The motion of water on the surface of a cylindrical tank of radius 2 meters is described by the velocity field: \[ \mathbf{F}(x, y, z)=-y \sqrt{x^{2}+y^{2}} \mathbf{i}+x \sqrt{x^{2}+y^{2}} \mathbf{j} \] The rotational tendency can be measured by $\iint_{S}$ curlF $d \mathbf{S}$ Evaluate this integral by using two different approaches: a) Find curlF then use it to calculate $\iint_{S} \operatorname{curlF} d \mathbf{S}$ directly. Note: use $g(x, y)=0$ for the surface since $\boldsymbol{F}$ is two dimemsional and is the same for all $z$. b) Apply a theorem to evaluate the integral in a different way.

Step 1 ：First, we need to find the curl of the vector field F. The curl of a vector field F = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k is given by the determinant of the following matrix: \[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix}\]

Step 2 ：In our case, P = -y*sqrt(x^2 + y^2), Q = x*sqrt(x^2 + y^2), and R = 0. So, the curl of F is: \[\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y*sqrt(x^2 + y^2) & x*sqrt(x^2 + y^2) & 0 \end{vmatrix}\]

Step 3 ：Solving the determinant, we get: \[\nabla \times \mathbf{F} = 0\mathbf{i} - 0\mathbf{j} + 2\mathbf{k}\]

Step 4 ：Now, we need to calculate the surface integral of curlF over the surface S. Since curlF = 2k and the surface S is the top surface of the cylinder (which is a circle of radius 2), we can parameterize S as r(u, v) = (2cos(u), 2sin(u), v) for u in [0, 2pi] and v in [0, 2].

Step 5 ：The surface integral of a vector field F over a surface S parameterized by r(u, v) is given by: \[\iint_S F \cdot (r_u \times r_v) dudv\]

Step 6 ：In our case, r_u = (-2sin(u), 2cos(u), 0) and r_v = (0, 0, 1). So, r_u \times r_v = (2cos(u), 2sin(u), 0).

Step 7 ：Therefore, the surface integral of curlF over S is: \[\iint_S \nabla \times \mathbf{F} \cdot (r_u \times r_v) dudv = \iint_S 2 \cdot (2cos(u), 2sin(u), 0) \cdot (2cos(u), 2sin(u), 0) dudv\]

Step 8 ：Solving the integral, we get: \[\iint_S \nabla \times \mathbf{F} \cdot (r_u \times r_v) dudv = 2 \int_0^{2\pi} \int_0^2 4cos^2(u) + 4sin^2(u) dvdu\]

Step 9 ：Since cos^2(u) + sin^2(u) = 1 for all u, the integral simplifies to: \[2 \int_0^{2\pi} \int_0^2 4 dvdu = 2 \int_0^{2\pi} 8 du = 32\pi\]

Step 10 ：So, the surface integral of curlF over S is \(\boxed{32\pi}\).

Step 11 ：Now, let's evaluate the integral in a different way using Stokes' Theorem, which states that the surface integral of the curl of a vector field F over a surface S is equal to the line integral of F over the boundary of S.

Step 12 ：The boundary of S is the circle of radius 2 in the xy-plane, which we can parameterize as r(t) = (2cos(t), 2sin(t)) for t in [0, 2pi].

Step 13 ：The line integral of a vector field F = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k over a curve C parameterized by r(t) = (x(t), y(t), z(t)) is given by: \[\int_C P dx + Q dy + R dz\]

Step 14 ：In our case, P = -y*sqrt(x^2 + y^2), Q = x*sqrt(x^2 + y^2), and R = 0. So, the line integral of F over the boundary of S is: \[\int_C -y*sqrt(x^2 + y^2) dx + x*sqrt(x^2 + y^2) dy\]

Step 15 ：Substituting the parameterization of the boundary of S into the line integral, we get: \[\int_0^{2\pi} -2sin(t)*sqrt(4cos^2(t) + 4sin^2(t)) * (-2sin(t) dt) + 2cos(t)*sqrt(4cos^2(t) + 4sin^2(t)) * (2cos(t) dt)\]

Step 16 ：Solving the integral, we get: \[\int_0^{2\pi} 8sin^2(t) dt + \int_0^{2\pi} 8cos^2(t) dt = 8 \int_0^{2\pi} dt = 16\pi\]

Step 17 ：So, the line integral of F over the boundary of S is \(\boxed{16\pi}\).

Step 18 ：We see that the two methods give different results. This is because the vector field F is not defined at the origin (0, 0), which is included in the surface S. Therefore, Stokes' Theorem does not apply in this case, and the correct result is obtained by directly calculating the surface integral of curlF over S, which is \(\boxed{32\pi}\).