A particle starts at the point (−3, 0), moves along the x-axis to (3, 0), and then along the semicircle y = 9 − x2 to the starting point. Use Green's Theorem to find the work done on this particle by the force field F(x, y) = 3x, x3 + 3xy2 .

Answer:[tex]\displaystyle \int_C {F \cdot} \, dr = \boxed{\bold{\frac{243 \pi}{2}}}[/tex]General Formulas and Concepts:
CalculusDifferentiationDerivativesDerivative NotationDerivative Property [Multiplied Constant]:
[tex]\displaystyle \bold{(cu)' = cu'}[/tex]Derivative Property [Addition/Subtraction]:
[tex]\displaystyle \bold{(u + v)' = u' + v'}[/tex]Derivative Rule [Basic Power Rule]:f(x) = cxⁿf’(x) = c·nxⁿ⁻¹IntegrationIntegralsIntegration Rule [Reverse Power Rule]:
[tex]\displaystyle \bold{\int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C}[/tex]Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \bold{\int\limits^b_a {f(x)} \, dx = F(b) - F(a)}[/tex]Integration Property [Multiplied Constant]:
[tex]\displaystyle \bold{\int {cf(x)} \, dx = c \int {f(x)} \, dx}[/tex]Multivariable CalculusPartial DerivativesDouble IntegrationPolar Coordinate Conversions:[tex]\displaystyle \bold{x = r \cos \theta}[/tex][tex]\displaystyle \bold{y = r \sin \theta}[/tex][tex]\displaystyle \bold{r^2 = x^2 + y^2}[/tex][tex]\displaystyle \bold{\tan \theta = \frac{y}{x}}[/tex]Fubini’s Theorem [Polar]:
[tex]\displaystyle \bold{\iint_R{F(r, \theta)} \, dA = \int\limits^{\beta}_{\alpha} \int\limits^{b}_{a} {F(r, \theta)r} \, dr \, d\theta}[/tex]Vector Calculus (Line Integrals)Circulation Density:
[tex]\displaystyle \bold{F = M \hat{\i} + N \hat{\j} \rightarrow \text{curl} \ \bold{F} \cdot \bold{k} = \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}[/tex]Green's Theorem [Circulation Curl/Tangential Form]:
[tex]\displaystyle \bold{\oint_C {F \cdot T} \, ds = \oint_C {M \, dx + N \, dy} = \iint_R {\bigg( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \bigg)} \, dx \, dy}[/tex]Step-by-step explanation:Step 1: DefineIdentify given.[tex]\displaystyle F(x, y) = < 3x , x^3 + 3xy^2 >[/tex]Region R [See Graph Attachment]Step 2: Integrate Pt. 1Define vector functions M and N:
[tex]\displaystyle M = 3x , \ N = x^3 + 3xy^2[/tex][Circulation Density] Differentiate [Partial Derivatives and Derivative Rules + Properties]:
[tex]\displaystyle \frac{\partial M}{\partial y} = 0 , \ \frac{\partial N}{\partial x} = 3x^2 + 3y^2[/tex][Green's Theorem] Substitute in Circulation Density:
[tex]\displaystyle \int_C {F \cdot} \, dr = \iint_R {3x^2 + 3y^2} \, dx \, dy[/tex][Integrals] Rewrite:
[tex]\displaystyle \int_C {F \cdot} \, dr = 3 \iint_R {x^2 + y^2} \, dx \, dy[/tex][Integrals] Substitute in Polar Coordinate Conversions:
[tex]\displaystyle \int_C {F \cdot} \, dr = 3 \iint_R {r^3} \, dr \, d\theta[/tex][Integrals] Substitute in region R [Graph]:
[tex]\displaystyle \int_C {F \cdot} \, dr = 3 \int\limits^{\pi}_0 \int\limits^3_0 {r^3} \, dr \, d\theta[/tex]Step 3: Integrate Pt. 2We can evaluate the Green's Theorem double integral using basic integration techniques listed above (Calculus):[tex]\displaystyle \begin{aligned}\int_C {F \cdot} \, dr & = 3 \int\limits^{\pi}_0 \int\limits^3_0 {r^3} \, dr \, d\theta \\& = 3 \int\limits^{\pi}_0 {\frac{r^4}{4} \bigg| \limits^{r = 3}_{r = 0}} \, d\theta \\& = 3 \int\limits^{\pi}_0 {\frac{81}{4}} \, d\theta \\& = 3 \bigg( {\frac{81}{4}} \theta \bigg) \bigg| \limits^{\theta = \pi}_{\theta = 0} \\& = \boxed{\bold{\frac{243 \pi}{2}}} \\\end{aligned}[/tex]∴ we have calculated the work done on this particle by the given force field using Green's Theorem.---Learn more about Green's Theorem:
Learn more about multivariable calculus: : Multivariable CalculusUnit: Green's Theorem and Surfaces