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 four-leafed 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 right-hand leaf, but from to you trace out the left-half of the lower leaf. You have many choices for a range of which
sweep out one leaf: [ to ] sweeps out the lower leaf; [ to 5 ] sweeps out the left-hand leaf; [ 5 to 7 ] sweeps out the upper-hand leaf; [ 7 to 9 ] sweeps out the right-hand leaf. We can also sweep out the right-hand 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 8 + 8 = #. 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 right-hand 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
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 = 5 7 7 cos d d + cos d d = sin + 5 5 7 sin 7 = sin sin + sin 7 sin 5 7 5 5 5. 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.
# 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 8 + 8. = 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
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 right-hand branch can be swept out by letting run from to, which sweeps out the top half of the right-hand piece. The top half of the left-hand 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
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