Find the area of the surface obtained by rotating the curve for y=x^4/16 +1/2x^2 about the y axis

also needs to be done for the x axis as well

What is the domain for integration (x values)?

In the mean time...
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Assuming that you mean y = x^4/16 + 1/(2x^2):
ds = √(1 + (dy/dx)^2) dx
.....= √[1 + (x^3/4 - 1/x^3)^2] dx
.....= √[1 + (x^6/16 - 1/2 + 1/x^6)] dx
.....= √(x^6/16 + 1/2 + 1/x^6) dx
.....= √(x^3/4 + 1/x^3)^2 dx
.....= (x^3/4 + 1/x^3) dx, assuming that x > 0.

Revolving the curve about the y-axis yields
A = ∫ 2πx ds = ∫ 2πx * (x^3/4 + 1/x^3) dx.

Revolving the curve about the x-axis yields
A = ∫ 2πy ds = ∫ 2π(x^4/16 + 1/(2x^2)) * (x^3/4 + 1/x^3) dx.

This is straightforward (with whatever the bounds of integration are).