Help with a line integral please

Click For Summary
The discussion revolves around a line integral expressed as ∫zds, which was initially complicated but simplified to ∫a^2cos(t)*(cos^2(2t)+5sin^2(t))^1/2 dt over the interval (0≤t≤pi/2). The user encountered difficulties in rewriting or integrating the expression. A suggestion was made regarding a correction in the derivative dy/dt, indicating it should be 2a sin t cos t = a sin(2t). Ultimately, the user successfully solved the integration using the substitution u=sin(t). The thread highlights the importance of accurate derivatives and substitutions in solving complex integrals.
Helloooo
Messages
6
Reaction score
0
Homework Statement
Show that the curve C given by
r=(a*cost*sint)i+(a*sin^2(t))j+(a*cos(t))k
, (0≤t≤pi/2)
lies on a sphere centred at the origin.
Find ∫zds of C
Relevant Equations
∫ds=∫F(r(t))·r'(t)dt
∫zds=∫acos(t)*( (acos(2t))^2+(2asin(t))^2+(-asin(t))^2 )^1/2 dt , (0≤t≤pi/2)
Simplified :
∫a^2cos(t)*(cos^2(2t)+5sin^2(t) )^1/2 dt , (0≤t≤pi/2)
However here i get stuck and i can´t find a way to rewrite it better or to integrate as it is.
Can i please get some help in this?
 
Physics news on Phys.org
I think you have dy/dt incorrect; it should be 2a \sin t \cos t = a \sin (2t).
 
pasmith said:
I think you have dy/dt incorrect; it should be 2a \sin t \cos t = a \sin (2t).
Oh okay!
I manged to solve the integration with the change and with the substitution u=sin(t)
Thank you!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
12
Views
2K
Replies
2
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
1
Views
1K