Question

Evaluate the intergal F•dr where F (x,y,z)=xyi+yzj+zxk and c is the twisted cubic given by x=t, y=t^2 ,z=t^3, 0<=t=<1

Answer 1

The integral can be evaluated by substituting everything you know:

$\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\int_C\mathbf F(x(t),y(t),z(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(t^3\,\mathbf i+t^5\,\mathbf j+t^4\,\mathbf k)\cdot(\mathbf i+2t\,\mathbf j+3t^2\,\mathbf k)\,\mathrm dt$

$=\displaystyle\int_0^1(t^3+5t^6)\,\mathrm dt$

$=\dfrac{t^4}4+\dfrac{5t^7}7\bigg|_0^1=\dfrac14+\dfrac57=\boxed{\dfrac{27}{28}}$