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((\partial(5y^2x^3))/(\partial x)-(\partial(-3y^5))/(\partial y)\right)\,\mathrm dx\,\mathrm dy`


`(\partial(5y^2x^3))/(\partial x)=15y^2x^2`

`(\partial(-3y^5))/(\partial y)=-15y^2`


`=\displaystyle\int_(x=-2)^(x=2)\int_(y=-√(4-x^2))^(y=√(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`