Surface Integrals of Flux

althaemenes · #1 $Old$ April 11th, 2009, 02:07 PM

Hi, am kinda having trouble figuring out the following problem? any help would be highly appreciated...

Evaluate

if

and the surface S is given by

for

. (Take S to have upward orientation.)

TheEmptySet · #2 $Old$ April 11th, 2009, 02:16 PM

Quote:

Originally Posted by althaemenes $View Post$

Hi, am kinda having trouble figuring out the following problem? any help would be highly appreciated...

Evaluate

if

and the surface S is given by

for

. (Take S to have upward orientation.)

if $\vec F(x,y,z)=P \vec i +Q \vec j + R \vec k$ and $z=g(x,y)$ then

$\iint_S \vec F \cdot d\vec S=\iint_D\left( -P \frac{\partial z}{\partial x}-Q \frac{\partial z}{\partial y}+R\right)dA$

This gives us

$\int \int (2xy\sin(8y)+4xy^2(8x\cos(8y))-2y(x\sin(8y)))dA$

althaemenes · #3 $Old$ April 11th, 2009, 03:12 PM

Quote:

Originally Posted by TheEmptySet $View Post$

if $\vec F(x,y,z)=P \vec i +Q \vec j + R \vec k$ and $z=g(x,y)$ then

$\iint_S \vec F \cdot d\vec S=\iint_D\left( -P \frac{\partial z}{\partial x}-Q \frac{\partial z}{\partial y}+R\right)dA$

This gives us

$\int \int (2xy\sin(8y)+4xy^2(8x\cos(8y))-2y(x\sin(8y)))dA$

okay, so I integrate

$\int \int (2xy\sin(8y)+4xy^2(8x\cos(8y))-2y(x\sin(8y)))dA$ [/quote]

with x = 0 to pi/16 and y = 0 to pi/16

and keep getting:

7.37119946071407E-05

I was wondering if I am using the right limits:

Please help..