- #1
PolarClaw
- 5
- 0
Hi,
I've been trying to do this one question: Let R be the region enclosed by the graph x=t^2-2 y=t^3-2t. Set up the integral for the area of R.
I know that if y is continuous function of x on an interval a ≤ x ≤ b where x=f(t) and y=g(t) then [tex]\int_{a}^{b}[/tex] y dx =[tex]\int_{t1}^{t2}[/tex] g(t)f'(t)dt provided that f(t1)=a and f(t2)=b and both g and f' are continuous on [t1,t2].
But everytime i equate and solve for t i get sqrt(2) and 1. This is where i get lost I can't seem to find the interval for parametric equations (any of them ). Other then the interval I do believe i have the rest of the integral:
[tex]\int[/tex] (t^3-2t)(2t)dt
Thanks in advance (sorry for long post as well as horrid use of that latex typing first time)
I've been trying to do this one question: Let R be the region enclosed by the graph x=t^2-2 y=t^3-2t. Set up the integral for the area of R.
I know that if y is continuous function of x on an interval a ≤ x ≤ b where x=f(t) and y=g(t) then [tex]\int_{a}^{b}[/tex] y dx =[tex]\int_{t1}^{t2}[/tex] g(t)f'(t)dt provided that f(t1)=a and f(t2)=b and both g and f' are continuous on [t1,t2].
But everytime i equate and solve for t i get sqrt(2) and 1. This is where i get lost I can't seem to find the interval for parametric equations (any of them ). Other then the interval I do believe i have the rest of the integral:
[tex]\int[/tex] (t^3-2t)(2t)dt
Thanks in advance (sorry for long post as well as horrid use of that latex typing first time)