1. ## Surface Integrals

Help!!!

Consider the surface given by the equation $2{x}^{2} + 2{y}^{2} + z = 6$, oriented upward, and let S be the part of this surface that is above the plane z = 4. Given the vector field:
$\vec{F} = x\vec{i} + y\vec{j} + 3\vec{k}$ ,
evaluate the surface integral
$\iint \vec{F} \cdot d\vec{S}$ .

2. Originally Posted by raven17
Help!!!

Consider the surface given by the equation $2{x}^{2} + 2{y}^{2} + z = 6$, oriented upward, and let S be the part of this surface that is above the plane z = 4. Given the vector field:
$\vec{F} = x\vec{i} + y\vec{j} + 3\vec{k}$ ,
evaluate the surface integral
$\iint \vec{F} \cdot d\vec{S}$ .
$\iint_S \mathbf{F}\cdot d\mathbf{S}=\iint_D\mathbf{F}\cdot \left(-\frac{\partial z}{\partial x}\mathbf{i}-\frac{\partial z}{\partial y}\mathbf{j} +\mathbf{k} \right)dA$

The area of integration can be found by the equation.

$2x^2+2y^2+4=6 \iff x^2+y^2+1$ so the domain of integration is the unit circle.

$\iint_S \mathbf{F}\cdot d\mathbf{S}=\iint_D\left( x\mathbf{i}+y\mathbf{j}+3\mathbf{k}\right)\cdot \left(4x\mathbf{i}+4y\mathbf{j} +\mathbf{k} \right)dA=$

$\iint 4(x^2+y^2)+3dA$

Now change to polar coordinates to finish.

3. Thanks sooo much! Just a question though... where did you get the "(4xi + 4yj + k)" from?

4. Originally Posted by raven17
Thanks sooo much! Just a question though... where did you get the "(4xi + 4yj + k)" from?
$-\frac{\partial z}{\partial x}\mathbf{i}-\frac{\partial z}{\partial y}\mathbf{j} +\mathbf{k}$

$z=6-2x^2-2y^2$

Now just take the partials and plug in.

5. Originally Posted by raven17
Thanks once again! I am sorry for asking again but how did you get -dz/dx (i) - dz/dy (j) + k?
If you solve the equation for $z=f(x,y)$

we can parametrize the surface as follows

$\mathbf{r}(x,y)=x\mathbf{i}+y\mathbf{j}+f(x,y) \mathbf{k}$

So $\mathbf{r}_x=\mathbf{i}+\frac{\partial f}{\partial x}\mathbf{k}$

and $\mathbf{r}_y=\mathbf{j}+\frac{\partial f}{\partial y}\mathbf{k}$

Now if you cross these two vectors you will get the formula