Surface Area of an Ellipsoid Next we’ll ﬁnd the surface area of the surface formed by revolving our elliptical curve: x = 2 sin t y = cos t about the y-axis. Remember that our surface area element dA is the area of a thin circular ribbon with width ds. The radius of this circle is x = 2 sin t, which is the distance between the ribbon and the y-axis. � dA = 2π (2 sin t) 4 cos2 t + sin2 tdt . �� x ds=arc length To ﬁnd the surface area we need to integrate dA between certain limits; what are they? t = 0 (0, 1) t= π 2 Figure 1: Elliptical path described by x = 2 sin t, y = cos t. By looking at Figure 1 we can see that we need to integrate from 0 to π. Remember that we only need to go from the top to the bottom of the ellipse to trace the right hand side; including the left hand side of the ellipse would double our result and give the wrong answer. � π � A= 2π(2 sin t) 4 cos2 t + sin2 tdt. 0 Notice that we’re integrating from the top of the ellipse to the bottom; if we think in terms of the y-variable we tend to think of going the opposite way. This integral turns out to be do-able but long. Start by using the substitution u = cos t, du = − sin t dt. � π � A = 2π(2 sin t) 4 cos2 t + sin2 tdt 0 � u=−1 −4π = � 3u2 + 1du. u=1 Next would be another trigonometric substitution to deal with the square root, and so on. 1