The area of the surface given by [tex]\vec r_1(u,v)[/tex] is 1. In terms of a surface integral, we have
[tex]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[/tex]
By multiplying each component in [tex]\vec r_1[/tex] by 5, we have
[tex]\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}[/tex]
and the same goes for the derivative with respect to [tex]v[/tex]. Then the area of the surface given by [tex]\vec r_2(u,v)[/tex] is
[tex]\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}[/tex]
