How to find the vector function $\vec{r}(u,v)$

Question

The lower half the ellipsoid $2x^2 + 4y^2 + z^2 = 1$ using spherical coordinates would equate the following:

$\sqrt{2}x = \sin\phi \cos\theta$

$2y = \sin\phi \sin\theta$

$z=\cos\phi$

Why are the coefficients being used here?

Answer 1

With the given parametrization of the surface your original equation $$2x^2 + 4y^2 + z^2 = 1$$ is satisfied.

Note that if you eliminate the coefficients then your original equation is not satisfied.

$$2x^2 = \sin^2\phi \cos^2\theta $$ $$ 4y^2 = \sin^2\phi \sin^2\theta $$

$$z^2=\cos^2\phi$$

Thus you have $$2x^2 + 4y^2 + z^2 = 1$$