2. Originally Posted by wik_chick88
I can't read it! Can you try to type out the question, using LaTeX?

--Chris

3. Originally Posted by Chris L T521
I can't read it! Can you try to type out the question, using LaTeX?

--Chris
haha well i can almost type it out using latex BUT i cant do the double integral symbol with a circle attaching them. the symbol also has an S down the bottom...it sort of looks like \oint but with 2 integral signs...can you help??

4. maybe... $\int\int\limits_{S}$
???? (dont know whether this will work or not...

5. Originally Posted by wik_chick88
maybe... $\int\int\limits_{S}$
???? (dont know whether this will work or not...
I'll keep in mind what you mean concerning the integral, but what is the question asking?

--Chris

6. Originally Posted by Chris L T521
I'll keep in mind what you mean concerning the integral, but what is the question asking?

--Chris
the question is:

evaluate

where $F$ = $4xz$i - $y^2$j + $yz$k and $S$ is the surface of the cube bounded by the planes $x$ = 0, $x$ = 1, $y$ = 0, $y$ = 1, $z$ = 0 and $z$ = 1/

7. Originally Posted by wik_chick88
the question is:

evaluate

where $F$ = $4xz$i - $y^2$j + $yz$k and $S$ is the surface of the cube bounded by the planes $x$ = 0, $x$ = 1, $y$ = 0, $y$ = 1, $z$ = 0 and $z$ = 1/
If your integral means the same thing as $\iint\limits_{\sigma}\bold F\cdot\bold n\,dS$, then we can make life a ton easier and apply the divergence theorem which states:

$\iint\limits_{\sigma}\bold F\cdot\bold n\,dS=\iiint\limits_G \text{div}\,\bold F \,dV$.

So in our case, $\iint\limits_{\sigma}\bold F\cdot\bold n\,dS=\iiint\limits_G \text{div}\,\bold F \,dV=\int_0^1\int_0^1\int_0^1\left(4z-y\right)\,dz\,dy\,dx$

--Chris

8. Originally Posted by Chris L T521
If your integral means the same thing as $\iint\limits_{\sigma}\bold F\cdot\bold n\,dS$, then we can make life a ton easier and apply the divergence theorem which states:

$\iint\limits_{\sigma}\bold F\cdot\bold n\,dS=\iiint\limits_G \text{div}\,\bold F \,dV$.

So in our case, $\iint\limits_{\sigma}\bold F\cdot\bold n\,dS=\iiint\limits_G \text{div}\,\bold F \,dV=\int_0^1\int_0^1\int_0^1\left(4z-y\right)\,dz\,dy\,dx$

--Chris
thanks so much Chris!! i got

$=\int_0^1\int_0^1\int_0^1\left(4z-y\right)\,dz\,dy\,dx$

= 1.5

is that right??

9. Originally Posted by wik_chick88
thanks so much Chris!! i got

$=\int_0^1\int_0^1\int_0^1\left(4z-y\right)\,dz\,dy\,dx$

= 1.5

is that right??
Yes!

--Chris