Area of A Sector of An Ellipse
Date: 02/28/98 at 08:29:29 From: Phill Brady Subject: Area of a sector of an ellipse I need a formula for the area of an ellipse sector. I have the semiminor and major axes and the angles of the 2 vectors bounding the sector. How do I work out the area?
Date: 02/28/98 at 16:13:08 From: Doctor Sam Subject: Re: Area of a sector of an ellipse Phill, You can find the area of a sector of an ellipse by representing the ellipse in polar coordinates and then integrating. If the ellipse has semi-major axes of A and B in the x- and y- directions, then the ellipse can be represented as x = AcosT and y = BsinT. Since x^2 + y^2 = r^2 we can get r^2 = A^2 cos^2 T + B^2 sin^2 T. Now, area in polar coordinates is given by an integral: (1/2) INT(r^2 dT) from T1 to T2 T1 and T2 are the angles of the vectors defining the sectors that you mention. So you need to evaluate area = (1/2) INT(A^2 cos^2 T + B^2 sin^2 T dT) from T1 to T2 = (1/2) INT(A^2 cos^2 T dT) from T1 to T2 + (1/2) INT(B^2 sin^2 T dT) from T1 to T2 = (1/4) A^2[ T2 - T1 + sinT2 cosT2 - sinT1 cosT1 ] + (1/4) B^2[ T2 - T1 - sinT2 cosT2 + sinT1 cosT1 ] This can be simplified a little by regrouping: area = (T2-T1)(A^2+B^2)/4 + (sinT2 cosT2 - sinT1 cosT1)(A^2-B^2)/4 I hope that helps. -Doctor Sam, The Math Forum Check out our web site http://mathforum.org/dr.math/
Date: 03/27/2000 at 11:57:24 From: Hume Peabody Subject: Area of ellipse sector I need to find the area of a sector of an ellipse between theta1 and theta2. The solution posted is incorrect. This can be shown by the simple fact that a full ellipse has an area of PI*A*B. The posted solution is: area = (T2-T1)(A^2+B^2)/4 + (sinT2 cosT2 - sinT1 cosT1)(A^2-B^2)/4 For T1 = 0 and T2 = 2*PI, this yields (A^2 + B^2)*PI/2 I can work the radius down as a function of theta, but cannot find a closed form solution from 0 to 2*PI for the integration. R = A*B/SQRT(B^2*cos^2(Theta) + A^2*sin^2(Theta)) I can integrate in rectangular coordinates and also find the area of the remaining triangle to get the sector area, but I was looking for a more elegant solution. Thanks for any help. Hume
Date: 03/28/2000 at 12:20:17 From: Doctor Peterson Subject: Re: Area of ellipse sector Hello, Hume. Thanks for letting us know about the error in the archives. It looks like Dr. Sam forgot that the variable T in the parametric form is not the angle, and therefore integrated with respect to the wrong variable. Replacing dT with A sec^2(T) dTheta = ------------ B + A tan(T) would give the correct integral; or instead we can use your integral in terms of Theta and integrate it by substituting u = tan(Theta). Eventually we get Area[0,Theta] = (AB/2) arctan(A/B tan(Theta)) so that Area[Theta1,Theta2] = (AB/2)[arctan(A/B tan(Theta2) - arctan(A/B tan(Theta1)] (In evaluating this, you have to choose the right branch of the arctan; be careful!) This can also be found without integration very easily. If we transform the ellipse to a circle of radius B by "compressing" in the X direction ************* ***** **** | /**** *** | /*** ** | /Theta ** ** |/T * *------------+------------* *------+------* ** | ** ** | * **** | **** *** | *** ************* ***** It turns out that the angle in the circle is exactly the T in the parametric equation, which is T = arctan(A/B tan(Theta)) The area of the sector between the x-axis and the angle Theta is b/a times the area of the sector between x-axis and angle T in the circle, which gives our formula. You can decide which is most elegant. For some related formulas, though not the area of a sector, see this page in World of Mathematics: http://mathworld.wolfram.com/Ellipse.html We'll be sure to correct the archives. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum