Area of Part of a Circle Given a circle of radius a, cut out a tab of height b. What is the area of this tab? (See Figure 1.) (0, b) (a, 0) Figure 1: Tab cut out of a circle. One way to compute the area would be split the area into vertical strips and integrate with respect to x: � Area = y dx. This is awkward, because near the end the height √ of the region changes from a constant y = b to the height of the circle y = a2 − x2 . What if we integrate with respect to y? That seems to work better; � there is a single simple expression for the length of each horizontal strip: x = a2 − y 2 . � b Area = x dy 0 b � � = a2 − y 2 dy 0 We don’t yet have a rule for integrating functions of this form. Considering that this integral arose from a question about a circle, it’s not surprising that trigonometry will play a role in its solution. When working with circles it often helps to use polar coordinates. In this case, note that the upper right hand corner of the region has polar coordinates (a cos θ0 , a sin θ0 ) where θ0 is the angle shown in Figure 2. In general, x = a cos θ and y = a sin θ. If we substitute y = a sin θ into our integrand we get: � x = a2 − y 2 1 (a cos θ0 , a sin θ0 ) θ0 Figure 2: Polar coordinates of a point. = = a cos θ = x � a2 − a2 sin2 θ � a 1 − sin2 θ √ a cos2 θ = Changing to polar coordinates made our integrand look much nicer; we’ve gone from an integrand with a square root and no trig functions to an integrand with trig functions and no square root. If we’re going to use the substitution y = a sin θ in our integral, we’ll also need to replace dy by something in polar coordinates. y = a sin θ dy = a cos θ dθ Plugging in, we get: � � a2 − y 2 dy = � (a cos θ)(a cos θ dθ) � a2 cos2 θ dθ = We computed the integral of cos2 x earlier in the lecture: � x sin 2x cos2 x dx = + + C. 2 4 Plugging this in, we get: � � a2 − y 2 dy = a2 � θ sin(2θ) + 2 4 � + c. Now we’d like to rewrite our solution in terms of the original variable y so that we can plug in the limits of integration. In order to do this, it’s helpful to rewrite sin(2θ) using the double angle formula sin(2θ) = 2 sin θ cos θ. 2 � � a2 − y2 dy � θ sin(2θ) = a + +c 2 4 � � θ sin θ cos θ 2 = a + +c 2 2 � 2 � a θ a sin θa cos θ = + + c. 2 2 2 � Next we solve y = a sin θ for y and plug in: �y� θ = arcsin . a � Since a sin θ = y and a cos θ = x = a2 − y 2 , we get: � � � � � a2 arcsin(y/a) y a2 − y 2 2 2 a − y dy = + + c. 2 2 � We’ve used trigonometric substitution to ﬁnd the indeﬁnite integral of a2 − y 2 . Whenever you see the square root of a quadratic in an integral you should think of trigonometry and sin2 θ + cos2 θ. Our original problem asked us to compute the value of a deﬁnite integral; let’s ﬁnish that. � � �b � � b� a2 arcsin(y/a) y a2 − y 2 �� 2 2 a − y dy = + � � 2 2 0 � � 0 √ a2 arcsin(b/a) b a2 − b2 = + −0 2 2 √ a2 arcsin(b/a) b a2 − b2 = + 2 2 Notice that θ0 = arcsin(b/a); we could rewrite this answer as: √ a2 θ0 b a 2 − b2 Area = + 2 2 Does this make sense? The ﬁrst term, θ20 a2 , is exactly the area of the sector of √ the circle swept out by angle θ0 . The second term, 12 b a2 − b2 , is the area of √ a triangle with base b and height a2 − b2 . In other words, it’s the area of the shaded triangle shown in Figure 2. Using some basic geometry, we’ve checked that our answer to this compli cated calculus problem is correct. 3