0
$\begingroup$

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 ???): $$\mathbf{r}(t) =(a\sqrt{2}sint, -\frac{a}{\sqrt{2}}sint, 2acost), t \in [0, \pi]$$

$\endgroup$
  • $\begingroup$ Your question is not clear. Are you saying that you have the formula for $r(t)$ and you want to find the interval for $t$? $\endgroup$ – Rory Daulton Jul 3 '16 at 0:38
  • $\begingroup$ Yes, the domain for t $\endgroup$ – JDoeDoe Jul 3 '16 at 8:01
1
$\begingroup$

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 $t \in \left[ \dfrac{\pi}{2}, \dfrac{3\pi}{2} \right]$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.