Let us have a circle of radius $1$, which is placed on $X$-$Y$ plane. The center of the circle is at $(0.5,0.5)$.
If I cut a slice of the circle at $x=0.3$, how to find the area of the slice?
Is there a formula which is a function of $x$ and radius?
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Your question is about the area but the tag is volume? – Juniven Acapulco Mar 23 '17 at 13:37
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Find the angle $\theta$ subtended by the slice at the center. Then the area of the slice is $\dfrac{1}{2}(\theta - \sin\theta)$. – ChargeShivers Mar 23 '17 at 13:41
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@Chris Varghese, what if the radius is r? – Dimitrios Mar 23 '17 at 13:50
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@Dimitrios just multiply by $r^2$, but note that $\theta$ itself depends on $r$. – ChargeShivers Mar 23 '17 at 21:38
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All points $(x,y)$ on the circle satisfy $$ (x - 1/2)^2 + (y - 1/2)^2 = 1^2 $$ Assuming the slice is the part with less area, we have e.g. $$ A = \int\limits_{-1/2}^{0.3} \,\,\,\int\limits_{y_-(x)}^{y_+(x)} dy \, dx $$ where $$ y_\pm(x) = (1/2) \pm \sqrt{1 - (x - 1/2)^2} $$
mvw
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You basically have a circle of radius $R$ and your cutting from a slice by a straight line segment that is $a$ units away from the center. The area of the slice of the area of the circular sector (like a pizza slice) minus the area of the triangle formed by the end points of the cutting line segment with the center of the circle.
Since the cutting line is $a$ units from the center then the chord length is
$ c = 2 \sqrt{ R^2 - a^2 } $
And half the angle of the circular sector mentioned above is
$ \theta = \cos^{-1} \dfrac{ a }{R } $
Therefore, the area of the cut out slice is
$ \text{Slice Area} = \dfrac{1}{2} R^2 (2 \theta ) - \dfrac{1}{2} R^2 \sin(2 \theta ) $
This simplifies to
$ \text{Slice Area} = R^2 \cos^{-1} \left(\dfrac{a}{R}\right) - a \sqrt{ R^2 - a^2 } $
Hosam Hajeer
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