MATH 141 Lecture 6: MATH 141 - Lecture 6- Sept 7

• School: University of Maryland
• Department: Mathematics
• Course: MATH 141
• Professor: Patrick Brosnan

Document Summary

Recall that if we have n masses m1,m2 mn at points (x1,y1) . (xn,yn), then the moments are: Mx = m1y1 + + mnyn about x and y axis (cid:1829)(cid:1857)(cid:1866)(cid:1872)(cid:1870)(cid:1867)(cid:1856) (cid:1871) (cid:4666)(cid:1876) ,(cid:1877) (cid:4667)=((cid:3052),(cid:3051)) (cid:1875) (cid:1857)(cid:1870)(cid:1857) =(cid:1871)(cid:1873)(cid:1865) (cid:1867)(cid:1858) (cid:1865)(cid:1853)(cid:1871)(cid:1871)(cid:1857)(cid:1871) Now let r be a region in r2 viewed as having density (cid:2025) (cid:1830)={(cid:4666)(cid:1876),(cid:1877)(cid:4667) (cid:2870) (cid:1853) (cid:1876) (cid:1854) (cid:1853)(cid:1866)(cid:1856) (cid:1859)(cid:4666)(cid:1876)(cid:4667) (cid:1877) (cid:1858)(cid:4666)(cid:1876)(cid:4667) Since the density is 1, mass is just a(d) = area of d. To compute my , break d into vertical strips centered at x = x0 and thickness dx. Mass of strip is area = height x dx (since density = 1) (cid:1856)={(cid:1858)(cid:4666)(cid:1876)0(cid:4667) (cid:1859)(cid:4666)(cid:1876)0(cid:4667)}(cid:1856)(cid:1876) Contribution of strip to my = x. da = (cid:1876)0((cid:1858)(cid:4666)(cid:1876)0(cid:4667) (cid:1859)(cid:4666)(cid:1876)0(cid:4667))(cid:1856)(cid:1876) (cid:3029) (cid:3028) (cid:2184) (cid:2183) (cid:2184) (cid:2183) (cid:3031) Y x area = y dx dy (since area of small horizontal part in a vertical strip is dx. dy) For mx , we break the vertical strip into horizontal strips.