1. **Problem statement:** Evaluate the line integral $$\int \mathbf{A} \cdot d\mathbf{r}$$ where $$\mathbf{A} = xy^2 \mathbf{i} + x^2 y^2 \mathbf{j}$$ along two paths: i) The straight line $$y = 2 - x$$ from $$(2,0)$$ to $$(0,2)$$. ii) The quarter circle $$x^2 + y^2 = 4$$ from $$(2,0)$$ to $$(0,2)$$. 2. **Recall the line integral formula:** $$\int_C \mathbf{A} \cdot d\mathbf{r} = \int_C (P dx + Q dy)$$ where $$\mathbf{A} = P \mathbf{i} + Q \mathbf{j}$$. 3. **Path i) Straight line:** - Parametrize the path using $$x$$ as parameter: $$y = 2 - x$$ $$dx = dx$$, $$dy = -dx$$. - Substitute into $$P$$ and $$Q$$: $$P = x y^2 = x (2 - x)^2$$ $$Q = x^2 y^2 = x^2 (2 - x)^2$$ - The integral becomes: $$\int_{x=2}^{0} P dx + Q dy = \int_2^0 x(2 - x)^2 dx + x^2 (2 - x)^2 (-dx)$$ $$= \int_2^0 \left[x(2 - x)^2 - x^2 (2 - x)^2\right] dx = \int_2^0 (2 - x)^2 (x - x^2) dx$$ 4. **Simplify the integrand:** $$ (2 - x)^2 = 4 - 4x + x^2$$ $$x - x^2 = x(1 - x)$$ So integrand: $$ (4 - 4x + x^2)(x - x^2) = (4 - 4x + x^2)(x - x^2)$$ Expand: $$= 4x - 4x^2 + x^3 - 4x^2 + 4x^3 - x^4 = 4x - 8x^2 + 5x^3 - x^4$$ 5. **Integral:** $$\int_2^0 (4x - 8x^2 + 5x^3 - x^4) dx = - \int_0^2 (4x - 8x^2 + 5x^3 - x^4) dx$$ 6. **Compute the integral:** $$\int_0^2 4x dx = 4 \cdot \frac{x^2}{2} \Big|_0^2 = 4 \cdot 2 = 8$$ $$\int_0^2 8x^2 dx = 8 \cdot \frac{x^3}{3} \Big|_0^2 = 8 \cdot \frac{8}{3} = \frac{64}{3}$$ $$\int_0^2 5x^3 dx = 5 \cdot \frac{x^4}{4} \Big|_0^2 = 5 \cdot 4 = 20$$ $$\int_0^2 x^4 dx = \frac{x^5}{5} \Big|_0^2 = \frac{32}{5}$$ 7. **Sum integrals:** $$8 - \frac{64}{3} + 20 - \frac{32}{5} = (8 + 20) - \frac{64}{3} - \frac{32}{5} = 28 - \frac{64}{3} - \frac{32}{5}$$ Find common denominator 15: $$28 = \frac{420}{15}, \quad \frac{64}{3} = \frac{320}{15}, \quad \frac{32}{5} = \frac{96}{15}$$ So sum: $$\frac{420}{15} - \frac{320}{15} - \frac{96}{15} = \frac{420 - 320 - 96}{15} = \frac{4}{15}$$ 8. **Apply negative sign from step 5:** $$- \frac{4}{15} = -\frac{4}{15}$$ So the value of the integral along path i) is $$-\frac{4}{15}$$. 9. **Path ii) Quarter circle:** - Parametrize the quarter circle: $$x = 2 \cos t, \quad y = 2 \sin t, \quad t \in [0, \frac{\pi}{2}]$$ - Compute $$dx$$ and $$dy$$: $$dx = -2 \sin t dt, \quad dy = 2 \cos t dt$$ - Substitute into $$P$$ and $$Q$$: $$P = x y^2 = (2 \cos t)(2 \sin t)^2 = 2 \cos t \cdot 4 \sin^2 t = 8 \cos t \sin^2 t$$ $$Q = x^2 y^2 = (2 \cos t)^2 (2 \sin t)^2 = 4 \cos^2 t \cdot 4 \sin^2 t = 16 \cos^2 t \sin^2 t$$ - The integral becomes: $$\int_0^{\pi/2} P dx + Q dy = \int_0^{\pi/2} 8 \cos t \sin^2 t (-2 \sin t) dt + 16 \cos^2 t \sin^2 t (2 \cos t) dt$$ $$= \int_0^{\pi/2} (-16 \cos t \sin^3 t + 32 \cos^3 t \sin^2 t) dt$$ 10. **Split the integral:** $$\int_0^{\pi/2} -16 \cos t \sin^3 t dt + \int_0^{\pi/2} 32 \cos^3 t \sin^2 t dt$$ 11. **Evaluate each integral separately:** - For $$I_1 = \int_0^{\pi/2} \cos t \sin^3 t dt$$: Use substitution $$u = \sin t$$, $$du = \cos t dt$$: $$I_1 = \int_0^1 u^3 du = \frac{u^4}{4} \Big|_0^1 = \frac{1}{4}$$ - For $$I_2 = \int_0^{\pi/2} \cos^3 t \sin^2 t dt$$: Use substitution $$u = \sin t$$, $$du = \cos t dt$$, rewrite $$\cos^3 t = \cos^2 t \cos t = (1 - \sin^2 t) \cos t = (1 - u^2) du$$: $$I_2 = \int_0^1 (1 - u^2) u^2 du = \int_0^1 (u^2 - u^4) du = \left( \frac{u^3}{3} - \frac{u^5}{5} \right)_0^1 = \frac{1}{3} - \frac{1}{5} = \frac{2}{15}$$ 12. **Combine results:** $$\int_0^{\pi/2} (-16 \cos t \sin^3 t + 32 \cos^3 t \sin^2 t) dt = -16 I_1 + 32 I_2 = -16 \cdot \frac{1}{4} + 32 \cdot \frac{2}{15} = -4 + \frac{64}{15} = -\frac{60}{15} + \frac{64}{15} = \frac{4}{15}$$ 13. **Final answers:** i) $$\int A \cdot dr = -\frac{4}{15}$$ ii) $$\int A \cdot dr = \frac{4}{15}$$