# Math Help - Line Integral for a sawtooth pattern

1. ## Line Integral for a sawtooth pattern

I have to evaluate the line integral:
$\int_Z(sin y + 2x) dx + (2y + x cos y)dy$

from the point (0,0) to (1,1) to (3,-1) to (4,0).

I evaluated the integrals for each of the 3 separate lines such that:
$\int_Z = \int_{C_1} + \int_{C_2} + \int_{C_3}$

I got the answer of $2cos(1) - 3sin(1) -7$ but I am pretty sure this is wrong. Can someone help?

2. ## Re: Line Integral for a sawtooth pattern

Originally Posted by tammyl
I have to evaluate the line integral:
$\int_Z(sin y + 2x) dx + (2y + x cos y)dy$

from the point (0,0) to (1,1) to (3,-1) to (4,0).

I evaluated the integrals for each of the 3 separate lines such that:
$\int_Z = \int_{C_1} + \int_{C_2} + \int_{C_3}$

I got the answer of $2cos(1) - 3sin(1) -7$ but I am pretty sure this is wrong. Can someone help?

Show your work on $\int_{C_1}$, so that we can see what you are doing.

3. ## Re: Line Integral for a sawtooth pattern

Thanks for the response.

So for $C_1$,

Parameterisation of straight line from (0,0) to (1,1)
$x = a(1-t) + ct = t$ where $a=0, c=1$
$y = b(1-t) + dt = t$ where $b=0, d=1$
Thus
$dx=dt$ and $dy=dt$ where $0\leq t \leq1$

So,
$\int_{C_1} (sin y + 2x)dx + (2y + x cos y)dt$
$= \int_0^1 (sin t + 2t)dt + (2t + t cos t)dt$
$= \int_0^1 [sin t + 4t + t cos t] dt$
$= [-cost + 2t^2 + t sin t + cos t]_0^1$
$= [2t^2 + t sin t]_0^1$
$= (2.(1)^2 - 2.(0)^2) + (1.sin (1) - 0.sin(0))$
$= 2 + sin (1)$

Repeating the procedure for the other two straight lines, I got
$\int_{Z} = \int_{C_1} + \int_{C_2} + \int_{C_3} = (2 + sin (1)) + (-16 - 4sin (1))+(7+2cos(1))= 2 cos(1) - 3 sin(1) -7$

I really am hoping my technique is right and its my integration that sucks.

Any help would be great.

4. ## Re: Line Integral for a sawtooth pattern

Originally Posted by tammyl
Thanks for the response.
So for $C_1$,
Parameterisation of straight line from (0,0) to (1,1)
$x = a(1-t) + ct = t$ where $a=0, c=1$
$y = b(1-t) + dt = t$ where $b=0, d=1$
Thus
$dx=dt$ and $dy=dt$ where $0\leq t \leq1$

So,
$\int_{C_1} (sin y + 2x)dx + (2y + x cos y)dt$
$= \int_0^1 (sin t + 2t)dt + (2t + t cos t)dt$
$= \int_0^1 [sin t + 4t + t cos t] dt$
$= [-cost + 2t^2 + t sin t + cos t]_0^1$
$= [2t^2 + t sin t]_0^1$
$= (2.(1)^2 - 2.(0)^2) + (1.sin (1) - 0.sin(0))$
$= 2 + sin (1)$

Repeating the procedure for the other two straight lines, I got
$\int_{Z} = \int_{C_1} + \int_{C_2} + \int_{C_3} = (2 + sin (1)) + (-16 - 4sin (1))+(7+2cos(1))= 2 cos(1) - 3 sin(1) -7$
Good, the above is correct.

If you are doing the other two carefully then the result should be correct.

The difficulty that I see is that both contours are more complicated.