LHS and RHS of Stokes' theorem not equal

Question:

I am trying to test my understanding of Stokes' theorem by calculating the left and right hand side of the theorems equality by way of example and seeing whether they equal each other. They don't, which leads me to question my understanding of the theorem and where I am making an error or whether the example does not satisfy the conditions to apply Stokes theorem?

Example Question:

To test Stokes' theorem, I'm using the following example: A surface $$S = \{(x, y, z) | z = 3 - x^2 - y^2; z >= 2\}$$ is oriented upwards which is located within a vector field $$F(x, y, z) = (-y, x, 1)$$.

Applicability of Stokes theorem: Surface S is smooth, oriented surface whose orientated boundary C is smooth, simple closed, curve. Furthermore, vector field F is smooth 3-dimensional vector field whose domain is a region in $$\mathbb R^3$$ that contains S. Therefore we can apply Stokes theorem:

$$\iint_S (\underline{curl} \underline{F}) \cdot \underline n dS$$ = $$\int_C \underline F \cdot d\underline r$$

Calculating the LHS of Stokes equation:

$$\underline{curlF} = \begin{vmatrix} \underline i & \underline j & \underline k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & 1 \\ \end{vmatrix} = (0, 0, 2)$$

Let $$G(x, y, z) = z - (3 - x^2 - y^2) = z - 3 + x^2 + y^2$$

$$\nabla G(x,y,z) = (2x, 2y, 1)$$

$$\underline n(x,y,z) = \frac{\nabla G(x,y,z)}{\lVert \nabla G(x,y,z)\rVert}$$

$$\underline n(x,y,z) = \frac{(2x, 2y, 1)}{\sqrt{4x^2 + 4y^2 + 1}}$$

$$\iint_S (\underline{curl} \underline{F}) \cdot \underline n dS = (0, 0, 2) \cdot \frac{(2x, 2y, 1)}{\sqrt{4x^2 + 4y^2 + 1}}$$

$$\iint_S (\underline{curl} \underline{F}) \cdot \underline n dS = \frac{2}{\sqrt{4x^2 + 4y^2 + 1}}$$

$$dS = \sqrt{\frac{\partial z}{\partial x}^2 + \frac{\partial z}{\partial y}^2 + \frac{\partial z}{\partial z}^2} dA$$

$$dS = \sqrt{4x^2 + 4y^2 + 1} dA$$

$$\iint_S (\underline{curl} \underline{F}) \cdot \underline n dS = \iint_R 2 dA$$

$$=\int_0^{2\pi}\int_0^1 2 r dr d\theta$$

$$=\int_0^{2\pi}[r^2]_0^1d\theta$$

$$=\int_0^{2\pi}1d\theta$$

$$=2\pi$$

Calculating the RHS of Stokes equation should also give 2Pi:

Parametising C gives us: $$(x,y,z) = (2\sin t, y = 2\cos t, z = 2), t \in [0, 2\pi]$$

$$\int_C \underline F \cdot d\underline r = \int_C -y dx + x dy + dz$$

where... $$dx = 2\cos t dt$$

$$dy = -2\sin t dt$$

$$dz = 0$$

$$=\int_0^{2\pi} (-2\cos t) (2 \cos t) dt + (2\sin t) (- 2\sin t) dt$$

$$=\int_0^{2\pi} -4\cos^2 t - 4\sin^2 tdt$$

$$=-4\int_0^{2\pi} \sin^2 t + \cos^2 tdt$$

$$=-4\int_0^{2\pi} 1dt$$

$$=-4[t]_0^{2\pi}$$

$$-8\pi$$

LHS doesnt equal RHS and I don't know why

• I think you did two mistakes for the RHS, specifically in parametrizing $C$: The orientation is the wrong way around (swap cos and sin), and the boundary is a circle of radius 1 (not 2) at height z=2. – Sebastian Schulz Apr 4 at 16:11
• Good catch, silly mistake for me making the radius 2. Okay that makes the magnitudes of both LHS and RHS equal (2*pi) but the signs are still opposite. Why is my parametisation orientation wrong? If I follow the curve C it follows the "right hand rule". i.e. if I walk along the curve counter clockwise as above then the surface points "up". – Dean P Apr 4 at 16:29
• Yes, but the parametrization you picked is clockwise! E.g. at $t=0$ you're at $(0,1)$ (i.e. at "12:00") and at $t= \pi /2$ you're at $(1,0)$ (i.e. at "3:00"). – Sebastian Schulz Apr 5 at 15:31
• That is to say, I think you have the completely right intuition, you just seem to have written down the wrong formula :) – Sebastian Schulz Apr 5 at 15:32
• Yes, that's right. Let me just point out that the way you drew the axes is called "left-handed" because it changes all of the hand-rules (such as the one from above or the one for the cross-product of vectors) ;) – Sebastian Schulz Apr 5 at 19:41