Troubleshooting Line Integrals: Finding Errors in Calculations

[hivesaeed4]

Let $${\tmmathbf\vec{F} = yx^2 \tmmathbf\hat{i} + \sin (\pi y) \tmmathbf\hat{j}}$$, and let $${C}$$ be the curve along the line segment starting at (0,2) and ending at (1,4).

$${\int_C \tmmathbf\vec{F} \cdot d \tmmathbf\vec{r} =}$$

My path comes out to be:

r=ti +2tj
dr=i+2j;
F=2t^3i+sin(2*pi*t)j
Now F.dr=2(t^3)+2(sin(2*pi*t))
so
∫2(t^3)+2(sin(2*pi*t))dt=(2t^4 - cos(2*pi*t)/pi ) where t goes from 0 to 1

Note: r,dr and F are vectors.

The answer from the above computation comes out to be 1/2.

I've checked the answer and it's wrong so I've made a mistake somewhere. Could someone identify it?

 

[hivesaeed4]

Let me rewrite the confusing part of my previous post:

Let F=yx^2 i + sin(pi*y) j;

and find

∫( on curve C) F.dr

Now:

My path comes out to be:

r=ti +2tj
dr=i+2j;
F=2t^3i+sin(2*pi*t)j
Now F.dr=2(t^3)+2(sin(2*pi*t))
so
∫2(t^3)+2(sin(2*pi*t))dt=(2t^4 - cos(2*pi*t)/pi ) where t goes from 0 to 1

Note: r,dr and F are vectors.

The answer from the above computation comes out to be 1/2.

I've checked the answer and it's wrong so I've made a mistake somewhere. Could someone identify it?

 

[LCKurtz]

r = ti + 2tj doesn't pass through (0,2).

 

[MostlyHarmless]

If I'm not mistaken

∫2(t^3)+2(sin(2*pi*t))dt=(1/2)(t^4)-(cos(2*pi*t)/pi)

you have =2t^4; by simply taking the derivative of this you would get 8t^3, not 2t^3.



How does r(t)=ti+2tj not pass through (0,2)? To find the vector of a line segment, it's just <x2-x1, y2-y1>, is it not?

 

[HallsofIvy]

Yes, but the vector is not the parametric equations.

The parametric equations of a line with direction vector <a, b> , passing through $(x_0, y_0)$ are $x= at+ x_0$ and $y= bt+ y_0$.

The line given by it+ 2tj or x= t, y= 2t, passes through (0, 0) and (1, 2), parallel to the line in this problem.

 

[DonAntonio]

Let me rewrite the confusing part of my previous post:

Let F=yx^2 i + sin(pi*y) j;

and find

∫( on curve C) F.dr

Now:

My path comes out to be:

r=ti +2tj
dr=i+2j;
F=2t^3i+sin(2*pi*t)j
Now F.dr=2(t^3)+2(sin(2*pi*t))
so
∫2(t^3)+2(sin(2*pi*t))dt=(2t^4 - cos(2*pi*t)/pi ) where t goes from 0 to 1

Note: r,dr and F are vectors.

The answer from the above computation comes out to be 1/2.

I've checked the answer and it's wrong so I've made a mistake somewhere. Could someone identify it?

Some points for you to consider:

1) Learn urgently to write with LaTeX in this forum

2) The path you wrote is not a straight line between (0,2) and (1,4)...in fact, it doesn't even pass through neither of these

two points! You must know how to produce the equation of a straight line between two points in the plane.

3) After the above is done, and if I didn't make any error, you must get the integral $$\int_0^1\left(2t^3+2t^2+2\sin (\pi(2t+2))\right)dt=\frac{7}{6}$$

DonAntonio

 

[hivesaeed4]

Thanks guys.