Center of Mass: 2-Dimensional Case The System’s Center of Mass is defined to be: My x= M n M =∑ m k i =1 n M y =∑m k xk i =1 n M x =∑m k yk i =1 y= ( x, y ) Mx M ⎛ x⎞ ⎛ x⎛ M = ∫ρ ⋅Aρea⎜ ⎟d ⎛ ⎛ ⎝ y⎠ ⎛ y⎛ a b ⎛ x⎞ ⎛ ⎛ x⎛ ⎛ ⎛ x⎛ M y = ∫ρ ⋅Aρea⎜ ⎟ ⎛ xk ⎛ ⎛ ⎛ d ⎛ ⎛ ⎝ y⎠ ⎛ ⎛ y⎛ ⎛ ⎛ y⎛ a b ⎛ x⎞ ⎛ ⎛ x⎛ ⎛ ⎛ x⎛ M x = ∫ρ ⋅Aρea⎜ ⎟ ⎛ yk ⎛ ⎛ ⎛ d ⎛ ⎛ ⎝ y⎠ ⎛ ⎛ y⎛ ⎛ ⎛ y⎛ a b
Find the center of mass of the the lamina R with density 1/3 in the region in the xy plane bounded by y = x2 and y = x + 2. Use slices perpendicular to the x-axis. [ −1, 2 ] Bounds: Top: x+2 Bottom: Each slice has balance point: x2 ⎛ ( x + 2 ) + x2 ⎞ x, ⎜ ⎟ 2 ⎝ ⎠ 2 3 ⎛1⎞ M = ∫⎜ ⎟ (( x + 2 ) −x2 ) dx = ⎝ 3⎠ 2 −1 2 3 ⎛1⎞ M y = ∫⎜ ⎟ (( x + 2 ) −x2 )( x) dx = ⎝ 3⎠ 4 −1 2 ⎛1 8⎞ ⎜⎝ , ⎟⎠ 2 5 2 ⎛ (x + 2 ) + x ⎛ 12 ⎛1⎞ M x = ∫⎜ ⎟ (( x + 2 ) −x2 )⎛ dx = ⎛⎛ ⎝ 3⎠ 2 5 ⎛ −1 2 To matching answers for Mx (with length) use the property: (a - b)(a + b) = a2 - b2 ⎛1⎞ 2 M x = ∫⎜ ⎟ (( x + 2 ) −x4 )dx ⎝6⎠ −1
Integration by Parts: “Undoing” the Product Rule for Derivatives ∫x ln(x)dx Consider: We have no formula for this integral. Notice that x and ln(x) are not related by derivatives, so we cannot use the substitution method.
Integration by Parts: “Undoing” the Product Rule for Derivatives Look at the derivative of a product of functions: d dv du uv = u +v ( ) dx dx dx Let’s use the differential form: d (uv ) =udv+ vdu And solve for udv udv =d (uv) −vdu Integrating both sides, we get: ∫udv = ∫d (uv) −∫vdu
Integration by Parts: “Undoing” the Product Rule for Derivatives Integrating both sides, we get: ∫udv = ∫d (uv) −∫vdu Or ∫udv =uv−∫vdu ∫vdu should be simpler that the original ∫udv The integral If two functions are not related by derivatives (substitution does not apply), choose one function to be the u (to differentiate) and the other function to be the dv (to integrate)
Integration by Parts Back to: ∫udv =uv−∫vdu ∫x ln(x)dx Choose u (to differentiate (“du”)) dv (to integrate (“v”)) u =ln(x) 1 du = dx x dv = xdx x2 v= 2 x2 ⎛ x 2⎞⎞ ⎛ 1 ⎞ ∫x ln(x)dx = 2 ln(x) −∫ ⎜⎝ 22⎟⎠⎟⎠dx⎜⎝ x ⎟⎠ dx x2 x2 = ln(x) − + C 2 4 This second integral is simpler than the first
Integration by Parts Evaluate: y 3 5 ∫y e ∫udv =uv−∫vdu dy Choose u (to differentiate (“du”)) dv (to integrate (“v”)) y 5 u =y 3 y 3 5 y 2 5 dv =ye dy 3y 2 5e 5 6y 25e 6 125e 0 625e y 5 y 5 y 5 y 5 y 5 y 5 5y e −75 y e + 750 ye −3750 e =e ( 5y 3 − 75y 2 + 750y − 3750 )
Integration by Parts ∫udv =uv−∫vdu Integrate: ∫cos (ln(x))dx ∫cos (ln(x))dx =x cos ( ln(x)) + ∫ sin ( ln(x)) dx u =sin (ln(x)) 1 du = cos(ln(x))dx x dv =dx v =x ∫cos (ln(x))dx =x cos ( ln(x)) +x sin ( ln(x)) −∫ cos ( ln(x)) dx + ∫ cos ( ln(x)) dx + ∫ cos ( ln(x)) dx 2 ∫cos (ln(x))dx =x cos ( ln(x)) + x sin ( ln(x)) 1 cos ln(x) dx = ( ) ( x cos ( ln(x)) + x sin ( ln(x))) ∫ 2
