I want to calculate the area of a sector of an ellipse. However the sector does not include the center of the ellipse as shown in the following image enter image description here

Here a,b are the minor and major axes,$\theta $ is given and h corresponds to the frame of reference of calculation of the integral which is denoted by the shaded area.

I have calculated the area as

$$\frac{1}{2}\int_{\alpha}^{\theta+\alpha}a^2cos^2(\phi )+(h-b(1+sin(\phi )))^2 d\phi$$ where $\alpha = sin^{-1}(\frac{h-b}{b})$ and $\theta$ is constant.

Is this the correct answer? Thanks in advance .

Edit: So I calculated the integral and it came out to be $$-\dfrac{\left(b^2-a^2\right)\sin\left(2\sin^{-1}\left(\frac{h-b}{b}\right)+2\theta\right)+\left(8b^2-8bh\right)\cos\left(\sin^{-1}\left(\frac{h-b}{b}\right)+\theta\right)+\left(a^2-b^2\right)\sin\left(2\sin^{-1}\left(\frac{h-b}{b}\right)\right)+\left(8h-8b\right)\sqrt{2bh-h^2}-4(\theta)h^2+8b(\theta)h+\left(-6b^2-2a^2\right)\theta}{8}$$

Here is my problem:If i plug h=b (or $\alpha$=0) and $\theta$ = $\frac{\pi}{2}$ then obviously the answer should be $\frac{\pi ab}{4}$.But the above integral does not return the same output.

Have I formulated the integral incorrectly or is the solution erroneous.Thanks.


1 Answer 1


We can take the x-axis line and integrate that to y axis to get area of P(shown as weavy shaded area in the image below). and then subtract the area of the triangle (shown in blue) as shown in the equation below.

$$\int_{h_{1}}^{h_{2}}\sqrt{\Big(1-\frac{y^{2}}{b^{2}}\Big)a^{2}}dy-\frac{1}{2}\sqrt{\Big(1-\frac{y^{2}}{b^{2}}\Big)\vert_{y=h_{2}}}$$ Now to calculate the integral:

$$\int_{h_{1}}^{h_{2}}\sqrt{\Big(1-\frac{y^{2}}{b^{2}}\Big)a^{2}}dy = \frac{a}{b}\int_{h_{1}}^{h_{2}}\sqrt{b^{2}-y^{2}}dy=\frac{a}{2b} \left[y\sqrt{b^{2}-y^{2}}]+b^{2}sin^{-1}\Big(\frac{y}{b}\Big)\right]_{h_{1}}^{h2}$$ $$= \frac{a}{2b}\left[h_{2}\sqrt{b^{2}-h_{2}^{2}}-h_{1}\sqrt{b^{2}-h_{1}^{2}}+b^{2}\left(sin^{-1}\Big(\frac{h_{2}}{b}\Big)-sin^{-1}\Big(\frac{h_{1}}{b}\Big)\right)\right] $$

This can be verified by putting $h_{1}=0\;h_{2}=b$ we get area of P $=\frac{1}{4}\pi a b$. And $4P = \pi ab$ which is area of the ellipse.

So the area of the sector within $h_{1}$ and $h_{2}$ can be calculated as: $$\frac{a}{2b}\left[h_{2}\sqrt{b^{2}-h_{2}^{2}}-h_{1}\sqrt{b^{2}-h_{1}^{2}}+b^{2}\left(sin^{-1}\Big(\frac{h_{2}}{b}\Big)-sin^{-1}\Big(\frac{h_{1}}{b}\Big)\right)\right]-\frac{1}{2}\sqrt{1-\frac{h_{2}^{2}}{b^{2}}}$$

Now you need to calculate $h_{1}$ in terms of b and your h (the h in the question) and $h_{2}$ in terms of $\theta$ and substitute.

enter image description here

  • 1
    $\begingroup$ that's a better approach to this problem (+1) $\endgroup$
    – G Cab
    Jun 28, 2018 at 21:00

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