Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = 5y cos(z) i + ex sin(z) j + xey k, S is the hemisphere x2 + y2 + z2 = 4, z ≥ 0, oriented upward. Step 1 Stokes' Theorem tells us that if C is the boundary curve of a surface S, then curl F · dS S = C F · dr Since S is the hemisphere x2 + y2 + z2 = 4, z ≥ 0 oriented upward, then the boundary curve C is the circle in the xy-plane, x2 + y2 = 4 Correct: Your answer is correct. seenKey 4 , z = 0, oriented in the counterclockwise direction when viewed from above. A vector equation of C is r(t) = 2 Correct: Your answer is correct. seenKey 2 cos(t) i + 2 Correct: Your answer is correct. seenKey 2 sin(t) j + 0k with 0 ≤ t ≤ 2π.

Stokes' Theorem relates the surface integral of the curl of a vector field over a surface \( S \) to the line integral of the vector field over the boundary curve \( C \) of the surface. The theorem states:

\[ \int\int_{S} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \int_{C} \mathbf{F} \cdot d\mathbf{r} \]

Given the vector field \( \mathbf{F}(x, y, z) = 5y\cos(z)\mathbf{i} + e^x\sin(z)\mathbf{j} + xe^y\mathbf{k} \) and \( S \) is the hemisphere \( x^2 + y^2 + z^2 = 4 \), \( z \geq 0 \) oriented upward, our goal is to compute \( \int\int_{S} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \) using Stokes' Theorem.

Firstly, we need to describe the boundary curve \( C \). Since \( S \) is a hemisphere, the boundary curve \( C \) will be the circle at the base of the hemisphere, in the xy-plane where \( z = 0 \).

The circle is given by \( x^2 + y^2 = 4 \), and it is oriented in the counterclockwise direction when viewed from above. A parametric equation of \( C \) can be written as:

\[ \mathbf{r}(t) = 2\cos(t)\mathbf{i} + 2\sin(t)\mathbf{j} \]

where \( t \) ranges from 0 to \( 2\pi \).

To proceed with Stokes' theorem, we must evaluate the line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) over \( C \).

The differential element on the curve \( C \) is:

\[ d\mathbf{r} = \frac{d\mathbf{r}}{dt}dt = (-2\sin(t)\mathbf{i} + 2\cos(t)\mathbf{j})dt \]

Next, we plug in \( x = 2\cos(t) \) and \( y = 2\sin(t) \) into \( \mathbf{F} \), remembering that \( z = 0 \) along \( C \):

\[ \mathbf{F}(t) = 5(2\sin(t))\cos(0)\mathbf{i} + e^{2\cos(t)}\sin(0)\mathbf{j} + (2\cos(t))e^{2\sin(t)}\mathbf{k} = 10\sin(t)\mathbf{i} + 2\cos(t)e^{2\sin(t)}\mathbf{k} \]

Now, we can calculate the dot product \( \mathbf{F}(t) \cdot d\mathbf{r} \):

\[ \mathbf{F}(t) \cdot d\mathbf{r} = (10\sin(t)\mathbf{i} + 2\cos(t)e^{2\sin(t)}\mathbf{k}) \cdot (-2\sin(t)\mathbf{i} + 2\cos(t)\mathbf{j})dt \] \[ = (-20\sin^2(t) + 0)dt \] \[ = -20\sin^2(t)dt \]

To obtain the line integral, we integrate this dot product over the interval \( t = 0 \) to \( t = 2\pi \):

\[ \int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{2\pi} -20\sin^2(t)dt \]

This integral can now be solved using the identity \( \sin^2(t) = \frac{1 - \cos(2t)}{2} \) and standard integration techniques.

Extra: Stokes' Theorem is a powerful tool in vector calculus, often used to simplify complex surface integrals by converting them into line integrals, which can be easier to evaluate. This theorem is particularly useful when the boundary of the surface is a simple closed curve, and the vector field is nicely behaved. Stokes' theorem is a generalization of Green's theorem in the plane to surfaces in three dimensions.

The theorem essentially says that the circulation of a vector field \( \mathbf{F} \) around a closed loop \( C \) is equal to the flux of the curl of \( \mathbf{F} \) across any surface \( S \) that is bounded by \( C \). This is assuming \( S \) is oriented consistently with the orientation of \( C \) (using the right-hand rule). For students learning this concept, visualizing the surface, its orientation, and the direction of the curve is essential for setting up and solving problems correctly.