Can someone check my work? The question was: find a parametric representation of the portion of the surface $x+3y-z=5$ with $x\geq0, y\geq0$, and $x^2+y^2\leq 1$.

I answered: $x=\cos\theta$, $y=\sin\theta$ and $z=\cos\theta + 3\sin\theta -5$ with $0\leq\theta\leq \frac{\pi}{2}$.

Thanks in advance.

  • $\begingroup$ Are you looking for the parametrization of the surface contained in the cylinder, or the parametrization of the intersection? What you have so far is the intersection of the cylinder and the plane. We know this because any point $(\cos \theta, \sin \theta, \cdot)$ lies on the unit circle. Therefore it does not represent the plane region contained inside the cylinder. $\endgroup$ – MathMajor Jan 2 '15 at 1:56

Based on the wording of your question. No, you are not correct. As I stated in the comments, any point $(\cos \theta, \sin \theta, \cdot )$ lies strictly on the unit circle on the $xy$-plane. What you want is the plane region entrapped in the cylinder $x^2 + y^2 \le 1$.

Here is a plot of both surfaces for $x \ge 0$, and $y \ge 0$:

enter image description here

Here is your parametrized curve $\vec r(\theta) = \langle \cos \theta, \sin \theta, \cos \theta +3 \sin \theta - 5 \rangle$:

enter image description here

It should be clear that your parametrization has only captured the intersection of these two surfaces.


A correct parametrization could be:

$$ z = r \cos \theta + 3r \sin \theta - 5$$

where $0 \le r \le 1$ and $0 \le \theta \le \frac{\pi}{2}$ in the cylindrical coordinate system.

  • $\begingroup$ So how would I parameterize the region bounded by that curve on the plane? $\endgroup$ – Gabriel Jan 2 '15 at 2:18
  • $\begingroup$ See my edited answer. $\endgroup$ – MathMajor Jan 2 '15 at 2:37

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