Stokes Theorem - double integral

**mybrohshi5** · December 8th 2010, 10:37 AM

Use Stokes’ Theorem to evaluate $\int_c^.F \cdot dr$

C: is the boundary of the portion of $z = x^2 + y^2$ below z = 4, oriented downward,

$F = <x^2, y^4 - x, z^2sin z>$

Can someone please check over my work

thanks!

First I found N --> $N = \frac{<2x, 2y, -1>}{\sqrt{4x^2 + 4y^2 + 1^2}}$

Then i found $\bigtriangledown \times F = <0, 0, -1>$

from using the formula $\bigtriangledown \times F = <F_{3y} - F_{2z}, -F_{3x} + F_{1z}, F_{2x} - F_{1y}>$

Now $\bigtriangledown \times F \cdot N = \frac{1}{\sqrt{4x^2 + 4y^2 + 1^2}}$

Then $dS = {\sqrt{4x^2 + 4y^2 + 1}}dA$

So putting it all together....

$\int\int_R^. \frac{1}{\sqrt{4x^2 + 4y^2 + 1^2}} \cdot \sqrt{4x^2 + 4y^2 + 1}$

The square roots cancel and reduce to....

$\int\int_R^. 1 dA$

This double integral of 1 dA is equal to the Area of the Region

And using $z = x^2 + y^2$ and $z = 4$ we come up with $r^2 = 4 .... r = 2$

$A(R) = \pi r^2 = \pi(2^2) = 4\pi$

Does $4\pi$ look correct for the answer?

Thanks for looking

**HallsofIvy** · December 8th 2010, 11:57 AM

I've always disliked the notation $\vec{n}dS$ because used that unthinkingly results in doing two square roots which, just as you see here, cancel!

Instead think of $d\vec{S}$ as the "vector differential of surface area" and calculate it directly. We can write the surface $z= x^2+ y^2$ in a vector equation with x and y as parameter: $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (x^2+ y^2)\vec{k}$ . The derivatives with respect to x and y give two tangent vectors, $\vec{i}+ 2x\vec{k}$ and $\vec{j}+ 2y\vec{k}$ . The cross product of those gives the "vector differential of surface area": $2x\vec{i}+ 2y\vec{j}- \vec{k}$ where I have chosen the order of multiplication to get a negative value for $\vec{k}$ , "oriented downward".

You are correct that $\nabla\times \vec{F}= -\vec{k}$ so that $\nabla\vec{F}\cdot\vec{n} dS= \nabla\vec{F}\cdot d\vec{S}$ [tex]= (0(2x)+ 0(2y)- 1(-1))dxdy= dxdy.

That gives $4\pi$ , not $-4\pi$ .

Your normal vector, $\frac{-2x\vec{i}-2y\vec{j}+ \vec{k}}{\sqrt{4x^2+ 4y^2+ 1}}$ was oriented upward, not downward.

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