Application of Stokes' Theorem

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Homework Help Overview

The discussion revolves around the application of Stokes' Theorem to evaluate a line integral defined over a curve C, which is an ellipse in three-dimensional space. The problem involves understanding the vector field and its curl, as well as the parametrization of the surface bounded by the curve.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the parametrization of the surface and the choice of normal vector. There is confusion regarding the setup for applying Stokes' Theorem and how to arrange the double integral. Some participants suggest that the integrand simplifies to a constant due to the nature of the curl and the normal vector.

Discussion Status

Several participants have provided insights into the nature of the normal vector and its implications for the integral. There is a recognition that the area of the ellipse and the constant value of the integrand are key to simplifying the problem. However, there is no explicit consensus on the final approach, as different interpretations of the normal vector and area calculations are being explored.

Contextual Notes

Participants are working within the constraints of applying Stokes' Theorem and are navigating the complexities of the geometry involved, including the relationship between the ellipse and the plane defined by the problem.

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Homework Statement


Solve the following question by using Stokes' Theorem.

(Line integral on C) 2zdx + xdy + 3ydz = ? where C is the ellipse formed by
z = x, x^2 + y^2 = 4.

Homework Equations





The Attempt at a Solution



We have the vector A=(2z,x,3y) which is cont. differentiable and
curl(A) = (3,2,1). Now we have to parametrize the smooth surface S whose boundary is C and obtain a normal. I'm confused here how to choose the parameters. I solved it using the usual way and got -8*Pi. Can you help me to arrange the double integral in order to solve it by using Stokes' Thm.? Thanks.
 
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The normal to the ellipse is the same as the normal to the plane x=z. So it's a constant vector. Your curl is also a constant vector. So the integrand of the double integral is just a constant. This means you only need to know the area of the ellipse and the value of the integrand. No need to actually integrate anything.
 
Okay, i got it. The normal to the plane is n=(1,0,-1) but we need an outward normal so we take n=(-1,0,1). We get (3,2,1)(-1,0,1) = -2. Projection onto xy-plane is a circle whose area is 4*Pi and multiplying it by -2 we obtain 8*Pi.
 
I guess I would normalize the normal by dividing by sqrt(2) and then multiply by the real area of the ellipse which is sqrt(2) times the area of the circle. But of course, you get the same thing.
 

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