Chapter 6: Q30MP (page 338)

$\int_{C} (x^{2} - y) dx + (x + y^{3}) dy$ : where $C$ is the parallelogram with vertices $(0, 0), (2, 0), (1, 1), (3, 1)$ .

Short Answer

Expert verified

The solution is $∮_{c} (x^{2} - y) dx + (x + y^{3}) dy = 4$

Step by step solution

Given Information.

The given information is, $∮_{C} (x^{2} - y) dx + (x + y^{3}) dy$

Definition of Green’s Theorem.

Green's theorem connects a line integral around a simple closed curve to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

Find the solution.

$∮_{c} (x^{2} - y) dx + (x + y^{3}) dy$

Use Green’s Theorem role="math" $\iint_{A} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dxdy = ∮_{\partial A} Pdx + Qdy$ , where is the boundary of the area $A$ .

It can be seen that

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

$= - 1$

$Q = x + y^{3} \to \frac{\partial Q}{\partial x}$

$= 1$

Since the contour is the parallelogram with the given vertices, the area $A$ is then the area of the paralleologram.

$∮_{c} (x^{2} - y) dx + (x + y^{3}) dy = 2 (2) (1)$

$= 4$

$\int V \times dr = 5 {(\sqrt{(1 - 0)^{2} + (0 - 1)^{2}})}^{2}$

= 10

Hence, the solution is $∮_{c} (x^{2} - y) dx + (x + y^{3}) dy = 4$ .

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$\int_{C} (x^{2} - y) dx + (x + y^{3}) dy$ : where $C$ is the parallelogram with vertices $(0, 0), (2, 0), (1, 1), (3, 1)$ .

Short Answer

Step by step solution

Given Information.

Definition of Green’s Theorem.

Find the solution.

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∫C (x2−y)dx+(x+y3)dy: where Cis the parallelogram with vertices (0,0),(2,0),(1,1),(3,1).

Short Answer

Step by step solution

Given Information.

Definition of Green’s Theorem.

Find the solution.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Particle Model of Matter

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Study anywhere. Anytime. Across all devices.

$\int_{C} (x^{2} - y) dx + (x + y^{3}) dy$ : where $C$ is the parallelogram with vertices $(0, 0), (2, 0), (1, 1), (3, 1)$ .