Find the length of the curve intersection of cone and cylinder

Question

Given a cone z = $\sqrt{x^2+y^2}$ which intersects with $x^2+y^2-2ay=0$ and $a>0$ . If C is the curve of that intersection, then set up the integral or integrals for finding the length of this curve. I thought about a line integral but I wasn't able to parametrize the curve

Answer 1

The given $x^2+y^2-2ay$ describes a upright vertical cylinder centered on the $y$-axis at $(0,a,0)$ with radius $a$. Being vertical means the axis of the cylinder is parallel to $z$-axis.

The natural parametrization of any point of the 2-dim surface of the cylinder is $(\text{angle}, \text{height})$:$$(a \sin t,\, a(1 - \cos t),\, z)$$ where the parameter $t$ is the angle made with $-\hat{y}\,$ such that $t = 0$ starts at the origin and rotates counter-clockwise viewed from above.

Now, the intersection of the cylinder with the cone $z = \sqrt{x^2 + y^2}$ conveniently gives $z = \sqrt{2ay}$ just by definition of $x^2 + y^2 - 2ay = 0$. Therefore, the intersection curve $\mathcal{C}$ can be parametrized as$$a(\sin t,\, 1 - \cos t,\, \sqrt{2(1- \cos t)})$$

Now you can set up the arc length integral as \begin{align*} \int\sqrt{\mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2} &= \int \sqrt{ \bigl( \frac{\mathrm{d}x}{\mathrm{d}t} \bigr)^2 + \bigl( \frac{\mathrm{d}y}{\mathrm{d}t} \bigr)^2 + \bigl( \frac{\mathrm{d}z}{\mathrm{d}t} \bigr)^2 }\,\mathrm{d}t \\ &= a \sqrt{ (\cos t)^2 + (\sin t)^2 + \frac{ \sin^2 t }{2(1 - \cos t) } } \,\mathrm{d}t \\ & = \frac{ a }{ \sqrt{2} } \int \sqrt{3 + \cos t} \,\mathrm{d}t \end{align*}

I guess this is a calculus exercise that only requires you to set up the integral. In case you haven't notice, the intersection $\mathcal{C}$ is an ellipse, and the this is an elliptic integral that has no closed form in elementary functions.

The complete curve length (integrating $t = 0$ to $2\pi$) is $4\sqrt{2}\,a\,E\left(\frac12\right)$, where $E(m)$ is the complete elliptic integral of the 2nd kind (if you prefer wiki, see this). Numerically, $E\left(\frac12\right) \approx 1.350643881\ldots$