So, I'm trying to get the surface integral of $$\iint_{S} 6\sqrt{4y-3} \,\,dS$$
when $S=${$(x,y,z)\in R^3|y=x^2+1, 1 \le x \le z, 1 \le z \le 2$}
Coincidentally, I know that the correct answer is 25, but I'm having tough time trying to visualize the shape of the surface this represents and how to actually calculate it.
I've tried starting parametrization from the boundaries with
$y(t)=t^2+1$ and integrating $\int_{1}^v\int_{1}^2 6\sqrt{4(t^2+1)-3} \,dt\,dv$, but as it gives the wrong answer, I'm thinking that I'm either missing some surfaces or the parametrization I've done is incorrect.
In your case, the surface element can be computed as $dS=\sqrt{dx^2+dy^2}dz=\sqrt{1+4x^2}dxdz$, so your integrale become: $$6\iint_S\sqrt{4y-3}dS=6\int_1^2dz\int_1^zdx(1+4x^2)$$ which gives the correct result $25$.