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
Unformatted text preview: 18.02A IAP 2010: Solutions Problem Set 7 Part II (30 points) Problem 1. We use cylindrical coordinates: x = r cos , y = r sin , and z , where in our case 0 z y/ 2 = ( r sin ) / 2, from which we get y 0, i.e., 0 . From x 2 + y 2 1 we get r 1. Then the mass of the wedge W is: ZZZ W p x 2 + y 2 + z 2 dV = ZZZ W r 2 dz dr d = Z Z 1 Z r sin 2 r 2 dz dr d = Z Z 1 r 3 sin 2 dr d = 1 2 Z sin d Z 1 r 3 dr = 1 2 [- cos ] =0 r 4 4 1 r =0 = 1 4 . Problem 2 We use spherical coordinates, in which we have a b . The density being proportional to the center means that ( x,y,z ) = p x 2 + y 2 + z 2 = . Then the mass of the spherical shell D is: ZZZ D ( x,y,z ) dV = ZZZ D 3 sin ddd = Z 2 Z Z b a 3 sin ddd = Z 2 d Z sin d Z b a 3 d = 2 2 b 4 4- a 4 4 = ( b 4- a 4 ) . Problem 3 Place the sphere so the point mass P is at the origin. Let Q be diametrically opposite to P (so it has coordinates (0 , , 2 a )), and form the right triangle with vertices P , Q , and R , where R is an arbitrary point on the surface of the sphere. By trigonometry, weis an arbitrary point on the surface of the sphere....
View Full Document
This note was uploaded on 05/14/2010 for the course 18.02A 18.02A taught by Professor Johnbush during the Winter '10 term at MIT.
- Winter '10