The problem asks you to compute the line integral of a vector field $F$ along a curve $\gamma$.

Problem summary:

• Vector Field: $F : \mathbb{R}^2 \to \mathbb{R}^2$, with $F(x, y) = (x^2, -xy)$
• Curve: $\gamma(t) = (\cos t, \sin t)$ for $t \in [0, \frac{\pi}{2}]$

Line integral to be computed:

$\int_{\gamma} F \cdot d\gamma$

Steps to solve:

1. Parameterize the vector field:

   • The curve $\gamma(t) = (\cos t, \sin t)$ is already parameterized. Therefore, for the vector field $F(x, y)$, substitute $x = \cos t$ and $y = \sin t$.

   $F(\gamma(t)) = F(\cos t, \sin t) = (\cos^2 t, -\cos t \sin t)$

2. Differentiate the curve $\gamma(t)$ with respect to $t$: $\gamma'(t) = \left( \frac{d}{dt} \cos t, \frac{d}{dt} \sin t \right) = (-\sin t, \cos t)$

3. Dot product $F(\gamma(t)) \cdot \gamma'(t)$: $F(\gamma(t)) \cdot \gamma'(t) = (\cos^2 t, -\cos t \sin t) \cdot (-\sin t, \cos t)$ Compute the dot product:

   = -\cos^2 t \sin t - \cos^2 t \sin t = -2 \cos^2 t \sin t$$

4. Set up the integral: Now integrate the result from $t = 0$ to $t = \frac{\pi}{2}$: $\int_0^{\frac{\pi}{2}} -2 \cos^2 t \sin t \, dt$

5. Solve the integral: This can be solved using a simple substitution:

   • Let $u = \cos t$, so $du = -\sin t \, dt$.
   • When $t = 0$, $u = 1$; when $t = \frac{\pi}{2}$, $u = 0$. Thus, the integral becomes: $\int_1^0 2u^2 \, du = -\int_0^1 2u^2 \, du = -\left[ \frac{2u^3}{3} \right]_0^1 = -\frac{2}{3}$

Final answer:

The value of the line integral is $-\frac{2}{3}$.

Do you want more details on any step? Let me know if you'd like additional questions or further explanations.

Here are 5 related questions you might explore:

1. What is the geometric meaning of a line integral of a vector field?
2. How does parameterizing a curve simplify calculating a line integral?
3. Could Green’s Theorem be used to solve this problem more efficiently?
4. What happens to the line integral if the direction of the curve is reversed?
5. How do line integrals apply to physics, particularly in work and energy calculations?

Tip: Always ensure the curve's parameterization matches the vector field's domain before computing the line integral.