Question

Calculating line integrals with parameterization

Answer 1

Can't we just integrate it? And then put in the co-ordinates for respective vectors.

Answer 2

we can

Answer 3

but i think we are suppossed to do it with respect to t

Answer 4

The parameteric equation of the line is
\[ r(t) =(x(t),y(t),z(t)= (1 + 4t, 0,0), 0\le t \le 1
\]
The field is
\[F(x,y,z)=(2x , 3 y,0)\\
F((x(t),y(t),z(t))=(2 + 8t,0,0)\\
F.dr= 8+ 16t\\
\int_0^1 (8+16 t)=24
\]

Answer 5

yes that i steh right answer

Answer 6

omg thanks

Answer 7

That was awesome. Wish I can give u a few medals :P

Answer 8

With pleasure @samjordon

Answer 9

@eliassaab ummm like i need help in explaining one step

Answer 10

∫ 1 0 (8+16t)=24
how did u get that 16?

Answer 11

The parameteric equation of the line is
\[ r(t) =(x(t),y(t),z(t)= (1 + 4t, 0,0), 0\le t \le 1
\]
The field is
\[F(x,y,z)=(2x , 3 y,0)\\
F((x(t),y(t),z(t))=(2 + 8t,0,0)\\
F.dr= 8+ 32t\\
\int_0^1 (8+32 t)=24
\]

Answer 12

Sorry this was 32. t was a misprint. The answer is not changed. Since I did the computation with 32