Question

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = tan−1(x2yz2)i + x2yj + x2z2k, S is the cone x = y2 + z2 , 0 ≤ x ≤ 3, oriented in the direction of the positive x-axis.

Answer 1

Stokes' Theorem relates a surface integral over a surface to a line integral around its boundary curve. The value of S ∮ curl F · dS over the given cone S is 9π.

Step-by-step explanation:

Stokes' Theorem relates a surface integral over a surface to a line integral around its boundary curve. For this problem, we're given F(x, y, z) = tan⁻¹(x²yz²)i + x²yj + x²z²k and the cone x = y² + z², 0 ≤ x ≤ 3, oriented in the direction of the positive x-axis. To apply Stokes' Theorem, we first need to calculate the curl of F.

The curl of F, denoted as curl F, is given by ∇ × F, where ∇ is the del operator. After computing the curl of F, we then perform the surface integral over the cone S by evaluating the dot product of curl F and the unit normal vector dS on the surface.

Given the curl of F as curl F = ⟨-2xzyz², 0, -x²y⟩, the surface integral S ∮ curl F · dS transforms into the line integral along the boundary curve of the cone. As per Stokes' Theorem, this line integral over the boundary curve x = y² + z², 0 ≤ x ≤ 3, is equivalent to evaluating the surface integral.

The boundary curve of the cone is a circle centered at the origin in the xz-plane, with radius 3. Setting up the line integral along this curve allows us to solve for the value, and after calculations, it equals 9π. This result represents the evaluation of S ∮ curl F · dS over the given cone S using Stokes' Theorem.

Answer 2

1. Using Stokes' Theorem, the value of
`\( \int_S \text{curl} \, \mathbf{F} \cdot d\mathbf{S} \)` for the given vector field
`\(\mathbf{F}(x, y, z) = \tan^(-1)(x^2yz^2)\mathbf{i} + x^2y\mathbf{j} + x^2z^2\mathbf{k}\)`over the cone Sis
`18\pi\)`.

Step-by-step explanation:

Stokes' Theorem relates a surface integral over a surface Sto a line integral around its boundary curve C Mathematically, it is given by:

`\[ \int_S \text{curl} \, \mathbf{F} \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r} \]`

In this case, the vector field
`\(\mathbf{F}(x, y, z) = \tan^(-1)(x^2yz^2)\mathbf{i} + x^2y\mathbf{j} + x^2z^2\mathbf{k}\)`and the surface S is the cone defined by
`\(x = y^2 + z^2\)`where
`\(0 \leq x \leq 3\)`and oriented in the direction of the positive x-axis.

To evaluate the line integral on the right-hand side, we need to find the curve C that bounds the surface. The boundary of the cone is the circle defined by x = 3, and C is the curve along this circle.

Using Stokes' Theorem, the line integral around this circle evaluates to
`\(18\pi\),` yielding the final result of
`\(18\pi\)`for
`\( \int_S \text{curl} \, \mathbf{F} \cdot d\mathbf{S} \).`

Calculation:

`\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^(2\pi) \tan^(-1)(27\sin^2\theta\cos^2\theta) \, d\theta = 18\pi \]`

Therefore, the value of
`\( \int_S \text{curl} \, \mathbf{F} \cdot d\mathbf{S} \) is indeed \(18\pi\).`