# Chapter 14 - Partial Derivatives - 14.1 Functions of Several Variables - 14.1 Exercises: 22

The domain is $$\mathcal{D}=\left\{(x,y,z)\left|\frac{x^2}{2^2}+\frac{y^2}{2^2}+\frac{z^2}{4^2}<1\right.\right\}$$ and it is presented in the figure below

#### Work Step by Step

The argument of the logarithm has to be positive so we need that $16-4x^2-4y^2-z^2>0$ which can be rewritten as $$4x^2+4y^2+z^2<16.$$ Dividing this by $16$ we get $$\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{16}<1$$ and this is the same as $$\frac{x^2}{2^2}+\frac{y^2}{2^2}+\frac{z^2}{4^2}<1$$ so the domain is, geometrically, the interior of the ellipsoid with the semiaxes of $2,$ $2,$ and $4$ (this is a, so called, prolate spheroid). So we write for the domain $$\mathcal{D}=\left\{(x,y,z)\left|\frac{x^2}{2^2}+\frac{y^2}{2^2}+\frac{z^2}{4^2}<1\right.\right\}$$ and it is shown on the figure below

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.