2 THE DEFINITE INTEGRAL (Plane Area - Polar).pdf - THE...

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Unformatted text preview: THE DEFINITE INTEGRAL PLANE AREA (POLAR COORDINATE) EM 1202 Lecture by: Engr. Stephanie Y. Canete Summer 2019 POLAR COORDINATE SYSTEM INTRODUCTION • Graph of a polar equation Application • Solving for Plane Areas ion: 3. • SEATWORK: r cos 3sin V. Sketch the graph of the polar equation: 1. r 2(1 sin ) 2. r 2(1 cos ) 3. r 2sin 4. r 1 2cos ) 5. r 6. r 2 25cos 2 1 1 2sin No. 1 No. 2 No. 3 No. 4 No. 5 No. 6 Section 3-8 : Area With Polar Coordinates In this section we are going to look at areas enclosed by polar curves. Note as well that we said “enclosed by” instead of “under” as we typically have in these problems. These problems work a little differently in polar coordinates. Here is a sketch of what the area that we’ll be finding in this section looks like. Area With Polar Coordinates (Single Curve) We’ll be looking for the shaded area in the sketch above. The formula for finding this area is, Notice that we use in the integral instead of so make sure and substitute accordingly Let’s take a look at an example. Example Example 1: 1 Determine the area of the inner loop of Calculus II - Area with Polar Coordinates 7/9/2019 . Calculus II - Area with Polar Coordinates Hide Solution For this problem we’ll also need to know the valuesCan of where you seethe why we needed to know the values of where the curve goes through th points curve goes through the origin. We can get these byThese setting the define where the inner loop starts and ends and hence are also the limit integration in the formula. equation equal to zero and solving. We graphed this function back when we first started looking at polar co ketch of this curve with the inner loop shaded in. So, the area is then, problem we’ll also Here need tosketch know thecurve values ofinnerwhere thein.curve goes th is the of this with the loop shaded can get these by setting the equation equal to zero and solving. tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx 7/9/2019 Calculus II - Area with Polar Coordinates So, that’s how we determine areas that are enclosed by a single curve, but what about situations like the following sketch where we want to find the area between two curves. Area With Polar Coordinates (Between Two Curves) In this case we can use the above formula to find the area enclosed by both and then the actual area is the difference between the two. The formula for this is, example of this. Let’s take a look at an example of this. e the area that lies inside and outside . Example 2 Determine the area that lies inside Example 2: and outside . Hide Solution e region that we are after. Here is a sketch of the region that we are after. Find the points of intersection by equating the two: 7/9/2019 a, we’ll need to know the values of Calculus II - Area with Polar Coordinates Here is a sketch of the figure with these angles added. for which the two curves intersect. We Note as well that we also acknowledged that another representation for the angle Let’s take a look at an example ofhere this. 7/9/2019 . This is important for this problem. In order to use the formula above the area must Calculus II - Area with Polar Coordinates Calculus II - Area with Polar Coordinates enclosed as we increase from Example 2: So, if we use Example 2 Determine the area that lies insidethe smaller to larger angle. and outside Hide Solution to . we w enclose the shaded area, instead we will enclose the bottom most of the three regions. However, if we use the angles to we will enclose the area that we’re after. So, the area is then, Here is a sketch of the with these angles added. Here is figure a sketch of the region that we h of the figure with these angles added. are after. Note as well here that we also acknowledged that another representation for the angle is re that we also that representation forthe theformula angle above is the area must be . Thisacknowledged is important for thisanother problem. In order to use Let’sabove work a the slight modification portant for this problem. In order to use the formula area must be of the previous example. enclosed as we increase from the smaller to larger angle. So, if we use to we will not 7/9/2019 n This time we’re looking for the following region. Let’s work a slight modification of the previous example. Example Example 3: 3 Determine the area of the region outside Calculus II - Area with Polar Coordinates Calculus II - Area with Polar Coordinates and inside . Hide Solution tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx This we’re looking e looking fortime the following region.for the following region. So, this is the region we region get by that using . The area So, thisthat is the wethe getlimits by usingtothe limits to for .this Thereg ar So, this is the region that we get by using the limits region that we get by using the limits to . The area for this region is, to . The area for this regio Notice that for“outer” this the the “outer” andfunction “inner” opposite! Notice that for Notice this area the and “inner” were were opposite! that for thisarea area “outer” and function “inner” function were opposite So, this is the region that we get by using the limits to . The area for this region is, Let’s do one final modification of this example. Example Example 4:4 Determine the area that is inside both and . Hide Solution Calculus II - Area with Polar Coordinates Here is the sketch for this example. We are not going to be able to do this problem in the same fashion that we did the previous two. There is no set of limits that will allow us to enclose this area as we increase from one to the other. Remember that as we increase the area we’re after must be enclosed. However, the only two ranges for that we can work with enclose the area from the previous two examples and not this region. In this case however, that is not a major problem. There are two ways to do get the area in this tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx problem. We’ll take a look at both of them. Solution 1 In this case let’s notice that the circle is divided up into two portions and we’re after the upper portion. Also notice that we found the area of the lower portion in Example 3. Therefore, the area is, e able to do this problem in the same fashion that we did the previous two. s that will allow us to enclose this area as we increase from one to the as we increase the area we’re after must be enclosed. However, the Let’s do one final only two ranges for that we can work with enclose the area from the previous two examples and not this region. modification of this example. In this case however, that is not a major problem. There are two ways to do get the area in this problem. We’ll take a look at both of them. Example Example 4:4 Determine the area that is inside both Hide Solution and . Solution 1 In this case let’s notice that the circle is divided up into two portions and we’re after the upper portion. Also notice that we found the area of the lower portion in Example 3. Therefore, the SOLUTION NO. 1 area is, Calculus II - Area with Polar Coordinates Here is the sketch for this example. Solution 2 SOLUTION NO. 2 In7/9/2019 this case we do pretty much the sameCalculus thing this time we’ll think of the area as the II - except Area with Polar Coordinates other portion of the limacon than the portion that we were dealing with in Example 2. We’ll also need to actually compute the area of the limacon in this case. So, the area using this approach is then, tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx e able to do this problem in the same fashion that we did the previous two. s that will allow us to enclose this area as we increase from one to the as we increase the area we’re after must be enclosed. However, the A slightly longer approach, but sometimes we are forced to take this longer approach. = 2∫ 5 + 18sin θ + 2 (1 − cos ( 2θ ) ) dθ 3π −0.3398 2 9 1 107 Make sure you can do=the trig manipulations required to do integrals. Most of th θ + θ + θ = these 42sin sin 2 42.0376 3.4814 ( ) ( ) ( ) 2 π 19 94 12 = 2will − 2 of cosmanipulation. ( 2θ ) dθ2 2 + 18sin ∫−0.3398 integrals this section involve this θ kind If you don’t recall how The area isinthen, them go back and take a1 look at the Integrals Involving3.4814 Trig Functions section. 19 9 θ4π−manipulations 18cos (θ ) − 4 sin 2θ ) ) to do=these 33.7074 ( Make sure you can do=the required integrals. Most of th 2 ( 2trig − 0.3398 2 2 3 1 1. Find theinarea thewill loop of . r = 3 − 8cos θ = ∫involve − − + Ainner 4 2 cos θ 6 2 cos θ integrals thisinside section this kind of manipulation. ( ) ( ) Ifdθyou don’t recall how t 2π 2 3 them go back and take a look at the Integrals Involving Trig Functions section. Ans. A = 15.2695 sq. units. Make sure you can do the trig 4 π manipulations required to do these integrals. Most of th 3 = ∫involve −10 −this θ dof θ manipulation. If you don’t recall how 20 cos integrals thisinside section kind 2. Find theinarea thewill graph 2 π of r = 7 + 3cos θ and to the left of the y-axis. Step 1 3 them go back and take a look at the the Integrals Involving Trig Functions section. Do not getisA too excited about all the second step above. Just because 4 π inwe Ans. =a42.0376 sq. units. First, here quick sketch of the graphminus of thesigns region are interested in. 3 −10θ θ ) )not = 13.6971 20sindoes ( ( terms have minus signs in=front of−them 2 π mean that we should get a negative va 3. Find the area that is inside r = 3 + 3sin θ and outside r = 2. 3 Step 1 out integral! is the “inner” graph. Problem SET # 8 Ans. is A =a33.7074 sq. units.of the graph of the region we are interested in. First, here quick sketch Do not the get area too excited aboutrall minus signs the integral. Just because all the term 4. Find that is inside and outside = 2the 3sin r =in3 + θ. minus signs in front of them does not mean that we should get a negative value from ou Ans. A = 3.8622 sq. units. 5. Find Step 1 the area that is inside r = 4 − 2 cos θ and outside r = 6 + 2 cos θ . Ans. is A =a13.6971 sq. units.of the graph of the region we are interested in. First, here quick sketch 6. Find the area that is inside both r = 1 − sin θ and r = 2 + sin θ . Step 1 #$ Ans. A = − 2 + 3 3 = 2.6578 . % First, here is a quick sketch of the graph of the region we are interested in. ...
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