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

In summary, Stoke's Theorem can be used to find the surface integral over the paraboloid ##z = 4 - x^2 - y^2## that lies above the plane ##z = 0##.
  • #1
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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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  • #2
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.
 
  • #3
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 ##
 
  • #4
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.
 
  • #5
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.
 

1. What is Stoke's Theorem?

Stoke's Theorem is a mathematical theorem that relates the surface integral of a vector field over a surface to the line integral of the same vector field along the boundary of the surface. It is a generalization of Green's Theorem in two dimensions.

2. What is a Paraboloid?

A paraboloid is a three-dimensional surface that is shaped like a bowl or a cup. It can be formed by rotating a parabola around its axis of symmetry. Examples of paraboloids include satellite dishes and some types of mirrors.

3. How do you find the surface integral of a Paraboloid using Stoke's Theorem?

To find the surface integral of a paraboloid using Stoke's Theorem, you need to first parameterize the surface and find the normal vector to the surface. Then, you need to calculate the curl of the vector field and set it equal to the surface integral. Finally, evaluate the line integral along the boundary of the surface to find the surface integral.

4. Why is Stoke's Theorem important in mathematics?

Stoke's Theorem is important in mathematics because it provides a powerful tool for calculating surface integrals. It allows for the conversion of a difficult surface integral into a simpler line integral, making calculations easier and more efficient. This theorem is also used in various fields such as physics, engineering, and fluid dynamics.

5. Are there any limitations to using Stoke's Theorem?

Yes, there are some limitations to using Stoke's Theorem. It can only be applied to surfaces that are smooth and have a well-defined boundary. Additionally, the vector field must be continuously differentiable in the region enclosed by the surface. If these conditions are not met, Stoke's Theorem cannot be used to find the surface integral of a paraboloid.

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