Question

Use the Green's Theorem to calculate the work done by the field [ F (x, y) = -3y^5 i + 5y^2x^3 j ] to move a particle along the circumference [C: x^2 + y^2 = 4] starting from the point (2;0) and arriving at the point (-2,0).

Answer 1

$\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\iint_R\left(\frac{\partial(5y^2x^3)}{\partial x}-\frac{\partial(-3y^5)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

$\dfrac{\partial(5y^2x^3)}{\partial x}=15y^2x^2$
$\dfrac{\partial(-3y^5)}{\partial y}=-15y^2$

$=\displaystyle\int_{x=-2}^{x=2}\int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}15y^2(x^2+1)\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates, the integral is equivalent to

$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}15r^3\sin^2\theta(r^2\cos^2\theta+1)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle15\int_0^{2\pi}\int_0^2\left(\frac{r^5}4\sin^22\theta+r^3\sin^2\theta\right)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\frac{15}4\left(\int_0^2r^5\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^22\theta\,\mathrm d\theta\right)+15\left(\int_0^2r^3\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^2\theta\,\mathrm d\theta\right)$
$=100\pi$