Math / Calculus

QuestionVerify the conclusion of Green's Theorem by evaluating both sides of each of the two forms of Green's Theorem for the field $F=6 x i-2 y j$ . Take the domains of integration in each case to be the disk $R: x^{2}+y^{2} \leq a^{2}$ and its bounding circle $C$ : $r=(a \cos t) i+(a \sin t) j, 0 \leq t \leq 2 \pi$ . i Click here for the two forms of Green's Theorem.
The flux is $\square$ (Type an exact answer, using $\pi$ as needed.)

Studdy Solution

STEP 1

Assumptions
1. The vector field is given by $F = 6x \, \mathbf{i} - 2y \, \mathbf{j}$ .
2. The domain of integration is the disk $R: x^2 + y^2 \leq a^2$ .
3. The bounding circle $C$ is parameterized by $r(t) = a \cos t \, \mathbf{i} + a \sin t \, \mathbf{j}$ , where $0 \leq t \leq 2\pi$ .
4. We will use Green's Theorem in both its circulation form and flux form.

STEP 2

Green's Theorem (Circulation Form) states that for a vector field $F = P \, \mathbf{i} + Q \, \mathbf{j}$ , the line integral around a simple closed curve $C$ is equal to the double integral over the region $R$ it encloses:
$\oint_C (P \, dx + Q \, dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA$

STEP 3

Identify $P$ and $Q$ from the vector field $F = 6x \, \mathbf{i} - 2y \, \mathbf{j}$ :
$P = 6x, \quad Q = -2y$

STEP 4

Calculate the partial derivatives needed for Green's Theorem (Circulation Form):
$\frac{\partial Q}{\partial x} = \frac{\partial (-2y)}{\partial x} = 0$
$\frac{\partial P}{\partial y} = \frac{\partial (6x)}{\partial y} = 0$

STEP 5

Substitute the derivatives into the double integral for the circulation form:
$\iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_R (0 - 0) \, dA = 0$

STEP 6

Green's Theorem (Flux Form) states that for a vector field $F = P \, \mathbf{i} + Q \, \mathbf{j}$ , the line integral around a simple closed curve $C$ is equal to the double integral over the region $R$ it encloses:
$\oint_C (Q \, dx - P \, dy) = \iint_R \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \, dA$

STEP 7

Calculate the partial derivatives needed for Green's Theorem (Flux Form):
$\frac{\partial P}{\partial x} = \frac{\partial (6x)}{\partial x} = 6$
$\frac{\partial Q}{\partial y} = \frac{\partial (-2y)}{\partial y} = -2$

STEP 8

Substitute the derivatives into the double integral for the flux form:
$\iint_R \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \, dA = \iint_R (6 - 2) \, dA = \iint_R 4 \, dA$

STEP 9

Convert the double integral over the disk $R$ to polar coordinates. The region $R$ is $0 \leq r \leq a$ and $0 \leq \theta \leq 2\pi$ .

STEP 10

Set up the double integral in polar coordinates:
$\iint_R 4 \, dA = \int_0^{2\pi} \int_0^a 4 \, r \, dr \, d\theta$

STEP 11

Evaluate the inner integral with respect to $r$ :
$\int_0^a 4r \, dr = 4 \left[ \frac{r^2}{2} \right]_0^a = 4 \cdot \frac{a^2}{2} = 2a^2$

STEP 12

Evaluate the outer integral with respect to $\theta$ :
$\int_0^{2\pi} 2a^2 \, d\theta = 2a^2 \left[ \theta \right]_0^{2\pi} = 2a^2 \cdot 2\pi = 4\pi a^2$

STEP 13

The flux through the region $R$ is $4\pi a^2$ .
The flux is $4\pi a^2$ .

Was this helpful?

Studdy Solution

STEP 1

STEP 2

STEP 3

Identify $P$ and $Q$ from the vector field $F = 6x \, \mathbf{i} - 2y \, \mathbf{j}$ :
$P = 6x, \quad Q = -2y$

STEP 4

Calculate the partial derivatives needed for Green's Theorem (Circulation Form):
$\frac{\partial Q}{\partial x} = \frac{\partial (-2y)}{\partial x} = 0$
$\frac{\partial P}{\partial y} = \frac{\partial (6x)}{\partial y} = 0$

STEP 5

Substitute the derivatives into the double integral for the circulation form:
$\iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_R (0 - 0) \, dA = 0$

STEP 6

STEP 7

Calculate the partial derivatives needed for Green's Theorem (Flux Form):
$\frac{\partial P}{\partial x} = \frac{\partial (6x)}{\partial x} = 6$
$\frac{\partial Q}{\partial y} = \frac{\partial (-2y)}{\partial y} = -2$

STEP 8

Substitute the derivatives into the double integral for the flux form:
$\iint_R \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \, dA = \iint_R (6 - 2) \, dA = \iint_R 4 \, dA$

STEP 9

Convert the double integral over the disk $R$ to polar coordinates. The region $R$ is $0 \leq r \leq a$ and $0 \leq \theta \leq 2\pi$ .

STEP 10

Set up the double integral in polar coordinates:
$\iint_R 4 \, dA = \int_0^{2\pi} \int_0^a 4 \, r \, dr \, d\theta$

STEP 11

Evaluate the inner integral with respect to $r$ :
$\int_0^a 4r \, dr = 4 \left[ \frac{r^2}{2} \right]_0^a = 4 \cdot \frac{a^2}{2} = 2a^2$

STEP 12

Evaluate the outer integral with respect to $\theta$ :
$\int_0^{2\pi} 2a^2 \, d\theta = 2a^2 \left[ \theta \right]_0^{2\pi} = 2a^2 \cdot 2\pi = 4\pi a^2$

STEP 13

The flux through the region $R$ is $4\pi a^2$ .
The flux is $4\pi a^2$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

STEP 3

Identify P P P and Q Q Q from the vector field F=6x i−2y j F = 6x \, \mathbf{i} - 2y \, \mathbf{j} F=6xi−2yj:P=6x,Q=−2yP = 6x, \quad Q = -2yP=6x,Q=−2y

STEP 4

STEP 5

Substitute the derivatives into the double integral for the circulation form:∬R(∂Q∂x−∂P∂y) dA=∬R(0−0) dA=0\iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_R (0 - 0) \, dA = 0∬R​(∂x∂Q​−∂y∂P​)dA=∬R​(0−0)dA=0

STEP 6

STEP 7

STEP 8

STEP 9

Convert the double integral over the disk R R R to polar coordinates. The region R R R is 0≤r≤a 0 \leq r \leq a 0≤r≤a and 0≤θ≤2π 0 \leq \theta \leq 2\pi 0≤θ≤2π.

STEP 10

Set up the double integral in polar coordinates:∬R4 dA=∫02π∫0a4 r dr dθ\iint_R 4 \, dA = \int_0^{2\pi} \int_0^a 4 \, r \, dr \, d\theta∬R​4dA=∫02π​∫0a​4rdrdθ

STEP 11

Evaluate the inner integral with respect to r r r:∫0a4r dr=4[r22]0a=4⋅a22=2a2\int_0^a 4r \, dr = 4 \left[ \frac{r^2}{2} \right]_0^a = 4 \cdot \frac{a^2}{2} = 2a^2∫0a​4rdr=4[2r2​]0a​=4⋅2a2​=2a2

STEP 12

Evaluate the outer integral with respect to θ \theta θ:∫02π2a2 dθ=2a2[θ]02π=2a2⋅2π=4πa2\int_0^{2\pi} 2a^2 \, d\theta = 2a^2 \left[ \theta \right]_0^{2\pi} = 2a^2 \cdot 2\pi = 4\pi a^2∫02π​2a2dθ=2a2[θ]02π​=2a2⋅2π=4πa2

STEP 13

The flux through the region R R R is 4πa2 4\pi a^2 4πa2.The flux is 4πa2 4\pi a^2 4πa2.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Identify $P$ and $Q$ from the vector field $F = 6x \, \mathbf{i} - 2y \, \mathbf{j}$ :
$P = 6x, \quad Q = -2y$

Substitute the derivatives into the double integral for the circulation form:
$\iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_R (0 - 0) \, dA = 0$

Convert the double integral over the disk $R$ to polar coordinates. The region $R$ is $0 \leq r \leq a$ and $0 \leq \theta \leq 2\pi$ .

Set up the double integral in polar coordinates:
$\iint_R 4 \, dA = \int_0^{2\pi} \int_0^a 4 \, r \, dr \, d\theta$

Evaluate the inner integral with respect to $r$ :
$\int_0^a 4r \, dr = 4 \left[ \frac{r^2}{2} \right]_0^a = 4 \cdot \frac{a^2}{2} = 2a^2$

Evaluate the outer integral with respect to $\theta$ :
$\int_0^{2\pi} 2a^2 \, d\theta = 2a^2 \left[ \theta \right]_0^{2\pi} = 2a^2 \cdot 2\pi = 4\pi a^2$

The flux through the region $R$ is $4\pi a^2$ .
The flux is $4\pi a^2$ .