Question

Verify that the line intergral and the surface integral of Stokes Theorem are equal for the following vector field, surface S and closed curve C. Assume that C has counterclockwise orientation and S has a consistent orientation.

F= < x,y,z>; S is the paraboloid z = 13 - x^2 - y^2, for 0 less than or equal z less than or equal 13 and C is the circle x^2 + y^2 = 13 in the xy plane.

Evaluate both integrals please showing steps of each

Answer 1

See steps below

Explanation:

We will verify Stokes' theorem for the vector field

F(x,y,z) = (x,y,z)

and the surface bounded by the paraboloid

`\large z = 13 - x^2 - y^2`

and the plane z = 0

(See picture)

Specifically, we must verify that

`\large \iint_(S)(curl\;F)\cdot nd\sigma=\int_(C)F\cdot dC`

First we compute the integral over the boundary curve, which is the circle C.

Now, we see that the force field F(x,y,z) is conservative since it is the gradient of the scalar function

`\large f(x,y,z)=(1)/(2)(x^2+y^2+z^2)`

That is,

`\large F=\triangledown f=((\partial f)/(\partial x),(\partial f)/(\partial y),(\partial f)/(\partial z))`

Hence the curve integral along any closed path equals 0

That is to say ,

`\large \int_(C)F\cdot dC=0`

On the other hand

`\large curl\;F=\begin{vmatrix}\hat i&\hat j&\hat k\\(\partial)/(\partial x)&(\partial)/(\partial y)&(\partial)/(\partial z)\\x&y&z\end{vmatrix}=((\partial z)/(\partial y)-(\partial y)/(\partial z),(\partial x)/(\partial z)-(\partial z)/(\partial x),(\partial y)/(\partial x)-(\partial x)/(\partial y))=(0,0,0)`

Hence

`\large \iint_(S)(curl\;F)\cdot nd\sigma=0`

and Stokes' theorem holds.