# Curve of intersection, value for parameter

This is for a line integral.

Parametrize the curve of intersection: \begin{align*} S_1: x^2+4y^2 + z^2 &= 4a^2, y<0\\ S_2: x+2y &= 0 \end{align*} Orientation from $(0,0,-2a)$ to $(0,0,2a)$.

Solution: $x=-2y$ \begin{align*} 8y^2+z^2 & = 4a^2 \\ \left(\dfrac{\sqrt{2}y}{a}\right)^2+\left(\dfrac{z}{2a}\right)^2&=1\\ \end{align*} The parameterization: \begin{align*} y &=\dfrac{a}{\sqrt{2}}\cos t\\ z &=2a\sin t\\ x &=-2y=-a\sqrt{2}\cos t\\ \mathbf{r}(t) &=(-a\sqrt{2}\cos t, \frac{a}{\sqrt{2}}\cos t, 2a\sin t) \end{align*} How should I find the value for the parameter t?

Two solutions in the book (I'm lost at both): $$\mathbf{r}(t) =(-a\sqrt{2}cost, \frac{a} {\sqrt{2}}cost, 2asint), t \in [\frac{\pi}{2}, \frac{3\pi}{2} ]$$ Alternative (how ???): $$\mathbf{r}(t) =(a\sqrt{2}sint, -\frac{a}{\sqrt{2}}sint, 2acost), t \in [0, \pi]$$

• Your question is not clear. Are you saying that you have the formula for $r(t)$ and you want to find the interval for $t$? – Rory Daulton Jul 3 '16 at 0:38
• Yes, the domain for t – JDoeDoe Jul 3 '16 at 8:01

You look at

$$\left(\dfrac{\sqrt{2}y}{a}\right)^2+\left(\dfrac{z}{2a}\right)^2=1$$

and you think of $\cos^2 t + \sin^2 t = 1$

From $\cos t = \dfrac{\sqrt{2}y}{a}$ you get $y = \dfrac{a}{\sqrt 2} \cos t$

From $\sin t = \dfrac{z}{2a}$ you get $z = 2a \sin t$

From $x = -2y$ you get $x = -\sqrt 2 a \cos t$

So your parameterized curve looks like

$r(t) = \left( -\sqrt 2 a \cos t,\; \dfrac{a}{\sqrt 2} \cos t,\; 2a \sin t \right)$

We still need to work out the domain of the parameter $t$.

We want r(t) to go from $(0,0,-2a)$ to $(0,0,2a)$ and we require $x \ge 0$ and $y \le 0$. (You wanted $y < 0$ but that won't work.) We know that $\cos t = 0$ when $t = \dfrac{\pi}{2}$ and when $t = \dfrac{3\pi}{2}$ and, by inspection, we see that this will work. that is $t \in \left[ \dfrac{\pi}{2}, \dfrac{3\pi}{2} \right]$