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suppose we have graph of $\sin(x)$ and a ellipse say $\frac{x^2}{10}+\frac{y^2}{2}=1$ it comes like following sin

ellipse

now when we intregrate $\int_{0}^{2\pi}\sin(x)dx$ it comes out to be zero , because area above $x$-axis is equal to area below $x$-axis so therefore when we compute area of ellipse it should also be equal to zero ? why is't it taken to be zero?

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    $\begingroup$ The integral gives signed area. Also, what's the function for the ellipse? $\endgroup$ – The Chaz 2.0 Nov 1 '13 at 5:54
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    $\begingroup$ for half-upper part $f(x)=\sqrt{2-\frac{2x^2}{10}}$ $\endgroup$ – Tesla Nov 1 '13 at 5:57
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    $\begingroup$ The ellipse above can be described by two functions, say $f_1$ and $f_2 = - f_1$. The area is given by $\int (f_1-f_2) = 2 \int f_1$. For the $\sin$ function, we have $f_1 = \sin$, $f_2 = 0$. $\endgroup$ – copper.hat Nov 1 '13 at 5:59
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As The Chaz said, computing the integral gives a signed integral; so, since your ellipse is centered, the result is zero. If you move the center of the same ellipse up or down, you will get a different result.

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  • $\begingroup$ that means i have to do what @copper.hat described above $\endgroup$ – Tesla Nov 1 '13 at 6:06

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