Let $M\subset \mathbb{R}^3$ be a compact $2$-dimensional manifold in $\mathbb{R}^3$,$\lambda >0$,$\lambda\in \mathbb{R}$ and $M':=\lambda M=\{\lambda x|x\in M\}$.
a)Prove that $M'$ is a $2$-dimensional manifold in $\mathbb{R}^3$.
b)Prove that: $area(M')=\lambda^2area(M)$.
I don't know how to approach this question...any help?
For any chart $g:\mathbb{R}^2\to M$, the function $\lambda g:\mathbb{R^2}\to \lambda M$ gives you a chart for $\lambda M$. The differentiability of compositions $g_1\circ g_2^{-1}$ gives us the differentiability of $(\lambda g_1)\circ (\lambda g_2)^{-1}=\lambda\circ g_1\circ g_2^{-1}\circ \lambda^{-1}$, where $\lambda$ is the multiplication by $\lambda$. We need to take into account that multiplication by $\lambda$ is differentiable in $\mathbb{R}^3$.
For (b) just compute the volume form on $M$ and do the change of variable. In a chart $g$ the form is $|g_x\times g_y|dxdy$. For $\lambda M$ it would be $|\lambda g_x\times \lambda g_y|dxdy=\lambda^2|g_x\times g_y|dxdy$