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: $$\int_{\gamma}\textrm {x.n(x)} ds\textrm{(x)}.$$
Evaluate: $\int_{\gamma}\textrm {x.n(x)} ds\textrm{(x)}$
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calculus
algebra-precalculus
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0Have you found n(x)? What did you get? – 2017-01-22
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0i have no idea about it – 2017-01-23
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0n(x)$=\Big(\frac{x}{2}\quad \frac{2y}{9} \quad1 \Big)^T$ – 2017-01-23
1 Answers
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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