Esercizi sulle Formule di Green

Esercizi sulle Formule di Green

+Esercizi sulle Formule di Green

Versione italiana

Esercizi sulle Formule di Green

Le formule di Green sono strumenti fondamentali nell’analisi vettoriale e collegano l’integrale di una funzione su una regione del piano con l’integrale della sua derivata su un contorno di quella regione. Esistono due forme principali delle formule di Green: la forma per il campo scalare e la forma per il campo vettoriale.

Concetti Chiave

1. Teorema di Green: Se C$C$ è un contorno chiuso e D$D$ è la regione delimitata da C$C$, allora per un campo vettoriale \mathbf{F} = (P, Q)$\mathbf{F} = (P, Q)$ con P$P$ e Q$Q$ continue su D$D$ e con derivate parziali continue, si ha:

   \oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA $\oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$

2. Interpretazione Geometrica: L’integrale a sinistra rappresenta il lavoro fatto da un campo vettoriale lungo il contorno C$C$, mentre l’integrale a destra rappresenta il flusso del rotore del campo attraverso la regione D$D$.

3. Applicazioni: Le formule di Green sono utilizzate in fisica e ingegneria, ad esempio per calcolare il lavoro, il flusso e in problemi di elettromagnetismo.

Esercizi

Esercizio 1

Calcola l’integrale di linea:

\oint_C (x^2 \, dy + y^2 \, dx) $\oint_C (x^2 \, dy + y^2 \, dx)$

dove C$C$ è il contorno del quadrato con vertici (0,0)$(0,0)$, (1,0)$(1,0)$, (1,1)$(1,1)$, (0,1)$(0,1)$.

Soluzione:

Identifichiamo P = y^2$P = y^2$ e Q = x^2$Q = x^2$. Calcoliamo le derivate parziali:

\frac{\partial Q}{\partial x} = 2x, \quad \frac{\partial P}{\partial y} = 2y $\frac{\partial Q}{\partial x} = 2x, \quad \frac{\partial P}{\partial y} = 2y$

Quindi, il rotore è:

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2y $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2y$

Ora calcoliamo l’integrale doppio su D$D$:

\iint_D (2x - 2y) \, dA $\iint_D (2x - 2y) \, dA$

Dove D$D$ è il quadrato con 0 \leq x \leq 1$0 \leq x \leq 1$ e 0 \leq y \leq 1$0 \leq y \leq 1$:

\iint_D (2x - 2y) \, dA = \int_0^1 \int_0^1 (2x - 2y) \, dy \, dx $\iint_D (2x - 2y) \, dA = \int_0^1 \int_0^1 (2x - 2y) \, dy \, dx$

Calcoliamo l’integrale interno:

\int_0^1 (2x - 2y) \, dy = [2xy - y^2]_0^1 = 2x - 1 $\int_0^1 (2x - 2y) \, dy = [2xy - y^2]_0^1 = 2x - 1$

Ora calcoliamo l’integrale esterno:

\int_0^1 (2x - 1) \, dx = [x^2 - x]_0^1 = 1 - 1 = 0 $\int_0^1 (2x - 1) \, dx = [x^2 - x]_0^1 = 1 - 1 = 0$

Quindi:

\oint_C (x^2 \, dy + y^2 \, dx) = 0 $\oint_C (x^2 \, dy + y^2 \, dx) = 0$

Esercizio 2

Verifica il Teorema di Green per il campo vettoriale \mathbf{F} = (y, x)$\mathbf{F} = (y, x)$ su un cerchio di raggio 1$1$ centrato nell’origine.

Soluzione:

Identifichiamo P = y$P = y$ e Q = x$Q = x$. Calcoliamo le derivate parziali:

\frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = 1 $\frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = 1$

Quindi, il rotore è:

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - 1 = 0 $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - 1 = 0$

Ora calcoliamo l’integrale doppio su D$D$ (il cerchio di raggio 1$1$):

\iint_D 0 \, dA = 0 $\iint_D 0 \, dA = 0$

Quindi, secondo il Teorema di Green, abbiamo:

\oint_C (y \, dx + x \, dy) = 0 $\oint_C (y \, dx + x \, dy) = 0$

Questo è coerente con il fatto che il rotore del campo vettoriale è zero, il che implica che non c’è flusso attraverso il contorno C$C$.

Esercizio 3

Calcola l’integrale di linea:

\oint_C (3x^2 \, dy + 2y^2 \, dx) $\oint_C (3x^2 \, dy + 2y^2 \, dx)$

dove C$C$ è il contorno del triangolo con vertici (0,0)$(0,0)$, (1,0)$(1,0)$, (1,1)$(1,1)$, (0,1)$(0,1)$.

Soluzione:

Identifichiamo P = 2y^2$P = 2y^2$ e Q = 3x^2$Q = 3x^2$. Calcoliamo le derivate parziali:

\frac{\partial Q}{\partial x} = 6x, \quad \frac{\partial P}{\partial y} = 4y $\frac{\partial Q}{\partial x} = 6x, \quad \frac{\partial P}{\partial y} = 4y$

Quindi, il rotore è:

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 6x - 4y $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 6x - 4y$

Ora calcoliamo l’integrale doppio su D$D$ (il triangolo con vertici (0,0)$(0,0)$, (1,0)$(1,0)$, (1,1)$(1,1)$):

\iint_D (6x - 4y) \, dA $\iint_D (6x - 4y) \, dA$

Per calcolare l’integrale, possiamo usare i limiti di integrazione per il triangolo:

\iint_D (6x - 4y) \, dA = \int_0^1 \int_0^x (6x - 4y) \, dy \, dx $\iint_D (6x - 4y) \, dA = \int_0^1 \int_0^x (6x - 4y) \, dy \, dx$

Calcoliamo l’integrale interno:

\int_0^x (6x - 4y) \, dy = [6xy - 2y^2]_0^x = 6x^2 - 2x^2 = 4x^2 $\int_0^x (6x - 4y) \, dy = [6xy - 2y^2]_0^x = 6x^2 - 2x^2 = 4x^2$

Ora calcoliamo l’integrale esterno:

\int_0^1 4x^2 \, dx = \left[ \frac{4}{3} x^3 \right]_0^1 = \frac{4}{3} $\int_0^1 4x^2 \, dx = \left[ \frac{4}{3} x^3 \right]_0^1 = \frac{4}{3}$

Quindi, secondo il Teorema di Green, abbiamo:

\oint_C (3x^2 \, dy + 2y^2 \, dx) = \frac{4}{3} $\oint_C (3x^2 \, dy + 2y^2 \, dx) = \frac{4}{3}$

English version

Green’s Formulas Exercises

Green’s formulas are fundamental tools in vector analysis and connect the integral of a function over a region of the plane with the integral of its derivative over a contour of that region. There are two main forms of Green’s formulas: the scalar field form and the vector field form.

Key Concepts

1. Green’s Theorem: If C$C$ is a closed boundary and D$D$ is the region bounded by C$C$, then for a vector field \mathbf{F} = (P, Q)$\mathbf{F} = (P, Q)$ with P$P$ and Q$Q$ continuous on D$D$ and with continuous partial derivatives, we have:

\oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA $\oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$

2. Geometric Interpretation: The integral on the left represents the work done by a vector field along the boundary C$C$, while the integral on the right represents the flux of the field’s rotor through the region D$D$.

3. Applications: Green’s formulas are used in physics and engineering, for example to calculate work, flux and in problems of electromagnetism.

Exercises

Exercise 1

Calculate the line integral:

\oint_C (x^2 \, dy + y^2 \, dx) $\oint_C (x^2 \, dy + y^2 \, dx)$

where C$C$ is the boundary of the square with vertices (0,0)$(0,0)$, (1,0)$(1,0)$, (1,1)$(1,1)$, (0,1)$(0,1)$.

Solution:

Let P = y^2$P = y^2$ and Q = x^2$Q = x^2$. Let’s calculate the partial derivatives:

\frac{\partial Q}{\partial x} = 2x, \quad \frac{\partial P}{\partial y} = 2y $\frac{\partial Q}{\partial x} = 2x, \quad \frac{\partial P}{\partial y} = 2y$

So, the curl is:

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2y $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2y$

Now let’s calculate the double integral on D$D$:

\iint_D (2x - 2y) \, dA $\iint_D (2x - 2y) \, dA$

Where D$D$ is the square with 0 \leq x \leq 1$0 \leq x \leq 1$ and 0 \leq y \leq 1$0 \leq y \leq 1$:

\iint_D (2x - 2y) \, dA = \int_0^1 \int_0^1 (2x - 2y) \, dy \, dx $\iint_D (2x - 2y) \, dA = \int_0^1 \int_0^1 (2x - 2y) \, dy \, dx$

Let’s calculate the internal integral:

\int_0^1 (2x - 2y) \, dy = [2xy - y^2]_0^1 = 2x - 1 $\int_0^1 (2x - 2y) \, dy = [2xy - y^2]_0^1 = 2x - 1$

Now let’s calculate the external integral:

\int_0^1 (2x - 1) \, dx = [x^2 - x]_0^1 = 1 - 1 = 0 $\int_0^1 (2x - 1) \, dx = [x^2 - x]_0^1 = 1 - 1 = 0$

So:

\oint_C (x^2 \, dy + y^2 \, dx) = 0 $\oint_C (x^2 \, dy + y^2 \, dx) = 0$

Exercise 2

Verify Green’s Theorem for the vector field \mathbf{F} = (y, x)$\mathbf{F} = (y, x)$ on a circle of radius 1$1$ centered at the origin.

Solution:

Let’s identify P = y$P = y$ and Q = x$Q = x$. Let’s compute the partial derivatives:

\frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = 1 $\frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = 1$

So, the curl is:

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - 1 = 0 $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - 1 = 0$

Now let’s compute the double integral on D$D$ (the circle of radius 1$1$):

\iint_D 0 \, dA = 0 $\iint_D 0 \, dA = 0$

So, by Green’s Theorem, we have:

\oint_C (y \, dx + x \, dy) = 0 $\oint_C (y \, dx + x \, dy) = 0$

This is consistent with the fact that the curl of the vector field is zero, which implies that there is no flux through the boundary C$C$.

Exercise 3

Calculate the line integral:

\oint_C (3x^2 \, dy + 2y^2 \, dx) $\oint_C (3x^2 \, dy + 2y^2 \, dx)$

where C$C$ is the contour of the triangle with vertices (0,0)$(0,0)$, (1,0)$(1,0)$, (1,1)$(1,1)$, (0,1)$(0,1)$.

Solution:

Let P = 2y^2$P = 2y^2$ and Q = 3x^2$Q = 3x^2$. Let’s calculate the partial derivatives:

\frac{\partial Q}{\partial x} = 6x, \quad \frac{\partial P}{\partial y} = 4y $\frac{\partial Q}{\partial x} = 6x, \quad \frac{\partial P}{\partial y} = 4y$

So, the curl is:

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 6x - 4y $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 6x - 4y$

Now let’s calculate the double integral on D$D$ (the triangle with vertices (0,0)$(0,0)$, (1,0)$(1,0)$, (1,1)$(1,1)$):

\iint_D (6x - 4y) \, dA $\iint_D (6x - 4y) \, dA$

To calculate the integral, we can use the limits of integration for the triangle:

\iint_D (6x - 4y) \, dA = \int_0^1 \int_0^x (6x - 4y) \, dy \, dx $\iint_D (6x - 4y) \, dA = \int_0^1 \int_0^x (6x - 4y) \, dy \, dx$

Let’s calculate the internal integral:

\int_0^x (6x - 4y) \, dy = [6xy - 2y^2]_0^x = 6x^2 - 2x^2 = 4x^2 $\int_0^x (6x - 4y) \, dy = [6xy - 2y^2]_0^x = 6x^2 - 2x^2 = 4x^2$

Now let’s calculate the external integral:

\int_0^1 4x^2 \, dx = \left[ \frac{4}{3} x^3 \right]_0^1 = \frac{4}{3} $\int_0^1 4x^2 \, dx = \left[ \frac{4}{3} x^3 \right]_0^1 = \frac{4}{3}$

So, according to Green’s Theorem, we have:

\oint_C (3x^2 \, dy + 2y^2 \, dx) = \frac{4}{3} $\oint_C (3x^2 \, dy + 2y^2 \, dx) = \frac{4}{3}$