Line Integral Double 76D422

1. **Problem Statement:** Evaluate the line integral $$\int_C x^2y^2\,dx + 4xy^3\,dy$$ where $C$ is the piecewise path consisting of line segments from $(0,0)$ to $(0,3)$, $(0,3)$ to $(1,3)$, and $(1,3)$ 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 integral on each segment:** - Segment 1: $$\int_0^3 \left(x^2 y^2 \cdot dx + 4 x y^3 \cdot dy\right) = \int_0^3 (0 + 4 \cdot 0 \cdot t^3) dt = 0$$ - Segment 2: $$\int_0^1 \left(t^2 \cdot 3^2 \cdot dt + 4 t \cdot 3^3 \cdot 0\right) = \int_0^1 9 t^2 dt = 9 \int_0^1 t^2 dt = 9 \left[ \frac{t^3}{3} \right]_0^1 = 3$$ - Segment 3: Substitute $x=1-t$, $y=3-3t$: $$x^2 y^2 = (1 - t)^2 (3 - 3t)^2 = (1 - t)^2 9 (1 - t)^2 = 9 (1 - t)^4$$ $$4 x y^3 = 4 (1 - t) (3 - 3t)^3 = 4 (1 - t) 27 (1 - t)^3 = 108 (1 - t)^4$$ The integral is: $$\int_0^1 \left(9 (1 - t)^4 (-dt) + 108 (1 - t)^4 (-3 dt)\right) = \int_0^1 \left(-9 (1 - t)^4 - 324 (1 - t)^4\right) dt = \int_0^1 -333 (1 - t)^4 dt$$ Change variable $u = 1 - t$, $du = -dt$, when $t=0$, $u=1$, when $t=1$, $u=0$: $$\int_1^0 -333 u^4 (-du) = \int_0^1 333 u^4 du = 333 \left[ \frac{u^5}{5} \right]_0^1 = \frac{333}{5} = 66.6$$ 4. **Sum all segments:** $$0 + 3 + 66.6 = 69.6$$ --- 5. **(a) i. Evaluate $$\iint_D e^{-x-2y} dA$$ where $D$ is the region in the first quadrant with $y \leq x$:** - Region $D$: $x \geq 0$, $y \geq 0$, and $0 \leq y \leq x$ - Integral: $$\int_0^\infty \int_0^x e^{-x - 2y} dy dx$$ - Inner integral: $$\int_0^x e^{-x - 2y} dy = e^{-x} \int_0^x e^{-2y} dy = e^{-x} \left[ -\frac{1}{2} e^{-2y} \right]_0^x = e^{-x} \left( -\frac{1}{2} e^{-2x} + \frac{1}{2} \right) = \frac{e^{-x}}{2} (1 - e^{-2x})$$ - Outer integral: $$\int_0^\infty \frac{e^{-x}}{2} (1 - e^{-2x}) dx = \frac{1}{2} \int_0^\infty e^{-x} dx - \frac{1}{2} \int_0^\infty e^{-3x} dx$$ - Evaluate: $$\frac{1}{2} \left[ -e^{-x} \right]_0^\infty - \frac{1}{2} \left[ -\frac{1}{3} e^{-3x} \right]_0^\infty = \frac{1}{2} (1 - 0) - \frac{1}{2} \cdot \frac{1}{3} (1 - 0) = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$$ 6. **(a) ii. Evaluate $$\int_0^4 \int_{\sqrt{x}}^2 \frac{1}{1 + y^3} dy dx$$** - Change order of integration: - $x$ varies from $0$ to $4$, $y$ varies from $\sqrt{x}$ to $2$. - For $y$ fixed, $x$ varies from $0$ to $y^2$. - So integral becomes: $$\int_0^2 \int_0^{y^2} \frac{1}{1 + y^3} dx dy = \int_0^2 \frac{y^2}{1 + y^3} dy$$ - Simplify integrand: $$\frac{y^2}{1 + y^3}$$ - Use substitution $u = 1 + y^3$, $du = 3 y^2 dy$, so $y^2 dy = \frac{du}{3}$ - Change limits: - When $y=0$, $u=1$ - When $y=2$, $u=1 + 8 = 9$ - Integral becomes: $$\int_1^9 \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \int_1^9 \frac{1}{u} du = \frac{1}{3} [\ln u]_1^9 = \frac{1}{3} \ln 9 = \frac{2}{3} \ln 3$$ 7. **(b) Find the area of the region in the 1st quadrant between $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$:** - Area between two circles in first quadrant is quarter of the difference of areas: - Area of circle radius $r$ is $\pi r^2$ - Area between circles: $$\frac{1}{4} \pi (4^2 - 1^2) = \frac{1}{4} \pi (16 - 1) = \frac{15 \pi}{4}$$ 8. **(c) Find volume bounded by $z = 3 - 2y$, $y = 1 - x^2$, $y=0$, and $z=0$:** - Volume is double integral over region $R$ in $xy$-plane where $0 \leq y \leq 1 - x^2$, $z$ from $0$ to $3 - 2y$. - Volume: $$\iint_R (3 - 2y) dA = \int_{x=-1}^1 \int_{y=0}^{1 - x^2} (3 - 2y) dy dx$$ - Inner integral: $$\int_0^{1 - x^2} (3 - 2y) dy = \left[ 3y - y^2 \right]_0^{1 - x^2} = 3(1 - x^2) - (1 - x^2)^2$$ - Simplify: $$3(1 - x^2) - (1 - 2x^2 + x^4) = 3 - 3x^2 - 1 + 2x^2 - x^4 = 2 - x^2 - x^4$$ - Outer integral: $$\int_{-1}^1 (2 - x^2 - x^4) dx = \left[ 2x - \frac{x^3}{3} - \frac{x^5}{5} \right]_{-1}^1$$ - Evaluate: $$\left(2(1) - \frac{1}{3} - \frac{1}{5}\right) - \left(2(-1) - \frac{-1}{3} - \frac{-1}{5}\right) = \left(2 - \frac{1}{3} - \frac{1}{5}\right) - \left(-2 + \frac{1}{3} + \frac{1}{5}\right)$$ - Simplify: $$2 - \frac{1}{3} - \frac{1}{5} + 2 - \frac{1}{3} - \frac{1}{5} = 4 - \frac{2}{3} - \frac{2}{5} = 4 - \frac{10}{15} - \frac{6}{15} = 4 - \frac{16}{15} = \frac{60}{15} - \frac{16}{15} = \frac{44}{15}$$ **Final answers:** - (c) Line integral = $69.6$ - (a) i. Double integral = $\frac{1}{3}$ - (a) ii. Double integral = $\frac{2}{3} \ln 3$ - (b) Area = $\frac{15 \pi}{4}$ - (c) Volume = $\frac{44}{15}$