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Lecture 14, 15

Lecture 14, 15 - MATH 118 LECTURES 14& 15 POLAR AREAS 1...

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Unformatted text preview: MATH 118, LECTURES 14 & 15: POLAR AREAS 1 Polar Areas We recall from Cartesian coordinates that we could calculate the area under the curve by taking Riemann sums. We divided the region into subregions, approximated the area over that subregion by a rectangle, added all of the regions up, and then took the limit as the width of the subregions went to zero. Furthermore, we were able to show that this process was equivalent to the process of taking the definite integral! We can carry out a very similar procedure in polar coordinates; however, rather than approximating our areas with rectangles we will be approximat- ing them by wedges of a circle (see Figure 1). Otherwise, the same exact principle applies: we divide the region into subregions (of θ ), approximate the integral within the subregions by wedges, add these approximates up, and then take the limit as the width of the subregions goes to zero. x y A 1 A 2 t t 1 t 2 r=f(t ) r=f(t 1 ) r=f(t 2 ) Area Figure 1: We can approximate the area of a region as the sum of wedges. ( t = θ , t 1 = θ 1 , etc.) 1 The question remains as to how we can approximate the areas of the wedges, A i . We will start by considering the whole circle associated with a particular wedge. The area of the circle associated with the region A i is Area = πr 2 = πf ( θ i ) 2 . Now consider the wedge between, say, θ i and θ i +1 . This wedge represents a fraction of the whole area of the circle, but how do we quantify the fraction? We notice that we would recover the area for the entire circle if we took θ i = 0 and θ i +1 = 2 π , so the fraction we need is ( θ i- θ i +1 ) / (2 π ). This gives Area of wedge A i = θ i +1- θ i 2 π πr 2 = 1 2 ( θ i +1- θ i ) f ( θ i ) 2 . So we have a formula for the area of our wedge, which approximates the actual area bound between the two angles. If we divide the region in n wedges with widths Δ θ i = θ i +1- θ i and take the sum, we have n summationdisplay i =1 A i = n summationdisplay i =1 1 2 [ f ( θ i )] 2 Δ θ i . We now take the limit as the widths of the regions go to zero to get Area = lim bardbl Δ θ i bardbl→ n summationdisplay i =1 1 2 [ f ( θ i )] 2 Δ θ i . We recognize the right-hand side as the integral of (1 / 2)[ f ( θ )] 2 over θ . We take the bounds on θ to be α and β to get Area = 1 2 integraldisplay β α [ f ( θ )] 2 dθ. We can do all of the tricks we could do with definite integrals in Cartesian coordinates: measure areas between curves, divide the integral into separate integrals to be calculated separately, etc. We must remember, however, that we are dividing things in terms of θ rather than x . Example 1: Find the area of one of the “petals” of the function r = sin(2 θ ) ....
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• Spring '09
• Soares
• Polar coordinate system, subregions, xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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