Multiple integrals and vector calculus (PS4): Divergence and Stokes theorems

We wish to verify the result that \[ \oint_C \mb{A} \cdot \de{\mb{r}} = \int_S \nabla \times \mb{F} \cdot \de{\mb{S}}, \]

where $S$ is the hemisphere. Remember that if we choose the orientation of the surface such that, if the normal, $\mb{n}$ is pointing upwards, then the curve $C$ must be oriented counter-clockwise. That is, if we choose \[ \mb{n} = \frac{[x,y,z]}{\sqrt{x^2 + y^2 + z^2}}, \]

for the normal (the denominator is unity on the surface of the sphere). Then the curve $C$ is defined with \[ \gamma(\theta) = [\cos\theta, \sin\theta, 0]. \]

We first compute the line integral. This gives \begin{align*} \oint_C \mb{A} \cdot \de{\mb{r}} &= \int_0^{2\pi} \mb{A}(\gamma(\theta)) \cdot \gamma'(\theta) \, \de{\theta} \\ &= \int_0^{2\pi} [\sin\theta, -\cos\theta, 0] \cdot [-\sin\theta, \cos\theta, 0] \, \de{\theta} \\ &= \int_0^{2\pi} (-1) \, \de{\theta} -2\pi. \end{align*}

Now we compute the right hand side of Stokes' Theorem. Remember that this integral is performed via \[ \int_S \nabla \times \mb{F} \cdot \de{\mb{S}} = \int_S \nabla \times \mb{F} \cdot \mb{n} \, \de{S}, \]

where $\mb{n}$ is the oriented unit normal of the surface. In this case, the curl is \[ \nabla \times \mb{F} = [0, 0, -2]. \]

We wish to perform the integration in spherical coordinates. We introduce \[ x = \cos\theta\sin\phi, \quad y = \sin\theta\sin\phi, \quad z = \cos\phi, \]

on the surface of the sphere, with $0 \leq \theta \leq 2\pi$ and $0 \leq \phi \leq \pi/2$. Then the integration element transforms using the Jacobian \[ J = r^2 \sin\phi = \sin\phi \]

so on the surface of the sphere, $r = 1$, we have \[ \de{S} = \sin\phi \, \de{\theta} \de{\phi}, \]

and the integral becomes \begin{align*} \int_S \nabla \times \mb{F} \cdot \de{\mb{S}} &= \int_0^{\pi/2} \int_0^{2\pi} [0, 0, -2] \cdot \frac{[x, y, z]}{r} \, \sin\phi \, \de{\theta} \de{\phi} \\ &= \int_0^{\pi/2} \int_0^{2\pi} (-2\cos\phi) \sin\phi \, \de{\theta} \de{\phi} \\ &= \int_0^{\pi/2} \int_0^{2\pi} -\sin(2\phi) \, \de{\theta} \de{\phi} \\ &= -2\pi \end{align*}

which verifies Stokes' Theorem.

Note that this question is not exactly well defined. We will set an arbitrary contour around the cylinder as \[ \gamma(\theta) = [3 + \sqrt{2}\cos\theta, \sqrt{2}\sin\theta, z(\theta)], \]

and so this assumes that the contour has a $z$ component that can be expressed as a function of $\theta$. This, for example, excludes curves that intersect themselves. The line integral element is \[ \de{\mb{r}} = \gamma'(\theta) \, \de{\theta}, \]

so that the integral \begin{multline*} \int_0^{2\pi} [\sqrt{2}\sin\theta, -(3 + \sqrt{2}\cos\theta), 0] \\ \cdot [-\sqrt{2}\sin\theta, \sqrt{2}\cos\theta, z'(\theta)] \, \de{\theta} = -4\pi. \end{multline*}

Note that both these problems are vector line integrals. So there is a difference between $\de{\mb{r}}$ and $\de{s}$. For specifically, if you parameterize $C$ by $[x,y,z] = \gamma(t)$, then the regular line integral is given by \[ \int_C \phi \, \de{s} = \int_0^1 \phi(\gamma(t)) \lvert \gamma'(t) \rvert \de{t}, \]

whereas the vector line integral is given by \[ \int_C \phi \, \de{\mb{r}} = \int_0^1 \phi(\gamma(t)) \gamma'(t) \de{t}, \]

and note that since $\gamma'(t)$ is a vector, then the answer is a vector.

In this problem, we parameterize the curve by $\gamma(t) = [t^2, 2t, t^3]$. The line integral is then computed from \begin{align*} I_1 &= \int_0^1 \Bigl[2xyz^2\Bigr]_{\gamma(t)} \gamma'(t) \, \de{t} \\ &= \int_0^1 [2(t^2)(2t)(t^3)^2 [2t, 2, 3t^2] \, \de{t} \\ &= \int_0^1 [8t^{10}, 8t^9, 12t^11] \, \de{t} \\ &= [8/11, 4/5, 1]. \end{align*}

This time, the integral is computed using \[ I2 = \int_C \mb{F}(\gamma(t)) \times \gamma'(t) \, \de{t}. \]

Note that \begin{align*} \mb{F}(\gamma(t)) \times \gamma'(t) &= [2t^3, -t^3, t^4] \times [2t, 2, 3t^2] \\ &= [-2t^4 - 3 t^5, -4 t^5, 4t^3 + 2t^4], \end{align*}

so that \begin{align*} I_2 &= \int_0^1 \mb{F}(\gamma(t)) \times \gamma'(t) \, \de{t} \\ &= \int_0^1 [-2t^4 - 3 t^5, -4 t^5, 4t^3 + 2t^4] \, \de{t} \\ &= [-9/10, -2/3, 7/5]. \end{align*}