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Question

Verify Stokes’s Theorem by evaluating $\int_{C} \mathbf{F} \cdot d \mathbf{r}$ as a line integral and as a double integral. $\mathbf{F}(x, y, z)=x y z \mathbf{i}+y \mathbf{j}+z \mathbf{k}$, $S: 6 x+6 y+z=12$, first octant

Solution

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Given field ${\bf F}(x, y, z) = xyz{\bf i} + y{\bf j} + z{\bf k}$ and surface $S$ : $6x+6y+z=12$. The curve along the edge of surface is a circle $x^2 + y^2= 1$. The curve consists of $3$ smooth edges of the plane. Let $C_1$ lines on $6y+z=12$, $C_2$ lies on $6x+z=12$ and $C_3$ lies on $6x+6y=12$. On $C_1$, $x=0$ and $6y+z=12 \implies dz=-6dy$. The line $C_1$ runs from $(0, 2, 0) \to (0, 0, 12)$. This gives

\begin{align*} \int_{C_1} {\bf F}\cdot d{\bf r} &= \int_{C_1} ( xyz{\bf i} + y{\bf j} + z{\bf k}) \cdot (dx {\bf i} + dy {\bf j} + dz {\bf k} ) \\ &= \int_2^0 (0 + ydy + z dz) \\ &= \int_2^0 (y dy + (12 - 6y) (-6dy))\\ &= \int_2^0 (-72 + 37 y) dy \\ &= 70 \end{align*}

On $C_2$, $y=0$ and $6x+z=12 \implies dz=-6dx$. Line $C_2$ runs from $(0, 0, 12) \to (2, 0, 0)$. This gives

\begin{align*} \int_{C_2} {\bf F}\cdot d{\bf r} &= \int_{C_2} ( xyz{\bf i} + y{\bf j} + z{\bf k}) \cdot (dx {\bf i} + dy {\bf j} + dz {\bf k} ) \\ &= \int_{12}^0 (0 + 0 + z dz) \\ &= -72 \end{align*}

On $C_2$, $z=0$ and $6x+6y=12 \implies dx=-dy$. Line $C_3$ runs from $(2, 0, 0) \to (0, 2,0)$ This gives

\begin{align*} \int_{C_1} {\bf F}\cdot d{\bf r} &= \int_{C_1} ( xyz{\bf i} + y{\bf j} + z{\bf k}) \cdot (dx {\bf i} + dy {\bf j} + dz {\bf k} ) \\ &= \int_0^2 (0 + y dy + 0) \\ &= 2 \end{align*}

This gives

$\oint_C {\bf F}\cdot d{\bf r} = 70-72+2 =0$

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