ltumechanicaltronics
New member
- Joined
- Dec 12, 2020
- Messages
- 2
I need help finding the area between the parametric curve y=(t)^(1/2) x=(t^2)-2t and the Y axis. I found dx/dt=2t-2. I just guessed and checked in my ti-84 to find the interval.
I got S [0, 2pi/3] t^(1/2)*(t-2) (dt) which I integrated by parts to be (2/3)*(2t-2)*(t^(3/2))-(8/15)*(t^(5/2)) and plugged in 2pi/3 to get the answer
(8pi/9-4/3)(2pi/3)^(3/2)-(8/15)*(2pi/3)^(5/2). I've tried 3 different versions of this answer, and I have one last attempt to answer this problem. Is anything visibly wrong?
I got S [0, 2pi/3] t^(1/2)*(t-2) (dt) which I integrated by parts to be (2/3)*(2t-2)*(t^(3/2))-(8/15)*(t^(5/2)) and plugged in 2pi/3 to get the answer
(8pi/9-4/3)(2pi/3)^(3/2)-(8/15)*(2pi/3)^(5/2). I've tried 3 different versions of this answer, and I have one last attempt to answer this problem. Is anything visibly wrong?