Question

If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −2≤u≤2,−3≤v≤3, has surface area equal to 4, what is the surface area of the parametric surface given by r2(u,v)=4r1(u,v) with −2≤u≤2,−3≤v≤3?

Answer 1

The area of the first surface is given to be

$\displaystyle\iint_{\mathcal S_1}\left\|\frac{\partial\mathbf r_1}{\partial u}\times\frac{\partial\mathbf r_1}{\partial v}\right\|\,\mathrm du\,\mathrm dv=4$

Recall that for any scalar $k$ and vectors $\mathbf a,\mathbf b\in\mathbb R^3$,

$(k\mathbf a)\times\mathbf b=\mathbf a\times(k\mathbf b)=k(\mathbf a\times\mathbf b)$

which means

$\left\|\dfrac{\partial\mathbf r_2}{\partial u}\times\dfrac{\partial\mathbf r_2}{\partial v}\right\|=\left\|4\dfrac{\partial\mathbf r_1}{\partial u}\times\dfrac{4\partial\mathbf r_1}{\partial v}\right\|$
$=16\left\|\dfrac{\partial\mathbf r_1}{\partial u}\times\dfrac{\partial\mathbf r_1}{\partial v}\right\|$

So it follows that the area of the $\mathcal S_2$ is 16 times that of $\mathcal S_1$.