Find the area of the region in the first quadrant that is below the parametric curve?
the parametric curves are
x = t^3 + 6t
y = 4t - t^2
any help would be greatly appreciated. Thanks!
First of all, here is a graph of the curve
http://www.wolframalpha.com/input/?i=plot+x+%3D+t%5E3+%2B+6t%2C++y+%3D+4t+-+t%5E2+
As you can see, when t = 0 , x = 0, y = 0
when t = 4, x = 88, y = 0
so our parameter will run form t = 0 to t = 4
in general the area would be
integral ( y ) dx from left boundary to right boundary
we want to sub values for y and for dx
x = t^3 + 6t
dx/dt = 3t^2 + 6
dx = (3t^2 + 6) dt
so Area = ∫y dx from t=0 to t=4
= ∫(4t-t^2)(3t^2 + 6) dt from 0 to 4
= ∫(12t^3 + 24t - 3t^4 - 6t^2) dt
= [ 3t^4 + 12t^2 - (3/5)t^5 - 2t^3 | from 0 to 4
= (768 + 192 - 614.4 - 128) - 0
= 217.6
check my arithmetic and work
Here is a youtube of a similar questions, although the author makes a silly error
https://www.youtube.com/watch?v=GDLZYp2U9g8
To find the area of the region below the parametric curve in the first quadrant, you can use the integral of the y-coordinate with respect to the x-coordinate.
First, let's find the range of values for t that correspond to the first quadrant. Since we want x and y to be positive, we can set up the inequalities:
0 ≤ x = t^3 + 6t
0 ≤ y = 4t - t^2
Solving the first inequality:
t^3 + 6t ≥ 0
t(t^2 + 6) ≥ 0
The value of t that satisfies this inequality is t ≥ 0.
Solving the second inequality:
4t - t^2 ≥ 0
t(4 - t) ≥ 0
The values of t that satisfy this inequality are 0 ≤ t ≤ 4.
So, we need to integrate the y-coordinate with respect to the x-coordinate over the range of x values:
∫[x₁, x₂] y dx
To find the limits of integration, we need to express x as a function of y. From the given parametric equations:
x = t^3 + 6t
t = (x - x^3/6)^(1/3)
Now, we can find the limits of integration by substituting the range of t values into the x expression:
x₁ = (0^3 + 6*0) = 0
x₂ = (4^3 + 6*4) = 88
So, the integral becomes:
∫[0, 88] y dx
Now, we need to express y as a function of x using the given parametric equations:
y = 4t - t^2
y = 4(x - x^3/6)^(1/3) - (x - x^3/6)^(2/3)
We can now integrate this expression with respect to x over the range [0, 88]:
Area = ∫[0, 88] (4(x - x^3/6)^(1/3) - (x - x^3/6)^(2/3)) dx
Evaluating this integral will give you the area of the region below the parametric curve in the first quadrant.