Line Integral and Vector Field Problem

zm500
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Homework Statement


Find the work done by the force field F(x,y) = x sin(y)i + yj on a particle that moves along on the parabola y = x^2 from (-1,1) to (2,4).

Homework Equations


Work = line integral of the dot product of Field vector and change in the path
The path is parabola equation.

The Attempt at a Solution


I tried to integrate with respect to x and y instead of t, because I don't know how to find the path vector r(t).
So, i got two integrals for X-direction and y - direction.
Then, I can just use the Pythagorean theorem to find the total work.

Srry, I didn't write any mathematical equations.
 
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The "path vector" is always a position vector:
\vec{r}(x)= x\vec{i}+ y\vec{j}= x\vec{i}+ x^2\vec{j}
so that
d\vec{r}= (\vec{i}+ 2x\vec{j})dx

If the force field is not conservative, the work done to move along two legs of a right triangle might have nothing to do with the work required to move along the hypotenuse. I don't believe that using the "Pythagorean Theorem" will work here.
 
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HallsofIvy said:
The "path vector" is always a position vector:
\vec{r}(x)= x\vec{i}+ y\vec{j}= x\vec{i}+ x^2\vec{j}
so that
d\vec{r}= (\vec{i}+ 2x\vec{j})dx

If the force field is not conservative, the work done to move along two legs of a right triangle might have nothing to do with the work required to move along the hypotenuse. I don't believe that using the "Pythagorean Theorem" will work here.

Thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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