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Let $\textrm{x}=(x,y)\in\mathbb{R^2} $. Let $\textrm{n(x)}$ denotes the unit outward normal to the ellipse $\gamma$ whose equation is given by $$\frac{x^2}{4}+\frac{y^2}{9}=1$$ at point $\textrm{x}$ on it. Evaluate: $$\int_{\gamma}\textrm {x.n(x)} ds\textrm{(x)}.$$

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  • $\begingroup$ Have you found n(x)? What did you get? $\endgroup$ – Paul Jan 22 '17 at 13:13
  • $\begingroup$ i have no idea about it $\endgroup$ – Subhash Chand Bhoria Jan 23 '17 at 1:13
  • $\begingroup$ n(x)$=\Big(\frac{x}{2}\quad \frac{2y}{9} \quad1 \Big)^T$ $\endgroup$ – Subhash Chand Bhoria Jan 23 '17 at 1:50
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You don't need to calculate the normal vector , use Gauss divergence theorem in plane. Your integral equals $$\int \int _{E} \text{div}(x)\,dx= \int \int_E 2 \,dx=2(\text{area of ellipse})=12\pi$$ where $E$ stands for ellipse

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