Evaluate the line integral   int(ydx+x^2dy) where C is the curve x=t , y=(1/2)t^2   where o=<t=<2 

Expert Answers
hala718 eNotes educator| Certified Educator

intg(ydx + x^2 dy)    x=t, y=(1/2)t^2      0=<t=<2

x=t  ==>   dx= dt

y=(1/2)t^2 ==> dy= t dt

Now substitute:

int(ydx +x^2 dy) = int((1/2)t^2 (dt) + t^2 (tdt)

                           = intg((1/2)t^2 dt + t^3 dt

                          = (1/2)( t^3/3) + t^4 / 4

                          = (t^3)/6  + (t^4) /4

Now we have 0=<t=<2

Substitute:

8/6 + 16/4 -0-0 = 4/3 + 4= 16/3

neela | Student

To find Int {y dx +x^2dy} where C is the curve x=t and y = (1/2) t^2.

Solution:

x= t. So dx = dt

y =(1/2)t^2. Therefore dy = 1/2* 2tdt  or dy = tdt. Substituting in the integral,

Int (y dx+x^2dy) = Int { (1/2)t^2 dt +t^2 tdt)

= (1/2)t^(2+1) / (2+1) + t^(3+1) / (3+1)

= (1/6)t^3 + t^4/4

=[ t^3(2+3t)/12]. Now for  t= 0 to 2 the integral value is:

= {2^3(2+3*2)/12  -  0 }

= 64/12

= 16/3