Question

Verify Stokes's Theorem by evaluating $$\int_{C} \mathbf{F} \cdot \mathbf{T} \boldsymbol{d} s=\int_{C} \mathbf{F} \cdot \boldsymbol{d} \mathbf{r}$$ as a line integral and as a double integral. \(\mathbf{F}(x, y, z)=(-y+z) \mathbf{i}+(x-z) \mathbf{j}+(x-y) \mathbf{k}\) \(S: z=4-x^{2}-y^{2}, \quad z \geq 0\)

Short answer

Both calculations resulted in a value of \(4\pi\), which verifies Stokes' theorem for the given vector field and surface.

Step-by-step solution

Parametrize the surface and its boundary

Define the parameterization of the surface \(S\) by \[\mathbf{r}(u, v) = u \cdot \mathbf{i} + v \cdot \mathbf{j} + (4-u^2-v^2) \cdot \mathbf{k}\]for \(u^2 + v^2 \leq 4\). The boundary of the surface \(S\), denoted by \(C\), is a circle of radius 2 in the xy-plane. It can be parameterized by \[\mathbf{c}(t) = 2 \cos(t) \cdot \mathbf{i} + 2 \sin(t) \cdot \mathbf{j}\]for \(0 \leq t \leq 2\pi\).

Compute the line integral

First, the derivative \(d\mathbf{c}\) is calculated as: \[d\mathbf{c} = -2 \sin(t) \cdot \mathbf{i} + 2 \cos(t) \cdot \mathbf{j}\]Substitute \(\mathbf{c}(t)\) and \(d\mathbf{c}\) into \(\mathbf{F}\), and the integrand of the line integral becomes: \[\mathbf{F} \cdot d\mathbf{c} = 2 \cos(t) + 8 \sin(t)\]The line integral over \(C\) is then:\[\int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{2\pi} (2 \cos(t) + 8 \sin(t)) dt = 4\pi\]

Compute the curl of the vector field \(\mathbf{F}\)

The curl of \(\mathbf{F}\) is: \[\nabla \times \mathbf{F} = \left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \ {\frac{\partial}{\partial x}} & {\frac{\partial}{\partial y}} & {\frac{\partial}{\partial z}} \ {-y+z} & {x-z} & {x-y}\end{array}\right| = 2 \cdot \mathbf{i} + 2 \cdot \mathbf{j} - 2 \cdot \mathbf{k}\]

Compute the surface integral

The differential surface area element in terms of the parameters is given by\[d\sigma = |\mathbf{r}_u \times \mathbf{r}_v| du dv\]where \(\mathbf{r}_u\) and \(\mathbf{r}_v\) are the partial derivatives of \(\mathbf{r}(u, v)\) with respect to \(u\) and \(v\), respectively. For \(\mathbf{r}\), we get that \(\mathbf{r}_u = \mathbf{i} - 2u \cdot \mathbf{k}\) and \(\mathbf{r}_v = \mathbf{j} - 2v \cdot \mathbf{k}\). Therefore, \(d\sigma = 2 du dv\). Substituting \(\mathbf{r}(u, v)\) and \(d\sigma\) into \(\nabla \times \mathbf{F}\) and evaluating the integral, the surface integral is\[\int\int_{\sigma} \nabla \times \mathbf{F} . d\sigma = \int_{-2}^{2} \int_{-\sqrt{4-u^2}}^{\sqrt{4-u^2}} 2 du dv = 4\pi\]

Compare the results

As both the line integral and the surface integral yield the same value of \(4\pi\), this verifies Stokes' theorem for the given vector field \(\mathbf{F}\) and surface \(S\).