Line Integral in second quadrant of Unit Circle

Question

If I am asked to compute

$$\int_c F . dr$$

Where

$$F(x,y) = <d/dx f(x, y), d/dy f(x,y)>$$ and $$f(x,y) =\sin(x^3 + y^3)$$

and C is the portion of the unit circle in the second quadrant, oriented counterclockwise, how would I go about doing that?

Answer 1

Let $r=(\cos t,\sin t)$ with $t\in[\dfrac{\pi}{2},\pi]$ be the parametrization of $C$. Then \begin{align} \int_C F.dr &= \int_{\frac{\pi}{2}}^{\pi}\left(3\cos^2t\cos(\cos^3t+\sin^3t),3\sin^2t\cos(\cos^3t+\sin^3t)\right)(-\sin t,\cos t)\ dt \\ &= \int_{\frac{\pi}{2}}^{\pi}\left(-3\cos^2t\sin t+3\sin^2t\cos t\right)\cos(\cos^3t+\sin^3t)\ dt \\ &= \sin(\cos^3t+\sin^3t)\Big|_{\frac{\pi}{2}}^{\pi} \\ &= \color{blue}{-2\sin1} \end{align}

Answer 2

You need the gradient theorem: $$\int_\gamma (\nabla f) \cdot d\mathbf r = f(\mathbf r_2) - f(\mathbf r_1) = f(-1, 0) - f(0, 1) = -2 \sin 1.$$