find the coordinates of the centroid bounded by y=25x^2 and y=4. the region is covered by a thin flat plate

Answer 1

Last time I did this type of question was 55 years ago, so I hope I can remember.
I think we first have to find the area
intercepts between y = 25x^2 and y = 4
25x^2 = 4
x = ± 2/5
by symmetry we can go from 0 to 2/5 and double
area = 2(integral) (4 - 25x^2) dx from 0 to 2/5
= 2[ 4x - (25/3)x^3] from 0 to 2/5
= 2[ 8/5 - (25/3)(8/125) - 0 ]
= 2[ 16/15]
= 32/15

check my arithmetic

the x-coordinate
= (1/(32/15)) integral x(f(x)) dx from -2/5 to 2/5
= (15/32) integral 25x^3 dx from -2/5 to 2/5
= (15/32)[(25/4)x^4] from -2/5 to 2/5
= (15/32) ((25/4)(16/625) - (25/4)(16/625))
= 0 , ahhh, of course, look at the symmetry

for the y-coordinate:
= (1/(32/15)) (integral) (4^2 - (f(x)^2) /2 dx from -2/5 to 2/5
= (15/32) [8x - 125x^5/2] from -2/5 to 2/5
= (15/32) (16/5 - 125(32/3125)/2 - (-16/5 - 125(-32/3125)/2)
= (15/32)(32/5 + 8000/6250)
= 18/5 or 3.6

the centroid is (0,3.6)
according to my sketch this looks reasonable, most of the mass is towards the line y = 4

that does not look right according to my sketch

Answer 2

forget the very last line, I had made corrections, but forgot to delete that line