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Use the Divergence Theorem to calculate the surface integral

S F · dS; that is, calculate the flux of F across S. F(x, y, z) = 3xy2i + xezj + z3k, S is the surface of the solid bounded by the cylinder

y2 + z2 = 1 and the planes

Verified Answer :

Space

Answer:

\(\displaystyle \int_S {\textbf{F} \cdot n} \, dS = \boxed{\frac{9 \pi}{2}}\)

General Formulas and Concepts:
Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
\(\displaystyle (cu)' = cu'\)

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Reverse Power Rule]:
\(\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C\)

Integration Rule [Fundamental Theorem of Calculus 1]:
\(\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)\)

Integration Property [Multiplied Constant]:
\(\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx\)

Multivariable Calculus

Partial Derivatives

Triple Integration

Cylindrical Coordinate Conversions:

  • \(\displaystyle x = r \cos \theta\)
  • \(\displaystyle y = r \sin \theta\)
  • \(\displaystyle z = z\)
  • \(\displaystyle r^2 = x^2 + y^2\)
  • \(\displaystyle \tan \theta = \frac{y}{x}\)

Integral Conversion [Cylindrical Coordinates]:
\(\displaystyle \iiint_T {f(r, \theta, z)} \, dV = \iiint_T {f(r, \theta, z)r} \, dz \, dr \, d\theta\)

Vector Calculus (Line Integrals)

Del (Operator):
\(\displaystyle abla = \hat{\i} \frac{\partial}{\partial x} + \hat{\j} \frac{\partial}{\partial y} + \hat{\text{k}} \frac{\partial}{\partial z}\)

Div

  • \(\displaystyle \text{div \bf{F}} = abla \cdot \textbf{F}\)

Divergence Theorem:
\(\displaystyle \iint_S {\big( abla \times \textbf{F} \big) \cdot \textbf{n}} \, d\sigma = \iiint_D { abla \cdot \textbf{F}} \, dV\)

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.

\(\displaystyle \textbf{F}(x, y, z) = 3xy^2 \hat{\i} + xe^z \hat{\j} + z^3 \hat{\text{k}}\)

\(\displaystyle \text{Region:} \left \{ {{\text{Cylinder: } y^2 + z^2 = 1} \atop {\text{Planes: } x = -1, 2}} \right\)

Step 2: Integrate Pt. 1

  1. Find div F:
    \(\displaystyle \text{div \bf{F}} = \frac{\partial}{\partial x}3xy^2 + \frac{\partial}{\partial y}xe^z + \frac{\partial}{\partial z}z^3\)
  2. [div F] Differentiate [Partial Derivatives]:
    \(\displaystyle \text{div \bf{F}} = 3y^2 + 3z^2\)
  3. [Divergence Theorem] Substitute in div F:
    \(\displaystyle \iint_S {\big( abla \times \textbf{F} \big) \cdot \textbf{n}} \, d\sigma = 3 \iiint_D {y^2 + z^2} \, dV\)

Step 3: Integrate Pt. 2

Convert region into cylindrical coordinates (r, θ, x) in the yz-plane.

\(\displaystyle \text{Region:} \left \{ {{\text{Cylinder: } y^2 + z^2 = 1} \atop {\text{Planes: } x = -1, 2}} \right \rightarrow \left \{ {{\text{Cylinder: } r^2 = 1} \atop {\text{Planes: } x = -1, 2}} \right\)

Identifying limits, we have the bounds:

\(\displaystyle \left\{ \begin{array}{ccc} 0 \leq \theta \leq \2 \pi \\ 0 \leq r \leq 1 \\ -1 \leq x \leq 2 \end{array}\)

Step 4: Integrate Pt. 3

  1. [Integrals] Convert [Integral Conversion - Cylindrical Coordinates]:
    \(\displaystyle \iint_S {\big( abla \times \textbf{F} \big) \cdot \textbf{n}} \, d\sigma = 3 \iiint_P {r^3} \, dr \, d\theta \, dx\)
  2. [Integrals] Substitute in region:
    \(\displaystyle \iint_S {\big( abla \times \textbf{F} \big) \cdot \textbf{n}} \, d\sigma = 3 \int\limits^2_{-1} \int\limits^{2 \pi}_0 \int\limits^1_0 {r^3} \, dr \, d\theta \, dx\)

We can evaluate the Divergence Theorem integral using basic integration techniques listed under "Calculus":

\(\displaystyle \begin{aligned}3 \int\limits^2_{-1} \int\limits^{2 \pi}_0 \int\limits^1_0 {r^3} \, dr \, d\theta \, dx & = \frac{3}{4} \int\limits^2_{-1} \int\limits^{2 \pi}_0 {r^4 \bigg| \limits^{r = 1}_{r = 0}} \, d\theta \, dx \\& = \frac{3}{4} \int\limits^2_{-1} \int\limits^{2 \pi}_0 {} \, d\theta \, dx \\& = \frac{3}{4} \int\limits^2_{-1} {\theta \bigg| \limits^{\theta = 2 \pi}_{\theta = 0}} \, dx \\& = \frac{3 \pi}{2} \int\limits^2_{-1} {} \, dx \\\end{aligned}\)

\(\displaystyle \begin{aligned}3 \int\limits^2_{-1} \int\limits^{2 \pi}_0 \int\limits^1_0 {r^3} \, dr \, d\theta \, dx & = \frac{3 \pi}{2} x \bigg| \limits^{x = 2}_{x = -1} \\& = \boxed{\frac{9 \pi}{2}} \\\end{aligned}\)

∴ \(\displaystyle \Phi = \boxed{\frac{9 \pi}{2}}\)

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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