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Unformatted text preview: Math 2220 Problem Set 7 Solutions Spring 2010 Section 1.7 : 20. (a) Graph the curve in R 2 having polar equation r = 2 a sin where a is a positive constant. Solution. There are two approaches to a question like this: Either try to graph the curve directly using polar coordinates, or try to convert the equation to xycoordinates and graph the xyequation. The second approach turns out to work nicely for the equation r = 2 a sin . We have r = radicalbig x 2 + y 2 and sin = y/r = y/ radicalbig x 2 + y 2 so we get the equation radicalbig x 2 + y 2 = 2 ay/ radicalbig x 2 + y 2 which is equivalent to x 2 + y 2 = 2 ay . This can be rewritten as x 2 + ( y a ) 2 = a 2 , so it is a circle of radius a with center at (0 , a ). (b) Graph the surface in R 3 having spherical equation = 2 a sin . Solution. It is possible to convert this equation into an equation in xyzcoordinates, but this equation is fairly complicated. Instead, notice that the equation = 2 a sin does not involve the variable , which means that is completely arbitrary, hence the surface has to be a surface of revolution about the zaxis, so all we need to figure out is what the curve is that is being rotated about the zaxis. The variable represents distance to the origin and is the angle down from the positive zaxis, so by part (a) the curve being rotated is a circle in the yzplane of radius a and center at (0 , a, 0). Thus the surface looks like a donut or bagel whose hole has shrunk to just a single point. 22. Graph the surface whose spherical equation is = 1 sin . Solution. Since the equation does not involve , this surface is a surface of revolution about the zaxis. To find out what the curve is that is being rotated, we look at the corresponding polar equation r = 1 sin . Converting this to xycoordinates is not very helpful, so we try to graph it directly. In rectangular coordinates the graph of r = 1 sin is shown on the left below, and this yields the graph on the right for the graph of r = 1 sin in polar coordinates. (The graph on the left tells you how far to go radially outward from the origin on each ray in the graph on the right.) This heartshaped curve is known as a cardioid. Math 2220 Problem Set 7 Solutions Spring 2010 For spherical coordinates we restrict to the interval [0 , ], so we only use the part of the cardioid above the xaxis. Rotating this about the zaxis gives the surface shown at the right. 34. Sketch the solid whose spherical coordinates ( , , ) satisfy 0 2 / cos , 0 / 4. Solution. The equation = 2 / cos is equivalent to 2 = cos = z so this equation defines the horizontal plane z = 2. The inequalities / 4 define the region inside a cone. The solid we want is the region inside the cone and below the plane z = 2....
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This note was uploaded on 04/21/2010 for the course MATH 2220 taught by Professor Parkinson during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 PARKINSON
 Math, Multivariable Calculus

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