Parameterize and find the arc length

Question

Parameterize the following curves in $\mathbb{R}^3$ and find their arc length:

• the intersection of the cylinder $x^2+y^2=1$ and the plane $x+y+z=1$
• the intersection of the sphere $x^2+y^2+z^2=1$ and the plane $x+y+z=1$

For the first part, I parameterized by $x = \sqrt{1-y^2}$ and $z = 1-x-y$. Then I substituted $y = \sin(t)$ for $0 \le t \le 2\pi$. So I got $r(t): \left<\cos(t),\sin(t),1-\cos(t)-\sin(t)\right>$. Is this a right way to approach to get the r(t)?

Also, when I am trying to use the arc length formula, how should I determine the "range" of an integral? We need some sort of range like from a to b in order to get arc length.

For second part, I just put $x^2 + y^2 + z^2 = x+y+z$ then made the equation $x(x-1) + y(y-1) + z(z-1)=0$, but then after here I am stuck on how to parameterize like the first part. Also, if I have to get the arc length for this curve, how should I set the "range"?

Thank you for your help!

Answer 1

• the intersection of the sphere $x^2+y^2+z^2=1$ and the plane $x+y+z=1$

Eliminate $z$

$$ x^2+y^2+(1-x-y)^2= 1\rightarrow x^2+y^2 +xy -x-y =0 \tag1 $$

Note that a surface is 2-parametered, while intersection is 1- parametered. So we find one parameter as a function of the other.

Solve for $y(x)$

$$ 2 y(x)= (1-x) \pm \sqrt {(1-x)(1-3x)} \tag2$$

suggested parametrization

$$ [x,2y,2z] =\big[t, (1-t)-\sqrt{(1-t)(1+3t)}, (1-t)+\sqrt{(1-t)(1+3t)}\big] \tag3$$

The parametrization can exist only if $ (1>t>-\dfrac13) $

Arc length can be found by integration between these limits.

$$ \int_0^{2 \pi}(\sqrt{x^{'2}+y^{'2}+z^{'2} })\,dt $$

The small circle circumference can be analytically evaluated.

Sketched the plane and sphere to visually verify the parametrization.

Same ( easier) procedure for cylinder/plane intersection.

Using direction cosines the slant plane works out as $ \cos^{-1}\frac{1}{\sqrt3}$. On flat development the intersection elliptic arc becomes a sine-curve $ z=\sqrt{3} \sin \, (s)$ whose analytical solution is $ a E(\theta, e)$ in terms of second order elliptic integrals.

Answer 2

The intersection of the cylinder $x^2+y^2=1$ with the plane $x+y+z=1$ is an ellipse $E$ going around the cylinder. As $E$ projects bijectively onto the unit circle in the $(x,y)$-plane we use the representation $$t\mapsto(\cos t,\sin t)\qquad(0\leq t\leq2\pi)$$ for this circle and then compute $$z=1-x-y=1-\cos t-\sin t\ .$$ This leads to your parametrization $$E:\quad t\mapsto{\bf r}(t)=(\cos t,\sin t, 1-\cos t-\sin t)\qquad(0\leq t\leq2\pi)\ .$$ For the length of $E$ we need $${\bf r}'(t)=(-\sin t,\cos t,\sin t-\cos t)\ ,$$ so that $$L(E)=\int_0^{2\pi}|{\bf r}'(t)|\>dt=\int_0^{2\pi}\sqrt{2-2\cos t\sin t}\>dt\ .$$ This is an elliptic integral. It cannot be evaluated in terms of elementary functions.