Question

Calc III:

Evaluate the line integral, where C is the given curve.

Answer 1

ok i will solve from (0,0) (2,1)
will you be able to do this from (2,1) to (3,0) ??

Answer 2

@calcIIIstudent ??

Answer 3

ok first of all find the equation of line

Answer 4

from (0.0) to (2,1)
m=(1-0)/(2-0)=1/2
now using point slope form
y-1=1/2(x-2)
y-1=x/2-1
let x=t
then
y=t/2
where 0<t<2
so
\[\Large r(t)=ti+t/2j\]
\[\Large r'(t)=1i+1/2j\]
so
line integral is given by
\[\Large \int\limits_{0}^{2} F(r(t)r'(t)dt\]
where F(t)=(x+2)i+y^2j (given)
so using x=t and y=t/2 write
F(r(t))=(t+t)i+t^2j
F(r(t)=2ti+t^2j
and
F(r(t)).r'(t)=(2ti+t^2j)(i+1/2j)=2t+t^2/2
integral becomes
\[\Large \int\limits_{0}^{2}(2t+\frac{t^2}{2})dt=\frac{16}{3}=5.33\]
i hope it is clear !

Answer 5

@calcIIIstudent did you get it ?

Answer 6

@calcIIIstudent waiting for your reply !

Answer 7

Sorry, my internet crashed

Answer 8

Yes, I got it, thanks!

Answer 9

ok.
so you got it now?

Answer 10

yw:)