Line Integral 0F5998

1. **State the problem:** Evaluate the line integral $$\int_C x^2 y^2 \, dx + 4xy^3 \, dy$$ where $C$ is the closed path consisting of three line segments: from $(0,0)$ to $(0,3)$, from $(0,3)$ to $(1,3)$, and from $(1,3)$ back to $(0,0)$. 2. **Parameterize each segment:** - Segment 1: from $(0,0)$ to $(0,3)$ - $x=0$, $y=t$, $t \in [0,3]$ - $dx=0$, $dy=dt$ - Segment 2: from $(0,3)$ to $(1,3)$ - $x=t$, $y=3$, $t \in [0,1]$ - $dx=dt$, $dy=0$ - Segment 3: from $(1,3)$ to $(0,0)$ - $x=1 - t$, $y=3 - 3t$, $t \in [0,1]$ - $dx = -dt$, $dy = -3 dt$ 3. **Evaluate the integral on each segment:** **Segment 1:** $$\int_0^3 \left(0^2 \cdot t^2 \right) \cdot 0 + 4 \cdot 0 \cdot t^3 \cdot dt = \int_0^3 0 \, dt = 0$$ **Segment 2:** $$\int_0^1 (t^2 \cdot 3^2) \, dt + \int_0^1 4 t \cdot 3^3 \cdot 0 = \int_0^1 9 t^2 \, dt = 9 \int_0^1 t^2 \, dt = 9 \left[ \frac{t^3}{3} \right]_0^1 = 9 \cdot \frac{1}{3} = 3$$ **Segment 3:** Substitute $x=1-t$, $y=3-3t$: $$x^2 y^2 = (1 - t)^2 (3 - 3t)^2 = (1 - t)^2 \cdot 9 (1 - t)^2 = 9 (1 - t)^4$$ $$4 x y^3 = 4 (1 - t) (3 - 3t)^3 = 4 (1 - t) \cdot 27 (1 - t)^3 = 108 (1 - t)^4$$ The integral on segment 3 is: $$\int_0^1 \left[ 9 (1 - t)^4 \cdot (-dt) + 108 (1 - t)^4 \cdot (-3 dt) \right] = \int_0^1 \left[ -9 (1 - t)^4 - 324 (1 - t)^4 \right] dt = \int_0^1 -333 (1 - t)^4 dt$$ Change variable $u = 1 - t$, $du = -dt$, when $t=0$, $u=1$, when $t=1$, $u=0$: $$\int_0^1 -333 (1 - t)^4 dt = -333 \int_0^1 (1 - t)^4 dt = -333 \int_1^0 u^4 (-du) = -333 \int_1^0 -u^4 du = -333 \int_0^1 u^4 du$$ Evaluate: $$-333 \left[ \frac{u^5}{5} \right]_0^1 = -333 \cdot \frac{1}{5} = -66.6$$ 4. **Sum the integrals over all segments:** $$0 + 3 - 66.6 = -63.6$$ 5. **Final answer:** $$\boxed{-63.6}$$