Let $T$ be a planar transformation from the $(u, v)-$plane to the $(x, y)-$plane that is a bijection, let $S$ be a domain in the $(u, v)-$plane, and let $R = T(S)$. If $\frac{d(x,y)}{d(u,v)} = 2$ at every point, and the area of $R$ is $8$, what is the area of $S$?
From what I gathered, $(x,y) = T(u,v)$ and the Jacobian of $T=2$. Knowing the formula : Area of $R = |f_u(x,y)g_v(x,y)-f_v(x,y)g_u(x,y)|$(Area of $R(u,v))$, how do we find Area of S without $f_u, g_v, f_v, g_u$?
What I understand from your question is that initially we have an integral to find the area of R which is
$$A = \iint_R dydx = 8$$
Then we made a transformation from $R(x, y)$ to $S(u, v)$.
Usually we make a transformation in order to simplify the integrand and/or the limits of integration. If the integral is to calculate the area, the result of integration before should be the same as after integration. Hence in this case after transformation we have
$$A = \iint_S J dvdu = \iint_S 2 dvdu = 8 $$
I would say the area is the same which is 8.
The expression in your formula |fu(x,y)gv(x,y)−fv(x,y)gu(x,y)| is in fact J which is given to be 2, hence there is no need to know the partial derivatives.
Unless the question is asking for $\iint_S dvdu $, then the answer will be 4.