Find the Surface integral of a Paraboloid using Stoke's Theorem

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Homework Help Overview

The problem involves finding the surface integral of a paraboloid defined by the equation z = 4 - x^2 - y^2, specifically the portion above the plane z = 0. The vector field under consideration is given as F = < z-y, x+z, -e^{xyz}cos y >, and the task is to apply Stoke's Theorem to compute the integral of the curl of F over the surface S.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss parameterizing the boundary curve C and calculating the line integral of F along C. There are attempts to verify calculations related to the derivatives and the evaluation of the integral. Some participants also suggest alternative methods to check the results, such as using a simpler surface with the same boundary.

Discussion Status

The discussion is ongoing, with participants providing feedback on calculations and suggesting alternative verification methods. There is acknowledgment of errors in the initial calculations, and adjustments are being made. Multiple approaches to the problem are being explored without a definitive consensus on the final outcome.

Contextual Notes

Participants are working within the constraints of applying Stoke's Theorem and are considering the implications of the surface and boundary conditions. There is a focus on ensuring the correctness of the parameterization and the evaluation of integrals.

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Homework Statement


Let S be the portion of the paraboloid ##z = 4 - x^2 - y^2 ## that lies above the plane ##z = 0## and let ##\vec F = < z-y, x+z, -e^{ xyz }cos y >##. Use Stoke's Theorem to find the surface integral ##\iint_S (\nabla × \vec F) ⋅ \vec n \,dS##.

Homework Equations


##\iint_S (\nabla × \vec F) ⋅ \, d\vec S = \int_C \vec F ⋅\,d\vec r ##

The Attempt at a Solution


Parameterize C as ##\vec r(t) = 2<\cos t, \sin t, 0> , 0≤t≤2π ##

##\int_C \vec F ⋅\,d\vec r = \int_0^{2π} \vec F (\vec r(t))⋅\vec r'(t) \,dt##

## \vec r' (t) = 2<\sin t, -\cos t, 0> ##

## \vec F (\vec r (t)) = <-\sin t , \cos t, -\cos (\sin t)> ##

## 4\int_0^{2π} <-\sin t , \cos t, -\cos (\sin t)>⋅<\sin t, -\cos t, 0> \, dt = 4\int_0^{2π} (-\sin ^2 t -\cos ^2 t) \, dt = -4 \int_0^{2π} \, dt = -8π ##

Just looking for someone to verify this was done correctly, it's been almost a year since I last took calculus and I don't remember if there's any way to check my work.
 
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Close. You made a sign error when calculating ##\vec{r}'##, and your expression for ##\vec{F}(\vec{r}(t))## is missing factors of 2 here and there.
 
that's embarrassing, thank you.
## \vec r' (t) = <-2\sin t, 2\cos t, 0> ##

## \vec F (\vec r (t)) = <-2\sin t , 2\cos t, -\cos (2\sin t)> ##

## \int_0^{2π} <-2\sin t , 2\cos t, -\cos (2\sin t)>⋅<-2\sin t, 2\cos t, 0> \, dt = \int_0^{2π} (4\sin ^2 t +4\cos ^2 t) \, dt = 4 \int_0^{2π} \, dt = 8π ##
 
Another way you can check your result is to do the surface integral over a simpler surface with the same boundary. In this case, try integrating over the circle of radius 2 with its center at the origin and lies in the xy plane. Since the normal points in the z direction, you only need to calculate the z-component of the curl.
 
vela said:
Another way you can check your result is to do the surface integral over a simpler surface with the same boundary. In this case, try integrating over the circle of radius 2 with its center at the origin and lies in the xy plane. Since the normal points in the z direction, you only need to calculate the z-component of the curl.

## \iint_S (∇×\vec F)⋅\vec n \,dS = \iint_S <P, Q,\frac ∂ {∂x} (x+z) - \frac ∂ {∂y} (z-y) >⋅<0, 0, 1> \, dS = \iint_S 2\,dS = 8π ##

That makes a lot of sense, thanks for the tip.
 

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