1. **State the problem:** Verify Stokes' theorem for the vector field $\vec{A} = (y+z)\hat{i} - xz\hat{j} + y^2\hat{k}$ over the surface $S$ in the first octant bounded by $2x + z = 6$ and $y=2$, excluding the planes (a) $xy$-plane, (b) plane $y=2$, (c) plane $2x + z = 6$. The curve $C$ is the boundary of $S$. 2. **Recall Stokes' theorem:** $$\oint_C \vec{A} \cdot d\vec{r} = \iint_S (\nabla \times \vec{A}) \cdot \hat{n} \, dS$$ where $\hat{n}$ is the unit normal to surface $S$. 3. **Calculate curl of $\vec{A}$:** $$\nabla \times \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y+z & -xz & y^2 \end{vmatrix}$$ Calculate components: - $\frac{\partial y^2}{\partial y} = 2y$ - $\frac{\partial (-xz)}{\partial z} = -x$ - $\frac{\partial (y+z)}{\partial z} = 1$ - $\frac{\partial y^2}{\partial x} = 0$ - $\frac{\partial (-xz)}{\partial x} = -z$ - $\frac{\partial (y+z)}{\partial y} = 1$ So, $$\nabla \times \vec{A} = \left(2y - (-x)\right)\hat{i} - \left(1 - 0\right)\hat{j} + \left(-z - 1\right)\hat{k} = (2y + x)\hat{i} - 1\hat{j} + (-z - 1)\hat{k}$$ 4. **Describe surface $S$:** $S$ is the part of the plane $2x + z = 6$ in the first octant with $0 \leq y \leq 2$, excluding the planes $xy$-plane ($z=0$), $y=2$, and $2x + z=6$. 5. **Parameterize surface $S$:** Express $z = 6 - 2x$, with $x \geq 0$, $z \geq 0 \Rightarrow 0 \leq x \leq 3$, and $0 \leq y \leq 2$. 6. **Find normal vector $\hat{n}$:** The plane $2x + z = 6$ has normal vector $\vec{n} = (2, 0, 1)$. Unit normal: $$\hat{n} = \frac{1}{\sqrt{2^2 + 0^2 + 1^2}}(2,0,1) = \frac{1}{\sqrt{5}}(2,0,1)$$ 7. **Surface integral:** $$\iint_S (\nabla \times \vec{A}) \cdot \hat{n} \, dS = \iint_D (\nabla \times \vec{A}) \cdot \hat{n} \sqrt{\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 + 1} \, dx dy$$ Since $z = 6 - 2x$, $\frac{\partial z}{\partial x} = -2$, $\frac{\partial z}{\partial y} = 0$, so $$dS = \sqrt{(-2)^2 + 0 + 1} \, dx dy = \sqrt{5} \, dx dy$$ 8. **Evaluate integrand:** At surface, $z = 6 - 2x$, so $$(\nabla \times \vec{A}) \cdot \hat{n} = (2y + x, -1, -z -1) \cdot \frac{1}{\sqrt{5}}(2,0,1) = \frac{1}{\sqrt{5}} \left[2(2y + x) + 0 + 1(-z -1)\right] = \frac{1}{\sqrt{5}} (4y + 2x - z -1)$$ Substitute $z = 6 - 2x$: $$4y + 2x - (6 - 2x) - 1 = 4y + 2x - 6 + 2x - 1 = 4y + 4x - 7$$ 9. **Integral becomes:** $$\iint_D \frac{1}{\sqrt{5}} (4y + 4x - 7) \sqrt{5} \, dx dy = \iint_D (4y + 4x - 7) \, dx dy$$ where $D = \{(x,y) | 0 \leq x \leq 3, 0 \leq y \leq 2\}$. 10. **Compute double integral:** $$\int_0^2 \int_0^3 (4y + 4x - 7) \, dx dy = \int_0^2 \left[4y x + 2x^2 - 7x\right]_0^3 dy = \int_0^2 (12y + 18 - 21) dy = \int_0^2 (12y - 3) dy$$ 11. **Integrate w.r.t $y$:** $$\left[6y^2 - 3y\right]_0^2 = 6(4) - 3(2) = 24 - 6 = 18$$ 12. **Line integral over $C$:** $C$ is the boundary of $S$ formed by the intersection of the planes and lines in the first octant. The boundary consists of three segments: - $C_1$: along $y=0$, $2x + z = 6$, $x$ from 0 to 3 - $C_2$: along $x=0$, $y$ from 0 to 2, $z=6$ - $C_3$: along $y=2$, $2x + z = 6$, $x$ from 3 to 0 Calculate $\oint_C \vec{A} \cdot d\vec{r} = \int_{C_1} + \int_{C_2} + \int_{C_3}$ 13. **Parametrize and compute $\int_{C_1}$:** $y=0$, $z=6 - 2x$, $x$ from 0 to 3. $$\vec{r}_1 = x \hat{i} + 0 \hat{j} + (6 - 2x) \hat{k}$$ $$d\vec{r}_1 = dx \hat{i} + 0 + (-2 dx) \hat{k}$$ Evaluate $\vec{A}$ on $C_1$: $$\vec{A} = (y+z)\hat{i} - xz \hat{j} + y^2 \hat{k} = (0 + 6 - 2x)\hat{i} - x(6 - 2x) \hat{j} + 0 \hat{k} = (6 - 2x)\hat{i} - (6x - 2x^2) \hat{j}$$ Dot product: $$\vec{A} \cdot d\vec{r}_1 = (6 - 2x) dx + 0 + 0 = (6 - 2x) dx$$ Integral: $$\int_0^3 (6 - 2x) dx = \left[6x - x^2\right]_0^3 = 18 - 9 = 9$$ 14. **Parametrize and compute $\int_{C_2}$:** $x=0$, $y$ from 0 to 2, $z=6$. $$\vec{r}_2 = 0 \hat{i} + y \hat{j} + 6 \hat{k}$$ $$d\vec{r}_2 = 0 + dy \hat{j} + 0$$ Evaluate $\vec{A}$ on $C_2$: $$\vec{A} = (y + 6) \hat{i} - 0 \hat{j} + y^2 \hat{k}$$ Dot product: $$\vec{A} \cdot d\vec{r}_2 = 0 - 0 + 0 = 0$$ Integral: $$\int_0^2 0 dy = 0$$ 15. **Parametrize and compute $\int_{C_3}$:** $y=2$, $z=6 - 2x$, $x$ from 3 to 0 (reverse direction). $$\vec{r}_3 = x \hat{i} + 2 \hat{j} + (6 - 2x) \hat{k}$$ $$d\vec{r}_3 = dx \hat{i} + 0 + (-2 dx) \hat{k}$$ Evaluate $\vec{A}$ on $C_3$: $$\vec{A} = (2 + 6 - 2x) \hat{i} - x(6 - 2x) \hat{j} + 4 \hat{k} = (8 - 2x) \hat{i} - (6x - 2x^2) \hat{j} + 4 \hat{k}$$ Dot product: $$\vec{A} \cdot d\vec{r}_3 = (8 - 2x) dx + 0 + 4(-2 dx) = (8 - 2x - 8) dx = -2x dx$$ Integral (note limits from 3 to 0): $$\int_3^0 -2x dx = - \int_0^3 -2x dx = \int_0^3 2x dx = \left[x^2\right]_0^3 = 9$$ 16. **Sum line integrals:** $$9 + 0 + 9 = 18$$ 17. **Conclusion:** Surface integral and line integral both equal 18, verifying Stokes' theorem. **Final answer:** $$\boxed{18}$$