find the area of the region f(y)=y/(16-y^2)^1/2, g(y)=0, y=3

To find the area of the region bounded by the functions f(y), g(y), and the lines y = 3, you need to set up an integral.

First, find the points of intersection between f(y) and the line y = 3. Set f(y) equal to 3, and solve for y:

3 = y/(16 - y^2)^(1/2)

To simplify the equation, square both sides:

9 = y^2/(16 - y^2)

Cross-multiply:

9(16 - y^2) = y^2

144 - 9y^2 = y^2

10y^2 = 144

y^2 = 14.4

y = ± √14.4

Since we are looking for the area between 0 and 3, we only consider the positive value, y = √14.4.

Next, we need to determine if f(y) or g(y) is the upper function. To do this, find the point of intersection between f(y) and g(y). Set f(y) equal to g(y), and solve for y:

y/(16 - y^2)^(1/2) = 0

This equation is satisfied when the numerator is zero, so y = 0.

Therefore, between y = 0 and y = √14.4, the function g(y) is the upper function, and f(y) is the lower function.

Now, we can set up the integral to find the area A:

A = ∫[√14.4, 3] [g(y) - f(y)] dy

Since g(y) = 0, the integral reduces to:

A = ∫[√14.4, 3] -f(y) dy

The expression -f(y) represents the negative curve of f(y), so the integral computes the area below the curve.

To evaluate this integral using calculus, you need to find the antiderivative of the function -f(y). The antiderivative of y/(16 - y^2)^(1/2) is arcsin(y/4). After finding the antiderivative, evaluate the integral using the limits of integration.

A = ∫[√14.4, 3] -f(y) dy
= -∫[√14.4, 3] y/(16 - y^2)^(1/2) dy
= -[arcsin(y/4)] [√14.4, 3]

Now, substitute the limits of integration into the antiderivative:

A = -[arcsin(3/4) - arcsin(√14.4/4)]

Finally, evaluate the expression to find the area A.