hw10sol - Math 53 Homework 10 Solutions 15.8 # 9: (a) z 2 =...

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Unformatted text preview: Math 53 Homework 10 Solutions 15.8 # 9: (a) z 2 = x 2 + y 2 (right angled cone centered on z-axis with vertex at origin) corresponds to = / 4 and also = 3 / 4 (since the equation z = radicalbig x 2 + y 2 actually describes two cones, one centered on the positive z-axis and the other one centered on the negative z-axis). Or: z 2 = x 2 + y 2 z 2 = r 2 2 cos 2 = 2 sin 2 cos 2 = sin 2 ( cos = sin = / 4 or 3 / 4.) (b) x 2 + z 2 = 9 ( sin cos ) 2 + ( cos ) 2 = 9 2 (sin 2 cos 2 + cos 2 ) = 9. 15.8 # 14: csc sin 1 r 1, or equivalently x 2 + y 2 1, which corresponds to the solid cylinder of unit radius centered on the z-axis. Moreover, 2 corresponds to the solid sphere of radius 2 centered at the origin. Hence, this solid is the intersection of the sphere and the cylinder. (The sphere and the cylinder intersect at the two circles r = 1, z = 3; so the boundary of the solid is given by the portion of the cylinder where 3 z 3, and spherical caps at the top and bottom). 15.8 # 15: The cone z = radicalbig x 2 + y 2 corresponds to z = r , i.e. cos = sin , i.e. = / 4. (See also 15.8 # 9(a)). Thus, the region above the cone corresponds to / 4. In spherical coordinates, the sphere x 2 + y 2 + z 2 = z (centered at (0 , , 1 2 ) and of radius 1 2 , since the equation rewrites as x 2 + y 2 + ( z 1 2 ) 2 = 1 4 ) has equation = cos . (This can be seen either geometrically on a slice by a vertical plane, or by manipulating the equation: x 2 + y 2 + z 2 = z becomes 2 = cos , which simplifies to = cos ). Hence, the solid is described by the inequalities cos , 0 / 4. (See Example 4 on page 1009 for a more detailed discussion and pictures). 15.8 # 19: In cylindrical coordinates: 0 z 2, 0 r 2, 0 / 2, so the integral is given by integraltext / 2 integraltext 3 integraltext 2 f ( r cos , r sin , z ) r dz dr d . In spherical coordinates: the top plane has equation z = cos = 2, i.e. = 2 sec . The cylinder corresponds to r = sin = 3, i.e. = 3 csc . They intersect when 2 sec = 3 csc , i.e. tan = 3 / 2. Therefore: integraltext / 2 integraltext tan- 1 (3 / 2) integraltext 2 sec f 2 sin ddd + integraltext / 2 integraltext / 2 tan- 1 (3 / 2) integraltext 3 csc f 2 sin ddd. 15.8 # 23: The spheres correspond to = 1 and = 2, and the first octant corresponds to 0 / 2, / 2. So integraltextintegraltextintegraltext E z dV = integraltext / 2 integraltext / 2 integraltext 2 1 ( cos ) 2 sin ddd ....
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hw10sol - Math 53 Homework 10 Solutions 15.8 # 9: (a) z 2 =...

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