# Chapter 16 - Vector Calculus - 16.6 Parametric Surfaces and Their Areas - 16.6 Exercises - Page 1160: 21

$x=\frac{\sqrt {4+4y^{2}+z^{2}}}{2},y=y,z=z$

#### Work Step by Step

Given:The part of the hyperboloid $4x^{2}-4y^{2}-z^{2}= 4$ that lies in front of the $yz$ plane. We need to use two parameters for the surface when we parametrize it. Since, it lies in the front of the $yz$ plane, we can let $y$ and $z$ be the independent variables and express $x$ in terms of $y$ and $z$. $4x^{2}-4y^{2}-z^{2}= 4$ $4x^{2}=4+4y^{2}+z^{2}$ $x^{2}=\frac{4+4y^{2}+z^{2}}{4}$ $x=\sqrt {\frac{4+4y^{2}+z^{2}}{4}}$ or $x=\frac{\sqrt {4+4y^{2}+z^{2}}}{2}$ We choose the positive square root (rather than negative square root) because it lies in the front of the $yz$ plane. Hence, the parametric representation of the hyperboloid is $x=\frac{\sqrt {4+4y^{2}+z^{2}}}{2},y=y,z=z$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.