MATH 2163 Midterm: MATH 2163 OK State PracticeExam3 sol

• School: OSU
• Department: Mathematics
• Course: MATH 2163
• Professor: All

Document Summary

0 (ey 1) dy = e 2. 6xyz dz dx dy =z 1 (3xy 3x3y) dx dy. =z 1 y3 r dr d =z /2. 2 sin d d d =z /2. 3 sin d d : d = {(r, )| 0 2 , 0 r 4 + 3 cos }, so. 0 z 4+3 cos da =z 2 . 3: solution 1: use a double integral, d is the intersection of (z = 3x2 + 3y2 z = 4 x2 y2. So d = {(r, )| 0 2 , 0 r 1}, and. V =zzd (top surface bottom surface) da. =z 2 ((4 r2) 3r2)r dr d = 2 . Solution 1: use triple integral in cylindrical coordinates, then. E = {(r, , z)| 0 2 , 0 r 1, 3r2 z 4 r2} and. E = {(x, y, z)| 0 x 1, 0 y 2 2x, 0 z 3(1 x y/2)}