Parametrization the curve of intersection of sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$.

Question

I've seen similar posts here but none of the answers helped me. I am trying to parametrize a curve of intersection of a (top half $z>0$) sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$. I tried $$x=\frac{1}{2}\cos t $$ $$y=\frac{1}{2}+\frac{1}{2}\sin t$$ $$z=\sqrt{5-\left(\frac{1}{2}\cos t\right)^2-\left(\frac{1}{2}+\frac{1}{2}\sin t\right)^2}=\sqrt{\frac{9}{2}-\frac{1}{2}\sin t}$$ for $t \in (0,2\pi)$, but I don't think it is correct. Even if it was, is there a better (more simple) approach?

Note: I need to find the circulation of a field $F=(y+z,x-z,0)$ over this curve so I need a good parametrizatian so that I was able to integrate it.

Answer 1

Sphere $$ x^2 +y^2 + z^2= 5 \tag1 $$

simplify given second Cylinder equation

$$ x^2+y^2 -y=0 $$

Subtract

$$z^2+y = 5 \; \tag2 $$ To satisfy this relation now introduce parameter $t$ $$z^2 = 5 \cos^2 t,\;\tag3 $$ so that
$$ y = 5 \sin ^2 t \tag4 $$

From (1) and (4)

$$ z^2= 5 -x^2 -25 \sin ^4 t \tag 5 $$

Equate RHS of $z^2$ from (3) and (5) simplify to find parametrization for $x$

$$ x = \sqrt{5 \sin^2 t (1-5 \sin^2 t )}\tag6$$ From(3) $$ z= \sqrt 5 \cos t \tag 7$$

So parametrization for three coordinates

$$ (x,y,z)=\big( \sqrt{5 \sin^2 t (1-5 \sin^2 t )}, 5 \sin ^2 t ,\sqrt 5 \cos t \big) \tag8$$

where positive sign is taken before radical signs for top hemisphere.

Answer 2

A third alternative is using Stokes' theorem, parameterize the surface enclosed by the curve and integrate... And at first it seems like you are going to end up with a nicer integral

\begin{eqnarray} && \int_0^\pi d\phi \int_0^{\arcsin\left[\frac{1}{\sqrt{5}}\sin(\phi)\right]} d\theta \, \left[5 \sin(\theta)\right] (\nabla \times F)\cdot(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) = \\ && \int_0^\pi d\phi \int_0^{\arcsin\left[\frac{1}{\sqrt{5}}\sin\phi\right]} d\theta \, \, 5 \sqrt{2} \sin^2 \theta \sin \left( \phi + \frac{\pi}{4}\right) \\ && \int_0^\pi d\phi \, \frac{1}{2} \left(\cos \phi + \sin \phi \right) \left\{ 5 \arcsin \left[\frac{1}{\sqrt{5}}\sin\phi\right] - \sin\phi \sqrt{5 - \sin^2\phi}\right\} \end{eqnarray}

but, at least at first sight, it does not seem specially simpler than the integral you get in the line integrals.

Answer 3

Use the fact that the right-circular cylinder of radius $r$ centered at $(h, k)$ in the $xy$-plane can be parametrized by $\mathbf G(t, z) = \langle h + r \cos t, k + r \sin t, z \rangle$ by the Pythagorean Identity, i.e., $\cos^2 t + \sin^2 t = 1.$

Considering that we are in the intersection of the right-circular cylinder $(x - 0)^2 + \bigl(y - \tfrac 1 2 \big)^2 = \tfrac 1 4$ and the sphere $x^2 + y^2 + z^2 = 5,$ both of these equations hold. Using the parametrization $\mathbf G,$ we have that $x = \tfrac 1 2 \cos t$ and $y = \tfrac 1 2 + \tfrac 1 2 \sin t,$ hence plugging these identities into the equation $x^2 + y^2 + z^2 = 5$ gives rise to the equation $\tfrac 1 4 \cos^2 t + \tfrac 1 4 + \sin t + \frac 1 4 \sin^2 t + z^2 = 5.$ Of course, we can now solve for $z$ as a function $z(t)$ of $t.$

Our original parametriation $\mathbf G(t, z)$ can now be viewed as a parametrization $\mathbf G(t) = \langle \tfrac 1 2 \cos t, \tfrac 1 2 + \tfrac 1 2 \sin t, z(t) \rangle.$

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One other way to parametrize the intersection of these surfaces is by spherical coordinates $x = \sqrt 5 \sin \phi \cos \theta,$ $y = \sqrt 5 \sin \phi \sin \theta,$ and $z = \sqrt 5 \cos \theta.$ By plugging these identities into the equation of the right-circular cylinder, we obtain the equation $$5 \sin^2 \phi \cos^2 \phi + 5 \sin^2 \phi \sin^2 \theta - \sqrt 5 \sin \phi \sin \theta + \frac 1 4 = \frac 1 4.$$ By the Pythagorean Identity, this simplifies to $5 \sin^2 \phi = \sqrt 5 \sin \phi \sin \theta.$ Considering that $0 \leq \phi \leq \pi,$ and $\sin \phi > 0$ whenever $0 \leq \phi < \pi,$ we can solve for $\sin \theta = \sqrt 5 \sin \phi$ whenever $0 < \phi < \pi.$ But this implies that $\theta = \arcsin(\sqrt 5 \sin \phi),$ and this is not desirable.

Answer 4

The cylinder is

$$x^2+y^2=y,$$ implying (by the equation of the sphere)

$$y+z^2=5.$$

In terms of $y$,

$$\begin{cases}x^2=y-y^2,\\z^2=5-y.\end{cases}$$

Anyway, I don't know what you expect from a "good parameterization". (Will be hard to find a continuous parameterization that describes the two disconnected curves.)