10.4 Areas in Polar Coordinates
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1 CHAPTER 0. PARAMETRIC AND POLAR Areas in Polar Coordinates Example. Find the area of the circle defined by r.5sin( ) (seeexample ). Solution. A Z 0.5 (.5sin( )) d sin( )cos( ).5 (.5/) (r where r.5/) 0 Example. Find the area contained in one loop for the following shape: r 4 cos( ) Solution. The hardest part of a lot of polar graph areas is figuring out the bounds of integration. For this, it usually helps to look at a graph, and then to solve some sort of equation. Here s the graph: graphics/three_pedaled_rose-eps-converted-to.pdf From the graph, we can see the following: So now we need to figure out the bounds for to describe one loop. You can also make a very rough guess for these bounds by the graph by imagining tangent lines that go through the origin and contain one loop:
2 CHAPTER 0. PARAMETRIC AND POLAR 9 graphics/single_polar_leaf_with_angles-eps-converted-to.pdf From this graph, you might guess that the bounds of the integral will by ±/4, or maybe ±/5, or ±/6. To be sure, we have to solve it algebraically. How do we solve for to get the bounds? The simplest way to do this is to set something equal to 0 and solve. There are three reasonable choices in this problem for what to set equal to 0: set y 0: y r sin( ) ) 04cos( )sin( ) set x 0: x r cos( ) ) 04cos( )cos( ) set r 0: r 4cos( ) ) 04cos( ) Any of these will work to some degree, but the last one works best: it gives only the values of where both x and y are 0. So, finally, let s solve this and integrate: set r 0: r 4cos( ) ) 04cos( ) cos( ) 0 cos(anything) 0 at anything ±/, ±/, ±5/,... ±/, ±/, ±5/,... ±/6, ±/6, ±5/6,... Thus, it appears that our bounds of integration are ±/6. confirm this, we graph the function between these bounds To
3 CHAPTER 0. PARAMETRIC AND POLAR 9 graphics/single_petal_polar_rose-eps-converted-to.pdf So, finally, we set up the integral and finish: A Z /6 /6 Z /6 /6 r d (4 cos( )) d (6) (6) Z /6 /6 cos ( ) d x + sin(6x) /6 /6 (6) () (/6) + sin(6(/6)) 4 Example. (a) Set up an integral to find the area of that part of the circle r sin( ) + 5 cos( ) that is in quadrant II. (b) Use your calculator/computer/online-math to evaluate the integral from part (a). Solution. We start by graphing the whole curve
4 CHAPTER 0. PARAMETRIC AND POLAR 94 graphics/area_in_qii_polar_circle-eps-converted-to.pdf As you can see, only a little part of the circle is in quadrant II. We need to solve for the values that just give us that part of the circle. The part of the circle that is in quadrant II is defined by where this graph crosses the y-axis, which means that the x-value is 0. So, we set x equal to 0 and then solve for : x r cos( ) 0(sin( )+5cos( )) cos( ) cos( ) 0) ±/, ±/, ±5/, ±7/,... or: sin( )+5cos( ) 0 sin( ) 5cos( ) sin( ) cos( ) 5 So, the integral is Z tan ( / 5/)+ tan( ) 5/ r d tan ( Z tan ( / 5/) + 5/)+ ( sin(theta)+5cos( )) d When we enter this in our calculator/computer/online-math we get tan ( 5/) 0.59 Can you tell if that looks about right? Yes: the part of the graph we want is part of a circle, with height and width about 0.5. If this were a rectangle, the area would be 0.5, but it s not the whole rectangle, so it s reasonable for the area to be 0.59.
5 CHAPTER 0. PARAMETRIC AND POLAR 95 Extra Examples Example 4. A chocolate company is coming out with a new chocolate product. It is going to make a mold which they can pour chocolate into. They need to know the volume of the mold, but basically this means they need to know area of the shape of the mold. The mold has shape given by the following r 6( /) +4 / +, apple apple +, units for r inches. (a) Use your calculator to calculate some points on the curve corresponding to, 0., 0.5, /6, 0.7, 0.9, /,.,.4 and /, and then plot them on the polar graph paper in Figure 0.. (b) Use your calculator to graph the polar function. (c) Find the area of the shape. Figure 0.: Polar graph paper for Example 4 /.9.75 /.5.4. / /6 / Solution. (a) We can put our calculators in polar mode, and enter the formula, and use the table function to get the following:
6 CHAPTER 0. PARAMETRIC AND POLAR 96 and the resulting points like like this graphics/points_on_heart_shaped_curve-eps-converted-to.pdf (b) When we graph the whole function with appropriate settings we get this graphics/heart_shaped_graph-eps-converted-to.pdf (c) To find the area we integrate Z + ( /) + / + d We perform a u-substitution to simplify this u / ) u du d +) u +
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