- #1
Rijad Hadzic
- 321
- 20
1. The problem statement, all variables and given/k
nown data
[itex] x = (sin(t))^2 [/itex] [itex] y = (cos(t))^2 [/itex] t goes from 0 to 3 pi
∫[itex] \sqrt{ {(dx/dt)}^2 + {(dy/dt)}^2 } dt[/itex]
∫[itex] \sqrt{ {(2sin(t)cost)}^2 + {(-2cos(t)sin(t))}^2 } dt[/itex]
∫[itex] \sqrt{ 4(sin(t)cos(t))^2 + {(-2cos(t)sin(t))}^2 } dt[/itex]
∫[itex] \sqrt{ 4(sin(t)cos(t))^2 + 4(cos(t)sin(t))^2 } dt[/itex]
∫[itex] \sqrt{ 8(sin(t)cos(t))^2 } dt[/itex]
[itex]2\sqrt 2 [/itex]∫[itex] \sqrt{ (sin(t)cos(t))^2 } dt[/itex]
[itex]2\sqrt 2 [/itex]∫[itex] { (sin(t)cos(t)) } dt[/itex]
keep in mind my bounds are from 0 to 3 pi right now
so I integrate this using u = sin (t) du = cos(t) dt
and get
(root 2)(sin(t))^2 from 0 to pi/3
now when I evaluate I get 0-0
where did my integration go wrong?
nown data
[itex] x = (sin(t))^2 [/itex] [itex] y = (cos(t))^2 [/itex] t goes from 0 to 3 pi
Homework Equations
∫[itex] \sqrt{ {(dx/dt)}^2 + {(dy/dt)}^2 } dt[/itex]
The Attempt at a Solution
∫[itex] \sqrt{ {(2sin(t)cost)}^2 + {(-2cos(t)sin(t))}^2 } dt[/itex]
∫[itex] \sqrt{ 4(sin(t)cos(t))^2 + {(-2cos(t)sin(t))}^2 } dt[/itex]
∫[itex] \sqrt{ 4(sin(t)cos(t))^2 + 4(cos(t)sin(t))^2 } dt[/itex]
∫[itex] \sqrt{ 8(sin(t)cos(t))^2 } dt[/itex]
[itex]2\sqrt 2 [/itex]∫[itex] \sqrt{ (sin(t)cos(t))^2 } dt[/itex]
[itex]2\sqrt 2 [/itex]∫[itex] { (sin(t)cos(t)) } dt[/itex]
keep in mind my bounds are from 0 to 3 pi right now
so I integrate this using u = sin (t) du = cos(t) dt
and get
(root 2)(sin(t))^2 from 0 to pi/3
now when I evaluate I get 0-0
where did my integration go wrong?