9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr


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1 9.9 #. Area inside the oval limaçon r = + cos. To graph, start with = so r =. Compute d = sin. Interesting points are where d vanishes, or at =,,, etc. For these values of we compute r:,,, and the values repeat. Hence, starting at = and rotating counterclockwise, we see the point moving in along the ray starting at until at = is has moved into. As the ray moves from = to =, the point move out along the ray starting a and finishing at. Just to get a good picture it is worthwhile to plug in = and = where r =. Hence the area we want is swept out once as rotates from to. From the formula in the book Area = r d = + cos d = d + cos d + cos d. Do the pieces: d = = ; cos d = sin = = ; cos + cos d = d = d + cosd. Pause to do cosd = sin = =. Hence cos d = =. Hence the Area is + + = 8. #. Area inside one leaf of the fourleafed rose r = cos. Begin with the graph, starting with =. d = sin which vanishes when =,,,,, etc. or when =,,,,, etc.: r itself vanishes when =,, 5, 7, etc. A remark which is apparent if you aw the graph but not if you just look at it is that from to you trace out the top half of the righthand leaf, but from to you trace out the lefthalf of the lower leaf. You have many choices for a range of which
2 sweep out one leaf: [ to ] sweeps out the lower leaf; [ to 5 ] sweeps out the lefthand leaf; [ 5 to 7 ] sweeps out the upperhand leaf; [ 7 to 9 ] sweeps out the righthand leaf. We can also sweep out the righthand leaf with [ to ] and this is the one we cos d = + cos d = choose. Hence the Area is r d = d + cos d. Do the pieces: cos d = sin d = # 7. Area shared by the circles r = cos and r = sin. = = + = ; = sin sin =. The Area is + = 8. By awing the graph, you see each circle can be swept out once by letting run from to. The polar coordinates of the intersection point can be found by solving cos = sin, or tan = or =. While there are many solutions to the equation tan =, they are all obtained by adding integer multiples of to and we see that the only one between and is. Hence the Area is sin d + cos d = sin d + cos d. Do the pieces: sin cos d = d = d cosd = sin = sin sin = = 8 ; cos + cos d = d = d + cosd = + sin = + sin sin = + =. Hence the 8 Area is = #. Inside the lemniscate r = cos and outside the circle r =. The lemniscate can be graphed as follows. It is actually two equations r = ± cos. Intervals where cos < are excluded: these intervals are,, 5, 7, etc. The righthand loop of the lemniscate is traced out by starting at and going to. The entire lemniscate can be described as the graph of r = cos where runs over the
3 intervals [, ] and [, 5 ]. Next we need to find the four points of intersection, so solve cos = r = or cos = so = + k, k an integer, or = + k. Hence = ± + k and the four points are = st quaant; th quaant; 7 rd quaant; and 5 nd quaant. The desired Area is cos d + cos d = cos d d + cos d d = sin sin 7 = sin sin + sin 7 sin Now sin = ; sin = sin. Since 7 = +, sin 7 = sin and 7 similarly, sin 5 = sin = sin. Hence Area= =. #. Find the length of the cardioid r = + cos. The graph is swept out once as runs from to. Length= r + r d. Compute as follows. d = sin, so r = sin so r + r = + cos + sin = + cos + cos + sin = + cos = + cos = + cos = cos. Hence Length = cos d = cos d = cos d cos d = sin sin = sin sin sin sin = 8.
4 # 5. Find the length of the curve r = cos. There is no need to graph this curve since we are told the limits of integration, but just for the record, here is the graph. cos cos Next compute d = cos sin +sin cos = cos cos = sin cos. Hence r + r = +sin = cos. Between and, >, so Length = r + r d = cos + cos d = d = d+ cos d = + sin = + sin sin = # 9. Find the surface area generated by revolving r = cos, about the y axis. Again the graph is not necessary but is included so you may practice if you wish. Compute as follows. d = sin cos sin cos = cos + sin = cos book we have the formula, SurfaceArea = cos cos cos d = = sin cos ; r + r = cos + cos. Hence r + r =. From the cos r cos r + r d = cos d = sin = sin
5 sin = = #. Find the surface area generated by revolving r = cos about the x axis. This time a graph is helpful in determining the limits of integration. The function cos vanishes at,,, 5, etc. so the curve can be described as r = cos for [, ] and [, 5 ]. A further problem occurs. To get the surface of revolution, we should only rotate the top half or the bottom half of the curve. The righthand branch can be swept out by letting run from to, which sweeps out the top half of the righthand piece. The top half of the lefthand piece needs to run from to. The next step is to compute r + r. We just did this in #9 and we got r + r =. cos Hence SurfaceArea = sin d + cos + cos cos d cos sin + cos sin d = cos + cos. Since cos and cos = we get SurfaceArea =. d cos sin = cos = cos = ; cos = ; cos = ; # 7. Find the centroid of the region enclosed by the cardioid r = a + cos. Look at # for the graph of the cardioid with a = and recall that the cure is swept out once as runs from to. d To find the centroid we first calculate the area: Area = a + cos = a + cos +cos d = a d+ cos d+ cos d = a d+ cos d + + cos d = a d + cos d + cos d. The integral involving cos is as is the integral involving cos so Area = a. Now we turn to the moments. 5
6 The moment about the x axis is Substitute u = + cos, du = sin d. r sin d = a + cos sin d = a + cos sin d. a u du =. Once can also argue from symmetry: there is as much of the curve above the x axis as below it. a The moment about the y axis is r cos d = a +cos cos d = + cos + cos +cos cos d = a cos + cos + cos +cos d =. We can do cos d by parts as follows. Let u = cos, dv = cos d. Then du = cos sin d and v = sin. Hence cos d = cos sin sin cos sin d = cos sin + cos sin d. Now write sin = cos so cos sin d = cos d cos d. Plug back in and solve for cos d: cos d = cos sin + cos d, or cos d = cos sin + cos d. A similar Integration by Parts and solving for the integral gives cos d = cos sin + cos d. Hence cos sin cos + cos + cos + cos d = + cos sin. Further cos d = and. Hence the moment about the y axis = a 5 = 5a Therefore, the x coordinate of the center of mass is the center of mass is.. 5a a cos + 5 cos d + cos d = = 5a + cos d =. The y coordinate of
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