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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.

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    $\begingroup$ 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) $\endgroup$
    – Triatticus
    Nov 29, 2017 at 9:21

1 Answer 1

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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 $25$.

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  • $\begingroup$ Thanks! Is there any simple way to figure out the surface this represents or is it often very hard/useless to try and figure with parametric statements like these? $\endgroup$
    – Grak
    Nov 29, 2017 at 9:42
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    $\begingroup$ 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. $\endgroup$ Nov 29, 2017 at 9:45
  • $\begingroup$ Alright, thanks again $\endgroup$
    – Grak
    Nov 29, 2017 at 9:48

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