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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online