# Thread: Line Integral along a straight line.

1. ## Line Integral along a straight line.

I can do the problem, but my final number is wrong.

$\displaystyle If $$C$$ is the line segment from $$(6,1)$$ to, $$(0,0)$$, find the value of the line integral:$
$\displaystyle $$\int_C (8 y^2\,\vec i + x\,\vec j)\cdot d\vec r =$$$

First we need to parametrize;
$\displaystyle \\r^{\rightarrow }(t)=(1-t)({p}_{1},{p}_{2})+t({q}_{1},{q}_{2})\\$
$\displaystyle r^{\rightarrow }(t)=(1-t)(6,1)+t(0,0)\\$
$\displaystyle r^{\rightarrow }(t)=(6-6t,1-t)+0t\\$
$\displaystyle r^{\rightarrow }(t)=<6-6t,1-t>\\$

This means that the individual parametric equations are,

$\displaystyle x=6-6t,y=1-t$

The next step is to integrate;

$\displaystyle work=\int_{C}^{}{F}^{\rightarrow }*d{r}^{\rightarrow }=\int_{t=0}^{t=1}{F}^{\rightarrow }({r}^{\rightarrow }(t))*({r'}^{\rightarrow }(t))dt$
The problem:
$\displaystyle \int_{t=0}^{t=1}<8(1-t)^{2},(6-6t)>\bullet <-6,-1>dt$
$\displaystyle \int_{t=0}^{t=1}48(1-t)^{2}+(-6+6t)dt$
$\displaystyle \int_{t=0}^{t=1}(48(t)^{2}-96t+48-6+6t)dt$
$\displaystyle \int_{t=0}^{t=1}(48(t)^{2}-102t+54)dt$
$\displaystyle (16(t)^{3}-51t^{2}+54t)\mid_{0}^{1}$

So if you do that last line you should end up with a...(19),using the FTC 'b-a'. Of course in this case b=1,a=0.

However, when I make it -19 I'm told I'm am correct; what have I done wrong to end up with the wrong sign.

2. Originally Posted by Zanderist
I can do the problem, but my final number is wrong.

$\displaystyle If $$C$$ is the line segment from $$(6,1)$$ to, $$(0,0)$$, find the value of the line integral:$
$\displaystyle $$\int_C (8 y^2\,\vec i + x\,\vec j)\cdot d\vec r =$$$

First we need to parametrize;
$\displaystyle \\r^{\rightarrow }(t)=(1-t)({p}_{1},{p}_{2})+t({q}_{1},{q}_{2})\\$
$\displaystyle r^{\rightarrow }(t)=(1-t)(6,1)+t(0,0)\\$
$\displaystyle r^{\rightarrow }(t)=(6-6t,1-t)+0t\\$
$\displaystyle r^{\rightarrow }(t)=<6-6t,1-t>\\$

This means that the individual parametric equations are,

$\displaystyle x=6-6t,y=1-t$

The next step is to integrate;

$\displaystyle work=\int_{C}^{}{F}^{\rightarrow }*d{r}^{\rightarrow }=\int_{t=0}^{t=1}{F}^{\rightarrow }({r}^{\rightarrow }(t))*({r'}^{\rightarrow }(t))dt$
The problem:
$\displaystyle \int_{t=0}^{t=1}<8(1-t)^{2},(6-6t)>\bullet <-6,-1>dt$
$\displaystyle \int_{t=0}^{t=1}48(1-t)^{2}+(-6+6t)dt$

Here it must be $\displaystyle -48 ...$

Tonio

$\displaystyle \int_{t=0}^{t=1}(48(t)^{2}-96t+48-6+6t)dt$
$\displaystyle \int_{t=0}^{t=1}(48(t)^{2}-102t+54)dt$
$\displaystyle (16(t)^{3}-51t^{2}+54t)\mid_{0}^{1}$

So if you do that last line you should end up with a...(19),using the FTC 'b-a'. Of course in this case b=1,a=0.

However, when I make it -19 I'm told I'm am correct; what have I done wrong to end up with the wrong sign.
.