1. ## Stokes' Theorem problem

I have a question about a problem involving Stokes' theorem

The question goes as follows:

Use Stokes' theorem to evaluate:

Double integral of the curl of F dot n ds

Where S is the part of the cylinder z = 1 - x^2 for 0<=x<=1, -2<=y<=2 and F(x, y, z) = xyi + yzj +xzk

So far I have the Curl of F is equal to -yi - zj - xk. I am now completely lost. Any help would be appreciated, thank you.

2. Originally Posted by allhalf425
I have a question about a problem involving Stokes' theorem

The question goes as follows:

Use Stokes' theorem to evaluate:

Double integral of the curl of F dot n ds

Where S is the part of the cylinder z = 1 - x^2 for 0<=x<=1, -2<=y<=2 and F(x, y, z) = xyi + yzj +xzk

So far I have the Curl of F is equal to -yi - zj - xk. I am now completely lost. Any help would be appreciated, thank you.
If you want to calculate $\displaystyle \iint \limits_S \nabla \times {\bf F} \cdot {\bf n}\, dS$ directly then your normal and dS are given by

$\displaystyle {\bf n} = \frac{<2x,0,1>}{\sqrt{1+4x^2}}$ and $\displaystyle dS = \sqrt{1 + 4x^2}dA$ so $\displaystyle \nabla \times {\bf F} \cdot {\bf n}\, dS = (- 2xy - x)\,dA$

and

$\displaystyle \int_0^1 \int_{-2}^2 \left(-2xy - x\right)\, dy\,dx = -2$

To use Stokes Thm then you want to evaluate

$\displaystyle \int \limits_C xydx + yz dy + xz dz$

There are four parts to the boundary of the surface (see the attached picture)

$\displaystyle C_1: x = 1, z = 0, y: -2 \to 2$ so $\displaystyle \int \limits_{C_1} xydx + yz dy + xz dz = 0$

$\displaystyle C_2: y = 2, \;\; \text{along}\; z = 1-x^2 \; \text{from}\; x = 1 \; \text{to}\; 0$ so $\displaystyle \int \limits_{C_2} xydx + yz dy + xz dz = \int_1^0 2 x\,dx + x (1-x^2)\,(-2x)\,dx = - \frac{11}{15}$

$\displaystyle C_3: x = 0, z = 1,\; y: 2 \to -2$ so $\displaystyle \int \limits_{C_3} xydx + yz dy + xz dz = 0$

$\displaystyle C_4: y = -2\;\; \text{along}\; z = 1-x^2 \; \text{from}\; x = 0 \; \text{to}\;1,$ so $\displaystyle \int \limits_{C_4} xydx + yz dy + xz dz = \int_0^1 -2x\,dx + x (1-x^2)\,(-2x)\,dx = - \frac{19}{15}$

giving that

$\displaystyle \int \limits_{C} xydx + yz dy + xz dz = 0 - \frac{11}{15} + 0 - \frac{19}{15} = -2$

3. Thank you!