# Prove that the area of $M$ is bigger than the area of $\mathbb{S}^2$

Given $$\mathbb{S}^2$$ the unit sphere in $$\mathbb{R}^3$$ and let $$f:\mathbb{S}^2\to \mathbb{R}$$ be a $$C^1$$ function such that $$f(x)\geq1$$ for every $$x\in\mathbb{S}^2$$ and define $$M=\{xf(x)|x\in\mathbb{S}^2\}$$
1.Prove that $$M$$ is smooth manifold with dimension of $$2$$
2.Prove that the area of $$M$$ is bigger than the area of $$\mathbb{S}^2$$

Attempt:

1. I proved it with composition of a map and f and showed that it is a regular parametrization.
2. I've thought of doing the following
$$r:U\to\mathbb{S}^2$$ map of the unit sphere
$$\int_M 1=\int_U \sqrt{\Gamma \left (\frac{d g}{dx_i} \right )}$$ where $$g(x_1,x_2)=r(x_1,x_2)f(r(x_1,x_2))$$ but got stuck here and somehow to use the fact that $$f(x)\geq1$$

any hint?

• 'area' of $M$ makes sense because $f$ is $C^1$ right?
– BCLC
Aug 11 at 13:32

Your idea can be made to work but it involves some calculations. Let's write vectors in $$\mathbb{R}^3$$ as column vectors and choose $$X \colon U \rightarrow \mathbb{R}^3$$ such that $$U \subseteq \mathbb{R}^2$$ is an open set and $$X$$ is a parametrization of almost all of $$\mathbb{S}^2$$. For example, $$X$$ can be defined using spherical coordinates but the specific form of $$X$$ does not matter. Then $$Y \colon U \rightarrow \mathbb{R}^3$$ given by $$Y(p) = \underbrace{f(X(p))}_{:=h(p)} X(p)$$ is a parametrization of almost all of $$M$$. First of all, note that $$\| X(p) \|^2 = X^T \cdot X = 1$$ for all $$p \in U$$. Differentiating this identity, we obtain $$X^T \cdot dX = 0$$ (where $$dX$$ is the $$3 \times 2$$ matrix representing the differential). Now, using the chain rule, we have $$dY = (X \cdot dh + h dX)$$ and so $$dY^T \cdot dY = \left( \nabla h \cdot X^T + h dX^T\right) \cdot \left(X \cdot \left( \nabla h \right)^T + h dX\right) = \nabla h \cdot \left( \nabla h \right)^T + h \left( \nabla h \cdot \underbrace{X^T \cdot dX}_{0} + \underbrace{dX^T \cdot X}_{0} \cdot \left( \nabla h \right)^T \right) + h^2 (dX^T \cdot dX) = (\nabla h) \cdot \left( \nabla h \right)^T + h^2 (dX^T \cdot dX).$$
Using the matrix determinant lemma we get $$\det \left( Y^T \cdot Y \right) = \left( 1 + \left( \nabla h \right)^T \left( h^2 \left( dX^T \cdot dX \right)\right)^{-1} \cdot \nabla h \right) h^4 \det \left( X^T \cdot X \right) = \left( h^4 + h^2 \left< \nabla h, \left( dX^T \cdot dX \right)^{-1} \nabla h \right>\right) \det \left( X^T \cdot X \right).$$
Note that $$\left( dX^T \cdot dX \right)^{-1}$$ is a positive definite symmetric matrix and so $$\left< \nabla h, \left( dX^T \cdot dX \right)^{-1} \nabla h \right> \geq 0$$. Together with the fact that $$h \geq 1$$, we can conclude that $$\det \left( Y^T \cdot Y \right) \geq \det \left(X^T \cdot Y \right).$$
Finally, $$\textrm{Area}(M) = \iint_{U} \sqrt{\det \left( Y^T \cdot Y \right)} \geq \iint_{U} \sqrt{\det \left( X^T \cdot X \right)} = \textrm{Area}(\mathbb{S}^2).$$
• You can even see from the calculation that if $f(q) > 1$ at some point $q \in \mathbb{S}^2$ then at that point, $\det(dY^T \cdot dY) > \det(dX^T \cdot dX)$ and so the area of $M$ is strictly larger than the area of $\mathbb{S}^2$, as expected. Aug 11 at 13:29