Find the area enclosed by the following parametric curves and the x−axis. Find its exact area, no decimals.
{x = 9e^t y= 6t-t^2
a = ∫y dx = ∫ g(t) f'(t) dt
y=0 at t=0,6 so
a = ∫[0,6] (6t-t^2) * 9e^t dt = 14595
To find the area enclosed by the parametric curves and the x-axis, we need to use the formula for the area between two curves in terms of a parameter.
The given parametric equations are:
x = 9e^t
y = 6t - t^2
To determine the bounds for the parameter 't', we need to find the values where the curves intersect with the x-axis. For that, we set y = 0:
0 = 6t - t^2
Rearranging the equation:
t^2 - 6t = 0
Factoring out t:
t(t - 6) = 0
So, t = 0 or t = 6.
Now, we need to find the integral of the absolute value of the y-coordinate with respect to t between t = 0 and t = 6. Since the curve lies below the x-axis between t = 0 and t = 6, we take the absolute value to find the area.
To calculate the area, we can use the integral:
Area = ∫ |y| dt
Substituting the given y-coordinate:
Area = ∫ |6t - t^2| dt
Now, we divide the interval [0, 6] into two parts where the y-coordinate is given by different expressions. In the first part, 0 ≤ t ≤ 6, the y-coordinate is positive, so the equation becomes:
Area = ∫ (6t - t^2) dt
= ∫ (6t - t^2) dt
= 6∫ t dt - ∫ t^2 dt
= 3t^2 - (t^3/3) + C
Evaluating the integral from t = 0 to t = 6:
Area = (3(6)^2 - (6^3/3)) - (3(0)^2 - (0^3/3))
= 108 - 0
= 108
Therefore, the exact area enclosed by the parametric curves and the x-axis is 108.