Find coordinates of the centroid of y = h(1 + x/L - 2x²/L²) curve enclosed by axes in 1st quadrant.
Answers
Answer:
Using intregation
Explanation:
y = h(1+ x/L-2 (x^2)/(l^2))
¯x=(∫_0^L▒〖x[h〗 (1+ x/L-2 (x^2)/(l^2))].dx )/(∫_0^L▒〖(1+ x/L-2 (x^2)/(l^2)) 〗.dx )
¯x=(〖h[(x^2)/2+ (x^3)/3L-(2x^3)/(L^2)]〗_0^L)/([x+ (x^2)/2-(2x^3)/(3L^2)]_0^L )
¯x=h[(L^2)/2+ (L^3)/3L-(Lx^3)/(L^2)]/[L+ (L^2)/2-(2L^3)/(3L^2)]
¯x= h[(L^2)/2-L/6 ]/[L+L/6 ]
¯x= 6h/7L [(L^2)/2-L/6 ]
¯x= 6h/7 [L/2-1/6 ]
¯y=(∫_a^b▒〖1/2 y^2〗.dx )/(∫_a^b▒〖y 〗.dx )
¯y=(∫_0^L▒〖h^2〗 (1+ x/L-2 (x^2)/(l^2)).dx )/(∫_0^L▒〖(1+ x/L-2 (x^2)/(l^2)) 〗.dx )
¯y=(〖(h〗^2*4L)/5)/(7L/6)
¯y= (24h^2)/35
y = h(1+ x/L-2 (x^2)/(l^2))
¯x=(∫_0^L▒〖x[h〗 (1+ x/L-2 (x^2)/(l^2))].dx )/(∫_0^L▒〖(1+ x/L-2 (x^2)/(l^2)) 〗.dx )
¯x=(〖h[(x^2)/2+ (x^3)/3L-(2x^3)/(L^2)]〗_0^L)/([x+ (x^2)/2-(2x^3)/(3L^2)]_0^L )
¯x=h[(L^2)/2+ (L^3)/3L-(Lx^3)/(L^2)]/[L+ (L^2)/2-(2L^3)/(3L^2)]
¯x= h[(L^2)/2-L/6 ]/[L+L/6 ]
¯x= 6h/7L [(L^2)/2-L/6 ]
¯x= 6h/7 [L/2-1/6 ]
¯y=(∫_a^b▒〖1/2 y^2〗.dx )/(∫_a^b▒〖y 〗.dx )
¯y=(∫_0^L▒〖h^2〗 (1+ x/L-2 (x^2)/(l^2)).dx )/(∫_0^L▒〖(1+ x/L-2 (x^2)/(l^2)) 〗.dx )
¯y=(〖(h〗^2*4L)/5)/(7L/6)
¯y= (24h^2)/35
y = h(1+ x/L-2 (x^2)/(l^2))
¯x=(∫_0^L▒〖x[h〗 (1+ x/L-2 (x^2)/(l^2))].dx )/(∫_0^L▒〖(1+ x/L-2 (x^2)/(l^2)) 〗.dx )
¯x=(〖h[(x^2)/2+ (x^3)/3L-(2x^3)/(L^2)]〗_0^L)/([x+ (x^2)/2-(2x^3)/(3L^2)]_0^L )
¯x=h[(L^2)/2+ (L^3)/3L-(Lx^3)/(L^2)]/[L+ (L^2)/2-(2L^3)/(3L^2)]
¯x= h[(L^2)/2-L/6 ]/[L+L/6 ]
¯x= 6h/7L [(L^2)/2-L/6 ]
¯x= 6h/7 [L/2-1/6 ]
¯y=(∫_a^b▒〖1/2 y^2〗.dx )/(∫_a^b▒〖y 〗.dx )
¯y=(∫_0^L▒〖h^2〗 (1+ x/L-2 (x^2)/(l^2)).dx )/(∫_0^L▒〖(1+ x/L-2 (x^2)/(l^2)) 〗.dx )
¯y=(〖(h〗^2*4L)/5)/(7L/6)
¯y= (24h^2)/35