# Parameterize the sphere of radius $5$ in $\mathbb{R}^3$ so that in all its points, the normal vector points outwards.

Parameterize the sphere of radius $$5$$ in $$\mathbb{R}^3$$ so that in all its points, the normal vector points outwards.

I've thought about doing the following but I do not know if I'm doing things right:

I thought about using spherical coordinates and I get that $$x=5\sin\theta\cos\phi, y=5\sin\theta\sin\phi, z=5\cos\theta$$ where $$0\leq \theta\leq\pi, 0\leq\phi\leq2\pi$$.

Is this fine? Is this the parameterization that works for me? How can I do to confirm that the normal vector points out? Thank you.

• What do you mean by "the normal vector points outwards"? Nov 17 '18 at 2:46

The parametrization is fine. The sphere is centered at the origin. The two partial derivatives would be $$\mathbf{u} =\left(\begin{array}{c}5\cos(\theta)\cos(\phi)\\5\cos(\theta)\sin(\phi)\\-5\sin(\theta)\end{array}\right)$$ and $$\mathbf{v} =\left(\begin{array}{c}-5\sin(\theta)\sin(\phi)\\5\cos(\phi)\sin(\theta)\\0\end{array}\right)$$ When you cross $$\mathbf{u}$$ and $$\mathbf{v}$$ you get the normal which comes from the origin and thus exits the sphere in the outward direction. To get it at a point on the sphere, you have to select a point, say using $$\theta$$ and $$\phi$$ then use the same values to compute u and v and then just add the cross product vector (i.e. the normal) to the calculated point. It will continue to point the same direction which will be away from the origin and thus outward.