Find the area enclosed by the x-axis and the parametric curve x=11t^3+8,y=t−t^2.Enter the exact answer as a decimal, but do NOT round. Hint: First find the intercepts of the curve with the x-axis.Show Work
when y=0, t = 0 or 1, so x = 8 or 19
That makes the area
∫[8,19] y dx = ∫[0,1] (t - t^2) (33t^2 dt) = 33/20
or, since t = ((x-8)/11)^(1/3)
the area is
∫[8,19] y dx = ∫[8,19] (((x-8)/11)^(1/3) - ((x-8)/11)^(2/3)) dx = 33/20
To find the area enclosed by the x-axis and the parametric curve, we need to integrate the absolute value of the y-coordinate with respect to t.
First, let's find the x-intercepts of the curve by setting y = 0 in the equation:
0 = t - t^2
t(1 - t) = 0
This equation has two solutions: t = 0 and t = 1.
Now, let's integrate the absolute value of the y-coordinate from t = 0 to t = 1:
∫ |y| dt = ∫ |t - t^2| dt
To evaluate this integral, we need to break it into two separate integrals based on the interval where y is positive or negative.
For t values between 0 and 1, the expression t - t^2 is positive, so the integral becomes:
∫ (t - t^2) dt
Integrating this expression gives us:
∫ (t - t^2) dt = (t^2/2) - (t^3/3) + C
Now, we need to subtract the integral of the negative portion of the curve from the positive portion.
For t values between 1 and 0, the expression t - t^2 becomes negative, so the integral becomes:
- ∫ (t - t^2) dt
Integrating this expression gives us:
- ∫ (t - t^2) dt = - (t^2/2) + (t^3/3) + C
Now, we can calculate the area by subtracting the negative integral from the positive integral:
Area = [(t^2/2) - (t^3/3)] evaluated from 0 to 1
= [(1^2/2) - (1^3/3)] - [(0^2/2) - (0^3/3)]
= (1/2) - (1/3) - 0
= 1/6
Therefore, the area enclosed by the x-axis and the parametric curve is 1/6.