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

What will be the value of $\displaystyle\int_{\gamma}x\cdot n(x)\,ds(x)\text{ ?}$

Matthew Leingang
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Pedrick Serre
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    Questions posted here should not be phrased in language suitable for assigning homework. One could wonder if you copied a question without understanding it, so that no actual question is present in your own mind. – Michael Hardy Nov 30 '16 at 16:59

1 Answers1

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Hint. Use the planar version of the Divergence Theorem: $$\int_{\partial D} (v_1,v_2) \cdot n \, ds=\int_D \left(\frac{\partial v_1}{\partial x} +\frac{\partial v_2}{\partial y} \right)\, dxdy.$$ In particular take a look to this example.

Robert Z
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  • How to cal calculate unit outward normal vector? – Pedrick Serre Dec 01 '16 at 03:20
  • @Pedrick Serre You need not calculate the outward normal. By the way, your notation $x=(x,y)$ is confusing. What is the given vector field in your exercise? $(x,y)$? $(x,0)$? $(0,x)$? – Robert Z Dec 01 '16 at 07:06
  • I know I dont require it. Just i want to know. $x=(x,y)$ – Pedrick Serre Dec 01 '16 at 08:25
  • @Pedrick Serre Take the gradient of $\frac{x^2} 4 +\frac{y^2} 9$ and normalize. See the answer here http://math.stackexchange.com/questions/1673378/determining-the-normal-of-an-ellipse?rq=1 – Robert Z Dec 01 '16 at 08:35