Question

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is the surface area of the parametric surface given by r2(u,v)=5r1(u,v) with −4≤u≤4,−4≤v≤4?

Answer 1

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$. Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$