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A vector field $\mathbf{G}(\mathbf{r}) = yz\mathbf{i}+\mathbf{j}+x^2\mathbf{k}$ fill all the space.

Evaluate the area integral $I=\int\mathbf{G}(\mathbf{r})\cdotp d\mathbf{A}$ over the rectangle in the $(x,y)$ plane with corners $ (0,0,0), (1,2,0), (1,0,0), (0,2,0). $

What I have tried so far:

$I=\int (yz\mathbf{i}+\mathbf{j}+x^2\mathbf{k})\cdotp (dx\mathbf{i}\cdotp dy\mathbf{j})$

I want to try and calculate the integral in terms of $x$ and have parametrised the diagonal from $(1,0,0)$ to $(0,2,0)$ as $y=2-2x$.

$dy=-2dx$

$I=\int ((2-2x)z\mathbf{i}+\mathbf{j}+x^2\mathbf{k})\cdotp (dx\mathbf{i}\cdotp -2dx\mathbf{j})$

I am not sure how to parametise $z$ in terms of $x$. Unless there is a different approach to this question? If possible diagrams would be helpful.

Stratos1
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  • You might try putting the vertices in the correct order for a rectangle: $(0,0)$, $(1,0)$, $(1,2)$, and $(0,2)$. $z=0$ throughout the problem, and if you're doing a flux integral you need a double-integral over the rectangle, not a line integral. – Ted Shifrin Apr 27 '17 at 22:53
  • We don't normally write area elements that way, it wouldn't make any sense as that scalar product is zero, the area element has a normal in the z direction so just $\hat{k}dx dy$ is all that's necessary – Triatticus Apr 28 '17 at 04:23

1 Answers1

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$$I=\int yz\mathrm{d}y\mathrm{d}z+\mathrm{d}z\mathrm{d}x+x^2\mathrm{d}x\mathrm{d}y=\int_0^1\mathrm{d}x\int_0^2 x^2\mathrm{d}y=2/3$$

xyz
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