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

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Answered 9 months ago
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Given field F(x,y,z)=xyzi+yj+zk{\bf F}(x, y, z) = xyz{\bf i} + y{\bf j} + z{\bf k} and surface SS : 6x+6y+z=126x+6y+z=12. The curve along the edge of surface is a circle x2+y2=1x^2 + y^2= 1. The curve consists of 33 smooth edges of the plane. Let C1C_1 lines on 6y+z=126y+z=12, C2C_2 lies on 6x+z=126x+z=12 and C3C_3 lies on 6x+6y=126x+6y=12. On C1C_1, x=0x=0 and 6y+z=12    dz=6dy6y+z=12 \implies dz=-6dy. The line C1C_1 runs from (0,2,0)(0,0,12)(0, 2, 0) \to (0, 0, 12). This gives

C1Fdr=C1(xyzi+yj+zk)(dxi+dyj+dzk)=20(0+ydy+zdz)=20(ydy+(126y)(6dy))=20(72+37y)dy=70\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 C2C_2, y=0y=0 and 6x+z=12    dz=6dx6x+z=12 \implies dz=-6dx. Line C2C_2 runs from (0,0,12)(2,0,0)(0, 0, 12) \to (2, 0, 0). This gives

C2Fdr=C2(xyzi+yj+zk)(dxi+dyj+dzk)=120(0+0+zdz)=72\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 C2C_2, z=0z=0 and 6x+6y=12    dx=dy6x+6y=12 \implies dx=-dy. Line C3C_3 runs from (2,0,0)(0,2,0)(2, 0, 0) \to (0, 2,0) This gives

C1Fdr=C1(xyzi+yj+zk)(dxi+dyj+dzk)=02(0+ydy+0)=2\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

CFdr=7072+2=0\oint_C {\bf F}\cdot d{\bf r} = 70-72+2 =0

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