Respuesta :
Answer:
[tex]\displaystyle \iint_S {\text{curl \bold{F}} \cdot} \, dS = \boxed{\bold{0}}[/tex]
General Formulas and Concepts:
Calculus
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]
Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Integration Methods: U-Substitution + U-Solve
Multivariable Calculus
Partial Derivatives
Triple Integrals
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
Surface Area Differential:
[tex]\displaystyle dS = \textbf{n} \cdot d\sigma[/tex]
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]
- [tex]\displaystyle \text{div \bf{F}} = \nabla \cdot \textbf{F}[/tex]
- [tex]\displaystyle \text{curl \bf{F}} = \nabla \times \textbf{F}[/tex]
Stokes’ Theorem:
[tex]\displaystyle \oint_C {\textbf{F} \cdot } \, d\textbf{r} = \iint_S {\big( \nabla \times \textbf{F} \big) \cdot \textbf{n}} \, d\sigma[/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]
Step-by-step explanation:
Step 1: Define
Identify given.
[tex]\displaystyle \textbf{F} (x, y, z) = x^2z^2 \hat{\i} + y^2z^2 \hat{\j} + xyz \hat{\text{k}}[/tex]
[tex]\displaystyle \text{Region:} \left \{ {{\text{Paraboloid:} \ z = x^2 + y^2} \atop {\text{Cylinder:} \ x^2 + y^2 = 16}} \right[/tex]
Step 2: Integrate Pt. 1
- Find div F:
[tex]\displaystyle \text{div } \textbf{F} = \frac{\partial}{\partial x} x^2z^2 + \frac{\partial}{\partial y} y^2z^2 + \frac{\partial}{\partial z} xyz[/tex] - [div F] Differentiate [Partial Derivatives]:
[tex]\displaystyle \text{div } \textbf{F} = 2xz^2 + 2yz^2 + xy[/tex] - [Divergence Theorem] Substitute in div F:
[tex]\displaystyle \iint_S {\text{curl } \textbf{F} \cdot} \, dS = \iiint_D {2xz^2 + 2yz^2 + xy} \, dV[/tex]
Step 3: Integrate Pt. 2
Convert region from rectangular coordinates to cylindrical coordinates.
[tex]\displaystyle \text{Region:} \left \{ {{\text{Paraboloid:} \ z = x^2 + y^2} \atop {\text{Cylinder:} \ x^2 + y^2 = 16}} \right \rightarrow \left \{ {{\text{Paraboloid:} \ z = r^2} \atop {\text{Cylinder:} \ r^2 = 16}} \right[/tex]
Identifying limits, we have the bounds:
[tex]\displaystyle \left\{ \begin{array}{ccc} 0 \leq z \leq r^2 \\ 0 \leq r \leq 4 \\ 0 \leq \theta \leq 2 \pi \end{array}[/tex]
Step 4: Integrate Pt. 3
- [Integral] Substitute in variables and region:
[tex]\displaystyle \iint_S {\text{curl } \textbf{F} \cdot} \, dS = \int\limits^{2 \pi}_0 \int\limits^4_0 \int\limits^{r^2}_0 {r \bigg( 2z^2r \cos \theta + 2z^2r \sin \theta +r^2 \cos \theta \sin \theta \bigg)} \, dz \, dr \, d\theta[/tex]
We evaluate the Stokes' Divergence Theorem Integral using basic integration techniques listed under "Calculus".
[tex]\displaystyle \begin{aligned}\iint_S {\text{curl } \textbf{F} \cdot} \, dS & = \int\limits^{2 \pi}_0 \int\limits^4_0 \int\limits^{r^2}_0 {r \bigg( 2z^2r \cos \theta + 2z^2r \sin \theta +r^2 \cos \theta \sin \theta \bigg)} \, dz \, dr \, d\theta \\& = \frac{1}{3} \int\limits^{2 \pi}_0 \int\limits^4_0 {zr^2 \bigg[ 2z^2 \big( \cos \theta + \sin \theta \big) + 3r \sin \theta \cos \theta \bigg] \bigg| \limits^{z = r^2}_{z = 0}} \, dr \, d\theta \\\end{aligned}[/tex]
[tex]\displaystyle \begin{aligned}\iint_S {\text{curl } \textbf{F} \cdot} \, dS & = \frac{1}{3} \int\limits^{2 \pi}_0 \int\limits^4_0 {r^5 \bigg[ 2r^3 \big( \cos \theta + \sin \theta \big) + 3 \sin \theta \cos \theta \bigg]} \, dr \, d\theta \\& = \frac{1}{54} \int\limits^{2 \pi}_0 {r^6 \bigg[ 4r^3 \big( \cos \theta + \sin \theta \big) + 9 \sin \theta \cos \theta \bigg] \bigg| \limits^{r = 4}_{r = 0}} \, d\theta \\\end{aligned}[/tex]
[tex]\displaystyle \begin{aligned}\iint_S {\text{curl } \textbf{F} \cdot} \, dS & = \frac{2048}{27} \int\limits^{2 \pi}_0 {\cos \theta \Big( 9 \sin \theta + 256 \Big) + 256 \sin \theta} \, d\theta \\& = \frac{-1024}{243} \bigg[ 4608 \cos \theta - \bigg( 9 \sin \theta + 256 \bigg)^2 \bigg] \bigg| \limits^{\theta = 2 \pi}_{\theta = 0} \\& = \boxed{\bold{0}}\end{aligned}[/tex]
∴ we have calculated the Stokes' Theorem integral with the given region and function using the Divergence Theorem.
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Topic: Multivariable Calculus
