Question

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!

Answer 1

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

Answer 2

https://www.youtube.com/watch?v=GDLZYp2U9g8