A 3-dimensional structure is obtained from rotating the parabola y=x^2 about the y-axis. Each second, 2π units^3 of water is being poured into the structure from the top. When 8π units^3 of water has been poured in the structure, the instantaneous change in water height level is a/b, where a and b are coprime positive integers. What is the value of a+b?
To find the value of a+b, we first need to determine the volume of the 3-dimensional structure obtained from rotating the parabola y=x^2 about the y-axis.
The volume of a solid of revolution can be calculated using the method of cylindrical shells. In this case, we rotate the parabola y=x^2 about the y-axis, so we integrate with respect to y.
The equation of the parabola is y = x^2. We need to express x in terms of y, so taking the square root, we get x = √y.
The limits of integration will be determined by the points of intersection of the parabola and the y-axis. Since the parabola is symmetric about the y-axis, the only point of intersection is at (0, 0).
To find the volume, we integrate the expression 2πx * h * dy, where h is the height of the parabola at y and x is the radius. In this case, since we're rotating about the y-axis, the radius is x = √y.
So, the integral to find the volume becomes:
V = ∫(0 to b) 2π√y * h * dy,
where b represents the height of the structure obtained from rotating the parabola.
To calculate the volume V, we need to find the height b such that V = 8π units^3.
Integrating the above expression will give us the volume in terms of b:
V = ∫(0 to b) 2π√y * h * dy = 8π.
Solving this integral will give us the value of b, which represents the height of the structure.
Once we have the value of b, we differentiate the equation with respect to time to find the rate of change of water height level, which is the instantaneous change.
Taking the derivative of both sides, we have:
dV/dt = (d/dt) ∫(0 to b) 2π√y * h * dy.
Since we are pouring water into the structure at a rate of 2π units^3 per second, we can substitute dV/dt with 2π.
Therefore:
2π = (d/dt) ∫(0 to b) 2π√y * h * dy.
The instantaneous change in water height level is given by:
(d/dt) ∫(0 to b) 2π√y * h * dy = 2π / dt.
To compute this derivative, we can apply the Leibniz rule for differentiating definite integrals with respect to time. This rule states that the derivative of a definite integral is equal to the integrand evaluated at the upper limit times the derivative of the upper limit function.
Using the Leibniz rule, the left-hand side becomes:
2π√b * (dh/dt) = 2π / dt.
Simplifying and solving for (dh/dt), we have:
(dh/dt) = (1 / (2π√b)).
Now we need to find the value of (dh/dt) when V = 8π units^3, which gives us the value of b.
We substitute V = 8π into the integral equation:
∫(0 to b) 2π√y * h * dy = 8π.
After solving this equation for b, we substitute the value of b into (dh/dt) to determine the instantaneous change in water height level.
Finally, we compute a + b to find the value of a+b.