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 $\mathbb{S}^2$


  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?

  • $\begingroup$ 'area' of $M$ makes sense because $f$ is $C^1$ right? $\endgroup$
    – BCLC
    Aug 11, 2021 at 13:32

1 Answer 1


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, $$ \textrm{Area}(M) = \iint_{U} \sqrt{\det \left( Y^T \cdot Y \right)} \geq \iint_{U} \sqrt{\det \left( X^T \cdot X \right)} = \textrm{Area}(\mathbb{S}^2). $$

  • $\begingroup$ 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. $\endgroup$
    – levap
    Aug 11, 2021 at 13:29

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