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I have to compute $\iint_S \sqrt{1+4x^2} dS$ and I'm looking a way to do it. The graph seems to be:

the surface $S$ is given by ($x=0, x=2, y=0, y=3$ and $z=x^2$)

enter image description here

Can anyone help me set up the integral to compute it?

Clayton
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Valent
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  • You have to parametrize $S$, what have u tried? – Marcos Dec 03 '21 at 13:20
  • @Marcos I'm not sure on the parametrization. Not seems to be easy for this surface – Valent Dec 03 '21 at 13:26
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    notice that the only reltion you have is $z=x^2$, so the only reasonable parametrization is $f(x,y)=(x,y,x^2)$. Can you continue from this? – Marcos Dec 03 '21 at 13:28
  • Initially I have $x,y,z$ so th parametrization reduce to $x,y$ but how about the limits? should be $\int_{0}^2\int_{0}^{x^2} \sqrt{1+4x^2}dy\ dz$ I cannot see it. – Valent Dec 03 '21 at 13:35
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    check this https://en.wikipedia.org/wiki/Surface_integral, there is a formula for the integral given a parametrization – Marcos Dec 03 '21 at 13:39
  • Notice also that this is a closed surface. If you're evaluating the surface integral, ensure you account for all faces/sides of the solid. – Clayton Dec 03 '21 at 13:43
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    Integrate[ Sqrt[1 + (2 x)^2]/({-2 x, 0, 1} . {0, 0, 1}/Norm[{2 x, 0, -1}]), {x, 0, 2}, {y, 0, 2}] should give you 76/3 – S L Dec 03 '21 at 13:46
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    @Valent your surface is not defined the way it is written. You need another bound of $z$ or if it is just the surface of the parabolic cylinder for given bounds of $x$ and $y$, then it must be stated that way. – Math Lover Dec 03 '21 at 14:04
  • I believe that we need variation in $x,y,z$ right? – weymar andres Dec 03 '21 at 14:04

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Usually, before I calculate surface area, I compute $\iint_S dS = \text{surface area}$. The area is easy to compute since you can compute the arc length and one side is constant i.e. $$ \int_0^2 \sqrt{ 1 + (\frac{dz}{dx} )^2} dx = \text{arc length} \implies \int_0^3 \int_0^2 \sqrt{ 1 + (\frac{dz}{dx} )^2} dx dy = \text{surface area} $$ This gives $$ \iint_S \sqrt{1 + (2x)^2} dS = \int_0^3 \int_0^2 \ (1 + (2x)^2) dx dy $$ Another way, note that $d\vec S \cdot \hat k = dydx \implies dS \hat n \cdot \hat k = dydx$. This gives $$ dS = \frac{dydx}{\hat n \cdot k} $$ where $\hat n$ is a unit normal vector of the surface which you can obtain by taking the gradient of $z - x^2$.

Another way, you can parameterize the surface in terms of $s$ and $t$. You may write it as $\vec r (s, t) = s \hat i + t \hat j + s^2 \hat k$ with $0 \le s, t \le 2$. Now, in your function replace $x \to s, y\to t, z \to s^2$ and you may use the formula $$ \iint_{T} f(\vec r(s, t)) \left \|\frac{\partial \vec{r}}{\partial s} \times \frac{\partial \vec{r}}{\partial t}\right \| \mathrm{d} s \mathrm{~d} t $$ All should give the same answer.

S L
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  • The question does not say that we need to find integral only over the surface $z = x^2$. What part in the question tells you that you are not supposed to integrate over planar surfaces $x = 0, x = 2, y = 0$ and $y = 3$? – Math Lover Dec 03 '21 at 16:29