1. ## Stoke's Theorem

Hi all,
I've been given a question based on Stoke's Theorem, and the question is as follows:
Given the surface, $\displaystyle S=x^2+y^2+2(z-1)^2=6, z\geq0$ and the vector field $\displaystyle \underline{F}=(xz-y^3cos(z))\underline{i}+(x^3e^z)\underline{j}+(xyz e^{x^2+y^2+z^2})\underline{k}$. Use stokes formula to evaluate the integral$\displaystyle \int\int_{S}curl\underline{F}.\underline{S}$.

Using Stokes Theorem: $\displaystyle \int\int_{S}curl\underline{F}.d\underline{S}=\int_ {C}\underline{F}.d\underline{r}$.
So, I need to find $\displaystyle d\underline{r}$. I can find a vector equation for S:
$\displaystyle \underline{r}(t)=2cos(t)\underline{i}+2sin(t)\unde rline{j}+2\underline{z}$.

Then, I can calculate $\displaystyle \underline{r}'(t)$ & $\displaystyle \underline{f}(\underline{r}(t))$. However, the latter leaves me with something that looks hideously ugly. I was just wondering if any could go over this, see whether I've done the right thing.

2. Originally Posted by Silverflow
Hi all,
I've been given a question based on Stoke's Theorem, and the question is as follows:
Given the surface, $\displaystyle S=x^2+y^2+2(z-1)^2=6, z\geq0$ and the vector field $\displaystyle \underline{F}=(xz-y^3cos(z))\underline{i}+(x^3e^z)\underline{j}+(xyz e^{x^2+y^2+z^2})\underline{k}$. Use stokes formula to evaluate the integral$\displaystyle \int\int_{S}curl\underline{F}.\underline{S}$.

Using Stokes Theorem: $\displaystyle \int\int_{S}curl\underline{F}.d\underline{S}=\int_ {C}\underline{F}.d\underline{r}$.
So, I need to find $\displaystyle d\underline{r}$. I can find a vector equation for S:
$\displaystyle \underline{r}(t)=2cos(t)\underline{i}+2sin(t)\unde rline{j}+2\underline{z}$.

Then, I can calculate $\displaystyle \underline{r}'(t)$ & $\displaystyle \underline{f}(\underline{r}(t))$. However, the latter leaves me with something that looks hideously ugly. I was just wondering if any could go over this, see whether I've done the right thing.

In your vector function for the curve of intersection of your surface with $\displaystyle z = 0$, why are you setting $\displaystyle z = 2$? i.e.

$\displaystyle \vec{r}(t)=2cos(t)\vec{i}+2sin(t)\vec{j}+2\vec{z}$.

3. Originally Posted by Silverflow
Hi all,
I've been given a question based on Stoke's Theorem, and the question is as follows:
Given the surface, $\displaystyle S=x^2+y^2+2(z-1)^2=6, z\geq0$ and the vector field $\displaystyle \underline{F}=(xz-y^3cos(z))\underline{i}+(x^3e^z)\underline{j}+(xyz e^{x^2+y^2+z^2})\underline{k}$. Use stokes formula to evaluate the integral$\displaystyle \int\int_{S}curl\underline{F}.\underline{S}$.

Using Stokes Theorem: $\displaystyle \int\int_{S}curl\underline{F}.d\underline{S}=\int_ {C}\underline{F}.d\underline{r}$.
So, I need to find $\displaystyle d\underline{r}$. I can find a vector equation for S:
$\displaystyle \underline{r}(t)=2cos(t)\underline{i}+2sin(t)\unde rline{j}+2\underline{z}$.

Then, I can calculate $\displaystyle \underline{r}'(t)$ & $\displaystyle \underline{f}(\underline{r}(t))$. However, the latter leaves me with something that looks hideously ugly. I was just wondering if any could go over this, see whether I've done the right thing.

Are you using "Calculus: A complete Course" by Robert A Adams? This is a question directly from the text.

Our surface $\displaystyle x^2 + y^2 + 2(z-1)^2 = 6$ is an ellipsoid of revolution about the z-axis, centred at (0,0,1). Setting z=0 we find our XY domain and it's the circle $\displaystyle x^2 + y^2 = 4$ but since our condition was $\displaystyle z \ge 0$ our domain becomes $\displaystyle x^2 + y^2 \le 4$

Noting that,

$\displaystyle \iint_S Curl F \cdot \hat N dS = \int_C F \cdot dr = \iint_D Curl F \cdot k dA$

We get,

$\displaystyle curl F \cdot k = [ \frac{ \partial }{ \partial x } x^3e^z - \frac{ \partial }{ \partial y} (xz-y^3 cosz) ]_{|z=0} = 3(x^2 + y^2)$

We are evaluating at z=0 because we transformed our stokes equation into 2D which is only in the XY plain!

Thus,

$\displaystyle \iint_D Curl F \cdot k dA = \iint_{x^2+ y^2 \le 4} 3(x^2 + y^2) dxdy$

Transforming to polar co-ordinates

$\displaystyle \iint_{x^2 + y^2 \le 4} 3(x^2 + y^2) dxdy = \int_0^{2 \pi } d \pi \int_0^2 3 r^2 r dr = 24 \pi$

4. Thanks guys!