Question

Use Stokes's Theorem to evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\). Use a computer algebra system to verify your results. In each case, \(C\) is oriented counterclockwise as viewed from above. \(\mathbf{F}(x, y, z)=z^{2} \mathbf{i}+x^{2} \mathbf{j}+y^{2} \mathbf{k}\) \(S: z=4-x^{2}-y^{2}, \quad z \geq 0\)

Short answer

The value of the line integral using Stokes' Theorem is 0

Step-by-step solution

Compute the Curl of \(\mathbf{F}\)

For a given vector field \(\mathbf{F}(x, y, z)\), the curl of \(\mathbf{F}\) is given by \(\nabla \times \mathbf{F}\). Applying this to our vector field, we get: \(\nabla \times \mathbf{F} = (2y - 0, 0 - 2z, 2x - 0) = 2y\mathbf{i}-2z\mathbf{j}+2x\mathbf{k}\).

Parameterize the Surface S

We are given the equation of the surface, so we need to translate this into a position vector. We are dealing with the upper half of a paraboloid, which can be parameterized as: \(\vec{r}(u,v) = (u\cos v)\mathbf{i} + (u\sin v)\mathbf{j} + (4 - u^2)\mathbf{k}\), where \(0 \leq u \leq 2\) and \(0 \leq v \leq 2\pi\).

Calculate the Surface Integral

We substitute the parameterization and the curl of \(\mathbf{F}\) into the surface integral and evaluate. This integral is: \(\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}\) where \(d\mathbf{S}\) is a small piece of surface. This gives us the integral: \(\iint_S (2y\mathbf{i}-2z\mathbf{j}+2x\mathbf{k}) \cdot (\vec{r}_u \times \vec{r}_v)\) du dv, where \(\vec{r}_u\) and \(\vec{r}_v\) are the partial derivatives of the position vector with respect to \(u\) and \(v\). After a series of calculations, we end up with an integral that simplifies to \(0\). Therefore, the surface integral of the curl of the vector field over the surface is \(0\).