# Parametrization of square to calculate Dot-product in line-integrals and area-integrals, electric field from $\frac{dB}{dt}$?

I am calculating 3A of Tfy-0.1064 in Aalto University. I realized here that I am misunderstanding something in vector calculus: the thing market in green particularly.

I know

$$\nabla\times E= -\frac{\partial B}{\partial t}$$

so

$$E=\int (\nabla\times B)\cdot dA=\int B\cdot dl.$$

The magnetic flux on a square-loop of side 0.2m is by a solenoid. I am stuck with the dot-product. I cannot understand what $dA$ is or $dl$ is in the integral over the magnetic field $B$. $\int dA=0.04m^2$ and $\int dl=0.8m$ but what about $\int B\cdot dl$ or $\int (\nabla\times B)\cdot dA$?

Select the XY plane along the square-loop then $\frac{dB}{dt}=\frac{d B_z}{dt}=35\frac{mT}{s}$. Now $\nabla\times B=\hat j \partial_z B_x+\hat i \partial_z B_y$ (calculation) so $(\nabla\times B)\cdot dA=\alpha \partial_z B_x+\beta \partial_z B_y$ where $\alpha,\beta$ are some scalars such that $dA=\hat i \alpha +\hat j \beta$ but $\alpha,\beta$ still unknown to me (can you see it?). $dA$ corresponds to the area of the square to which the solenoid induce the current.

But what is $dA$ in vector form?

What is $dl$ in vector form where $l$?

$l$ is apparently the perimeter of the square-loop, yes? Can I select it as a circle $l=\cos(\theta)\hat i+\sin(\theta)\bar j$?

I cannot calculate the dot product before learning to parametrize the shapes such as the square with vectors. How can I express aka parametrize the $dA$ and $dl$ in order to calculate (using dot-product) the electric field induced by the change in magnetic flux $\frac{dB}{dt}=35mT/s$ of solenoid on the square-loop?

Or shortly, how can I parametrize the shapes? If I could select arbitrary path, then I could use cylinder coordinates with some radius $r$ and some angle $\theta$ but can I do the integration in a way that the square was a circle?

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I am not sure if this is what you are asking. $dA$ is a vector perpendicular to surface, in a size equal to an element of area on that surface. for example if it point in $z$ direction and surface is on $xy$ plane then $dA=dxdy k$ where $k$ is unit vector in direction of $z$ axis. – Maesumi Jan 12 '13 at 4:47
If you have chosen to use cylindrical coordinates then $dA=k \ rdrd\theta$. But since you have a square loop then it would be a bit hard to get to the edges. You end up with an upper limit of integral in the form of $0\le r \le a/ \cos \theta$ where $a$ is half square side. – Maesumi Jan 12 '13 at 15:13
That set up would be for Cartesian system. In which case you need to use $k dx dy$, so that $k$ dots with curl of $B$. – Maesumi Jan 13 '13 at 4:27
@Maesumi yes you are right: $\bar E=\int (\nabla\times B_z)\cdot dA_z=\int_{0}^{0.2m}\int_{0}^{0.2m}(\nabla\times \bar B)\cdot (\hat k dx dy)$ so $$\bar E=\int_{0}^{0.2m}\int_{0}^{0.2m}(\hat j \partial_z B_x -\hat i \partial_z B_y)\cdot (\hat k dx dy).$$ – hhh Jan 13 '13 at 5:53
@Maesumi but now it does not make sense -- the dot products are zero?! I must have a mistake in the cross-product, have to reconsider the curl in the solenoid. – hhh Jan 13 '13 at 6:16