$ Your first answer definitely isn't right, since it's symmetric under a swap of $x$ and $y$ (i.e. reflection along the $y=x$) plane but your figure isn't. But your second one can't be right either, since $|\vec{r}|^2=1+2r\cos \theta\neq 1$ (your points aren't on the sphere). Your newest one, though, looks completely right. $\endgroup$ Aug 6, 2014 at 1:45
  • $\begingroup$ Yeah, those aren't right. I think the updated part is right, but I haven't taken into consideration that it's only the surface above the $z=0$ plane. I could always divide the answer by 2, but the $z$ part of $\vec r(r,\theta)$ has a square root. Does that mean that it's only considering the positive part of the $z$ axis (because it has to be positive)? $\endgroup$ Aug 6, 2014 at 2:05

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