I'll assume that when you say bounded by the y-axis, you mean bounded on the left. You've also stated that the function is valid for only a positive y. Therefore all we have to do is integrate the function from x=0 to x=4.
The integral of sqrt(16-x²) = 1/2*x*sqrt(16-x²) + 8*Arcsin(x/4)
At x=4, this is 8*(PI/2) = 4PI
At x=0, this is 8*0 = 0
4PI-0=4PI
The area is 4 PI.
You could also just take the area of the circle with radius 4 divided by 4 which would give you the same answer.
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