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Calculate $\int \!\!\! \int E\cdot \vec n d\sigma$ where S is the parametric surface $X(s,t)=[st,s^2,t^2]^T$, $0\le s\le t\le 1$, and the E is the vector field $E(x,y,z)=[3yz,zx,2xy]^T$

Do I just use the formula:
$\int\!\!\!\int E(x(s,t))\cdot (\frac{\partial x}{\partial t}\times \frac{\partial x}{\partial s})dsdt$

Leading me to: $\int_0^1\int_0^1 (3s^2t^2,st^3,2s^3t)\cdot (0,0,-s)=\int_0^1\int_0^1 -2s^4t=-\frac{1}{5}$

Is that correct?

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up vote 1 down vote accepted

Either your $\frac{\partial X}{\partial t}$ and $\frac{\partial X}{\partial s}$ are wrong, or you computed your cross product wrong. Here's what I get:

$$ X_t=[s,0,2t]^\intercal\\ X_s=[t,2s,0]^\intercal\\ X_t\times X_s=[-4st,2t^2,2s^2]^\intercal $$ Also, I think you misinterpreted what $0\leq s\leq t\leq 1$ means. That means that the integral bound for $t$ should go from $s$ to 1, not from 0 to 1. This also means we integrate with respect to $s$ last.

$$ \begin{align*} \int E\cdot\vec{n}d\sigma &= \int_0^1\int_s^1 (3s^2t^2,st^3,2s^2t)\cdot(-4st,2t^2,2s^2)dtds\\ & = \int_0^1\int_s^1 (-12s^3t^3+2st^5+4s^4t)dtds=-\frac{19}{140} \end{align*} $$

You might need to verify that the sign is correct, depending on which way $\vec{n}$ is oriented (in or out).

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