# Line Integral in second quadrant of Unit Circle

If I am asked to compute

$$\int_c F . dr$$

Where

$$F(x,y) = $$ 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?

• What is $F(x,y)$ ? – Nosrati Nov 3 '18 at 18:56
• I had trouble using MathJax. F(x, y) equals the partial derivatives with respect to x and y (in vector form) of the function sin(x^3 + y^3) – Joshua Nov 3 '18 at 18:59
• Can you use Green theorem? – Nosrati Nov 3 '18 at 19:01
• Technically not because Green's theorem, as far as I know, would be used for enclosed structures, not simple lines. – Joshua Nov 3 '18 at 19:03
• So do straightforward – Nosrati Nov 3 '18 at 19:04

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