Homework Help

Evaluate the line integral: int c (x+2y)dx +(x-y)dy where c is the curve x=2cost,...

user profile pic

owlstalk | Student, Undergraduate | eNoter

Posted January 24, 2011 at 12:09 AM via web

dislike 0 like

Evaluate the line integral:

int c (x+2y)dx +(x-y)dy where c is the curve

x=2cost, y=4sint

0≥t≤pi/4

 

1 Answer | Add Yours

user profile pic

kjcdb8er | Teacher | (Level 1) Associate Educator

Posted January 24, 2011 at 5:15 AM (Answer #1)

dislike 0 like

To answer this question, split up the line integral into two pieces:

intc (x + 2y)dx   and   intc (x - y)dy.

Our parameter is t,  0<=t<=pi/4    (I assume, because your problem statement gives inf <-- t < 0, which diverges )

We need to convert dx, dy  into dt:

x = 2 cos t   -->  dx = -2 sin t dt

y = 4 sin t  -->  dy = 4 cos t

Now,

intc (x + 2y)dx = int [ (2 cos t + 8 sin t) ( -2 sin t ) dt , 0<= t <=pi/4]

= int [ -4 costsint - 16sint^2, 0<= t <=pi/4 ]

And,

= intc (x - y)dy = int [ (2 cos t - 4 sin t) ( 4 cos t ) dt , 0<= t <=pi/4]

= int [ 8 cost^2 - 16 costsint, 0<= t <=pi/4 ]

So,

intc (x + 2y)dx  + intc (x - y)dy = int[ 8 cost^2 - 20 costsint - 16 sint^2, 0<= t <=pi/4 ]

=-4 t + 5 cos(2 t) + 6 sin(2 t) + C, evaluated from 0<=t<=pi/4

= 1 - pi = 2.14

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes