Surface integral of a parametric form function for f(x)

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.

• Your surface element is incorrect, $dS=\left | \frac{\partial \vec{r}(u,v)}{\partial u} \times \frac{\partial \vec{r}(u,v)}{\partial v} \right |du dv$ where $r(u,v)=x(u,v)\hat{i} + y(u,v)\hat{j} + z(u,v)\hat{k}$ is the vector that defines the surface (I'm using the notation in Stewart's calculus) Nov 29, 2017 at 9:21

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 $25$.
• In this case, is useful to note that the surface is contained in a cilinder whose axis is parallel to $z$. This is enough to derive the surface element $dS$ above. Nov 29, 2017 at 9:45