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### Section 6-2 : Parametric Surfaces

4. The portion of $$y = 4 - {x^2} - {z^2}$$ that is in front of $$y = - 6$$.

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Okay, the basic set of parametric equations in this case is pretty easy since we already have the equation in the form of “$$y$$ = ”.

The set of parametric equations that will give the full surface is just,

$\vec r\left( {x,z} \right) = \left\langle {x,y,z} \right\rangle = \left\langle {x,4 - {x^2} - {z^2},z} \right\rangle$

Remember that all we need to do to get the parametric equations is plug in the equation for $$y$$ into the $$y$$ component of the vector $$\left\langle {x,y,z} \right\rangle$$.

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Finally, all we need to do is restrict $$x$$ and $$z$$ to get only the portion of the surface we are looking for. That is pretty simple however since we are given that we only want the portion that is in front of $$y = - 6$$.

This is equivalent to requiring that $$y \ge - 6$$ and we do have the equation of the surface so all we need to do is plug that into the inequality and do a little rewrite. Doing this gives,

$4 - {x^2} - {z^2} \ge - 6\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,{x^2} + {z^2} \le 10$

In other words, we only want the points $$\left( {x,z} \right)$$ that are inside the disk of radius $$\sqrt {10}$$.

Putting all of this together gives the following set of parametric equations for the portion of the surface we are after.

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( {x,z} \right) = \left\langle {x,4 - {x^2} - {z^2},z} \right\rangle \,\,\,\,\,\,\,\,{x^2} + {z^2} \le 10}}$