1. **Problem:** Calculate the line integral \(\int_c \vec{F} \cdot d\vec{r}\) where \(\vec{F} = (x^2 + y) \vec{i} + (y^2 - x) \vec{j}\) and \(c\) is the quarter circle \(x^2 + y^2 = 4\) from \((2,0)\) to \((0,2)\). 2. **Formula and rules:** The line integral of a vector field \(\vec{F} = P\vec{i} + Q\vec{j}\) along a curve \(c\) is given by $$\int_c \vec{F} \cdot d\vec{r} = \int_c P\,dx + Q\,dy$$ where \(P = x^2 + y\) and \(Q = y^2 - x\). 3. **Parameterization:** Since \(c\) is a quarter circle of radius 2, parameterize as $$x = 2\cos t, \quad y = 2\sin t, \quad t \in [0, \frac{\pi}{2}]$$ Then, $$dx = -2\sin t\, dt, \quad dy = 2\cos t\, dt$$ 4. **Substitute into integral:** $$P = (2\cos t)^2 + 2\sin t = 4\cos^2 t + 2\sin t$$ $$Q = (2\sin t)^2 - 2\cos t = 4\sin^2 t - 2\cos t$$ Integral becomes $$\int_0^{\pi/2} \left(4\cos^2 t + 2\sin t\right)(-2\sin t) + \left(4\sin^2 t - 2\cos t\right)(2\cos t) \, dt$$ 5. **Simplify integrand:** $$= \int_0^{\pi/2} \left(-8\cos^2 t \sin t - 4\sin^2 t + 8\sin^2 t \cos t - 4\cos^2 t\right) dt$$ Group terms: $$= \int_0^{\pi/2} \left(-8\cos^2 t \sin t + 8\sin^2 t \cos t - 4\sin^2 t - 4\cos^2 t\right) dt$$ 6. **Use identity:** \(\sin^2 t + \cos^2 t = 1\), so $$-4\sin^2 t - 4\cos^2 t = -4$$ Integral is $$\int_0^{\pi/2} \left(-8\cos^2 t \sin t + 8\sin^2 t \cos t - 4\right) dt$$ 7. **Split integral:** $$= \int_0^{\pi/2} -8\cos^2 t \sin t \, dt + \int_0^{\pi/2} 8\sin^2 t \cos t \, dt - \int_0^{\pi/2} 4 \, dt$$ 8. **Evaluate each integral:** - For \(I_1 = \int_0^{\pi/2} -8\cos^2 t \sin t \, dt\), let \(u = \cos t\), \(du = -\sin t dt\), so $$I_1 = 8 \int_1^0 u^2 du = 8 \left[\frac{u^3}{3}\right]_1^0 = 8 \left(0 - \frac{1}{3}\right) = -\frac{8}{3}$$ - For \(I_2 = \int_0^{\pi/2} 8\sin^2 t \cos t \, dt\), let \(v = \sin t\), \(dv = \cos t dt\), so $$I_2 = 8 \int_0^1 v^2 dv = 8 \left[\frac{v^3}{3}\right]_0^1 = \frac{8}{3}$$ - For \(I_3 = \int_0^{\pi/2} 4 dt = 4 \times \frac{\pi}{2} = 2\pi$$ 9. **Sum results:** $$I = I_1 + I_2 - I_3 = -\frac{8}{3} + \frac{8}{3} - 2\pi = -2\pi$$ **Final answer:** $$\boxed{-2\pi}$$