Stoke's Theorem

**Silverflow** · May 9th 2010, 01:10 AM

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

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

Then, I can calculate $\underline{r}'(t)$ & $\underline{f}(\underline{r}(t))$ . 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.

Thank you in advance!

**Jester** · May 9th 2010, 07:32 AM

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, $S=x^2+y^2+2(z-1)^2=6, z\geq0$ and the vector field $\underline{F}=(xz-y^3cos(z))\underline{i}+(x^3e^z)\underline{j}+(xyz e^{x^2+y^2+z^2})\underline{k}$ . Use stokes formula to evaluate the integral $\int\int_{S}curl\underline{F}.\underline{S}$ .

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

Then, I can calculate $\underline{r}'(t)$ & $\underline{f}(\underline{r}(t))$ . 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.

Thank you in advance!

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

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

**AllanCuz** · May 9th 2010, 08:31 AM

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, $S=x^2+y^2+2(z-1)^2=6, z\geq0$ and the vector field $\underline{F}=(xz-y^3cos(z))\underline{i}+(x^3e^z)\underline{j}+(xyz e^{x^2+y^2+z^2})\underline{k}$ . Use stokes formula to evaluate the integral $\int\int_{S}curl\underline{F}.\underline{S}$ .

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

Then, I can calculate $\underline{r}'(t)$ & $\underline{f}(\underline{r}(t))$ . 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.

Thank you in advance!

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

Our surface $x^2 + y^2 + 2(z-1)^2 = 6$ 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 $x^2 + y^2 = 4$ but since our condition was $z \ge 0$ our domain becomes $x^2 + y^2 \le 4$

Noting that,

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

We get,

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

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

Thus,

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

Transforming to polar co-ordinates

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

**Silverflow** · May 12th 2010, 09:41 PM

Thanks guys!

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