y = 16 -x^2 find centroid bounded by x axis
x = (1/A) ∫ x(f(x)) dx and (1/A) (1/2)(f(x))^2 dx = y
The Attempt at a Solution
I just applied it. It is a weird because I would of thought that x would of been at 0. But I didn't get that I x = 9/16 and y = 8/5. Y might be OK but wouldn't x have to be at 0?
I calculated the area I got 256/3
So x = (3/256) ∫(x(16-x^2) dx = (3/256)∫ 16x - x^3 dx = (3/256) (8x^2 -(x^4/4)) between [-4,4] = 9/16
For y = (3/256) ∫ (1/2)(16-x^2)^2 dx = (3/256)∫ 256 - 32x^2 + x^4 dx = (3/512)(256x = (32/2)x^3 - (x^5/5)) between [-4,4] = 8/5
Like I said it seems that x would be at 0 but I'm not sure I think I did everthing OK.