Respuesta :
Answer:
[tex]\displaystyle \int_S {\textbf{F} \cdot n} \, dS = \boxed{\frac{9 \pi}{2}}[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]
Derivative Rule [Basic Power Rule]:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Integration
- Integrals
Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Multivariable Calculus
Partial Derivatives
Triple Integration
Cylindrical Coordinate Conversions:
- [tex]\displaystyle x = r \cos \theta[/tex]
- [tex]\displaystyle y = r \sin \theta[/tex]
- [tex]\displaystyle z = z[/tex]
- [tex]\displaystyle r^2 = x^2 + y^2[/tex]
- [tex]\displaystyle \tan \theta = \frac{y}{x}[/tex]
Integral Conversion [Cylindrical Coordinates]:
[tex]\displaystyle \iiint_T {f(r, \theta, z)} \, dV = \iiint_T {f(r, \theta, z)r} \, dz \, dr \, d\theta[/tex]
Vector Calculus (Line Integrals)
Del (Operator):
[tex]\displaystyle \nabla = \hat{\i} \frac{\partial}{\partial x} + \hat{\j} \frac{\partial}{\partial y} + \hat{\text{k}} \frac{\partial}{\partial z}[/tex]
Div
- [tex]\displaystyle \text{div \bf{F}} = \nabla \cdot \textbf{F}[/tex]
Divergence Theorem:
[tex]\displaystyle \iint_S {\big( \nabla \times \textbf{F} \big) \cdot \textbf{n}} \, d\sigma = \iiint_D {\nabla \cdot \textbf{F}} \, dV[/tex]
Generalization of Green's Theorem
Explanation:
*Note:
Your question is incomplete, but I have defined the portions of the questions that were missing below.
Step 1: Define
Identify given.
[tex]\displaystyle \textbf{F}(x, y, z) = 3xy^2 \hat{\i} + xe^z \hat{\j} + z^3 \hat{\text{k}}[/tex]
[tex]\displaystyle \text{Region:} \left \{ {{\text{Cylinder: } y^2 + z^2 = 1} \atop {\text{Planes: } x = -1, 2}} \right[/tex]
Step 2: Integrate Pt. 1
- Find div F:
[tex]\displaystyle \text{div \bf{F}} = \frac{\partial}{\partial x}3xy^2 + \frac{\partial}{\partial y}xe^z + \frac{\partial}{\partial z}z^3[/tex] - [div F] Differentiate [Partial Derivatives]:
[tex]\displaystyle \text{div \bf{F}} = 3y^2 + 3z^2[/tex] - [Divergence Theorem] Substitute in div F:
[tex]\displaystyle \iint_S {\big( \nabla \times \textbf{F} \big) \cdot \textbf{n}} \, d\sigma = 3 \iiint_D {y^2 + z^2} \, dV[/tex]
Step 3: Integrate Pt. 2
Convert region into cylindrical coordinates (r, θ, x) in the yz-plane.
[tex]\displaystyle \text{Region:} \left \{ {{\text{Cylinder: } y^2 + z^2 = 1} \atop {\text{Planes: } x = -1, 2}} \right \rightarrow \left \{ {{\text{Cylinder: } r^2 = 1} \atop {\text{Planes: } x = -1, 2}} \right[/tex]
Identifying limits, we have the bounds:
[tex]\displaystyle \left\{ \begin{array}{ccc} 0 \leq \theta \leq \2 \pi \\ 0 \leq r \leq 1 \\ -1 \leq x \leq 2 \end{array}[/tex]
Step 4: Integrate Pt. 3
- [Integrals] Convert [Integral Conversion - Cylindrical Coordinates]:
[tex]\displaystyle \iint_S {\big( \nabla \times \textbf{F} \big) \cdot \textbf{n}} \, d\sigma = 3 \iiint_P {r^3} \, dr \, d\theta \, dx[/tex] - [Integrals] Substitute in region:
[tex]\displaystyle \iint_S {\big( \nabla \times \textbf{F} \big) \cdot \textbf{n}} \, d\sigma = 3 \int\limits^2_{-1} \int\limits^{2 \pi}_0 \int\limits^1_0 {r^3} \, dr \, d\theta \, dx[/tex]
We can evaluate the Divergence Theorem integral using basic integration techniques listed under "Calculus":
[tex]\displaystyle \begin{aligned}3 \int\limits^2_{-1} \int\limits^{2 \pi}_0 \int\limits^1_0 {r^3} \, dr \, d\theta \, dx & = \frac{3}{4} \int\limits^2_{-1} \int\limits^{2 \pi}_0 {r^4 \bigg| \limits^{r = 1}_{r = 0}} \, d\theta \, dx \\& = \frac{3}{4} \int\limits^2_{-1} \int\limits^{2 \pi}_0 {} \, d\theta \, dx \\& = \frac{3}{4} \int\limits^2_{-1} {\theta \bigg| \limits^{\theta = 2 \pi}_{\theta = 0}} \, dx \\& = \frac{3 \pi}{2} \int\limits^2_{-1} {} \, dx \\\end{aligned}[/tex]
[tex]\displaystyle \begin{aligned}3 \int\limits^2_{-1} \int\limits^{2 \pi}_0 \int\limits^1_0 {r^3} \, dr \, d\theta \, dx & = \frac{3 \pi}{2} x \bigg| \limits^{x = 2}_{x = -1} \\& = \boxed{\frac{9 \pi}{2}} \\\end{aligned}[/tex]
∴ [tex]\displaystyle \Phi = \boxed{\frac{9 \pi}{2}}[/tex]
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Learn more about Divergence Theorem: https://brainly.com/question/21029519
Learn more about multivariable calculus: https://brainly.com/question/13933633
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Topic: Multivariable Calculus
Unit: Stokes' Theorem and Divergence Theorem
