Flux Integrals

1. Dec 7, 2014

dwn

1. The problem statement, all variables and given/known data

Attached Image

2. Relevant equations
this is not a simple plane curve or a close plane curve so I use the formula:
∫ F ⋅ dr/dt dt

3. The attempt at a solution
From the point (0,0) to (2,4)
Direction Vector v(t) = <2-0, 4-0>
Parametric Equation:
r(t) = (2t + 0) i + (4t + 0) j
r'(t) = 2i + 4j

∴ F(x(t),y(t)) = (4t)2i + 2(2t)(4t)j = (16t2)i + (16t2)j
F(x(t),y(t)) ⋅ r'(t) =( (16t2)i + (16t2)j ) ⋅ (2i + 4j) = 32t2 + 64t2

I feel my math is correct which leads me to believe my approach is incorrect. The final answer is supposed to be 32 evaluated from 0 to 2.

But if I finished my integral : ∫ 96t2dt = 32t3

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Last edited: Dec 7, 2014
2. Dec 7, 2014

SteamKing

Staff Emeritus
The parameter t should run from 0 to 1, rather than 0 to 2. Plug these limits into your flux integral and everything should work out.

Try the next path.

3. Dec 7, 2014

dwn

Thank you for your help. I had considered that but I'm still not confident on how interval is defined. Would you mind explaining that to me? All of the examples I see in the book simply set the interval from 0 to 1. Is this always the case?

4. Dec 7, 2014

SteamKing

Staff Emeritus
It doesn't have to be, but the interval from 0 to 1 is usually the simplest to evaluate.

The path over which you were evaluating the flux integral was y = 2x, where 0 ≤ x ≤ 2. By choosing the parameterization r(t) = 2t i + 4t j for this path, it is important to check and make sure that by plugging in the limits of t into the parametric equation, you obtain the same values of x and y for the limits of the actual path expression, hence the interval 0 ≤ t ≤ 1 when plugged into r(t) gives (0, 0) and (2, 4) for the end points of the path, which are identical to y = 2x for 0 ≤ x ≤ 2.

You could have specified r(t) = t i + 2t j, for 0 ≤ t ≤ 2, and obtained the same evaluation for the flux integral after making the appropriate integration.

5. Dec 8, 2014

HallsofIvy

Staff Emeritus
There are an infinite number of ways to set up parametric equations for a curve. Myself, since in both of these y is a function of x, I would have just used x itself as parameter. In the first, y= 2x so that $F(x, y)= y^2\vec{i}+ 2xy\vec{j}= (2x)^2\vec{i}+ 2x(2x)\vec{j}=4x^2\vec{i}+ 4x^2\vec{j}$. Since y= 2x, dy= 2dx and $d\vec{r}= dx\vec{i}+ dy\vec{j}= dx\vec{i}+ 2dx\vec{j}= (\vec{i}+ 2\vec{j})dx$. The integral becomes
$\int_0^2 (4x^2\vec{i}+ 4x^2\vec{j})\cdot(\vec{i}+ 2\vec{j})dx= \int_0^2 (4x^2+ 8x^2) dx= 12\int_0^2 x^2 dx$.
The limits of integration are 0 and 2, of course, because the variable of integration is x and x goes from 2 to 4.

Similarly, the second integral is over the curve $y= x^2$ so that $F(x, y)= y^2\vec{i}+ 2xy\vec{j}= (x^2)^2\vec{i}+ 2x(x^2)\vec{j}= x^4\vec{i}+ 2x^3\vec{j}$. Further, since $y= x^2$, $dy= 2xdx$ so that $d\vec{r}= dx\vec{i}+ dy\vec{j}= dx\vec{i}+ 2xdx\vec{j}= (\vec{i}+ 2x\vec{j})dx$. The integral becomes
$\int_0^2 (x^4\vec{i}+ 4x^2\vec{j})\cdot (\vec{i}+ 2x\vec{j})dx= \int_0^2( x^4+ 8x^3) dx$.

Again, the limits of integration are 0 to 2 because x is the variable of integration and x goes from 0 to 2.