1. Surface integral questions

I need to find the surface integrals for these various questions. Some I don't know where to begin, others I don't know how to continue. Help would be appreciated

I get: $\int \int F \cdot n dS$
$F = (x,y,2z)$
$n = \frac {2x}{\sqrt {4x^2 + 4y^2 + 1}} + \frac {2y}{\sqrt {4x^2 + 4y^2 + 1}} + \frac {1}{\sqrt {4x^2 + 4y^2 + 1}}$
$\int \int F \cdot n dS = \int \int 2 dA$

But not sure what the limits are to proceed

I'm guessing similar manner to question above, but do I get z on its own i.e. $z = 3-3x-\frac {3}{2}y$ then proceed. But once again, not sure what the limits would be

Using divF = 3 then $\int^1_0 \int^1_0 \int^1_0 3 dzdxdy = 3$ Is that right?

Need to use divergence again, but not sure how to proceed.

I could do the same as below, but not sure how to proceed

I got the curl to be i + j + k and $\sqrt { ( \frac {\partial f}{\partial x})^2 + ( \frac {\partial f}{\partial y})^2 + 1 } = \sqrt {3}$
Not sure how to proceed

2. I didn't read all your problems - way too many in one post! But I can tell you the first one is easily solved by the divergence theorem. Find $\mbox{div } F$ and integrate $\mbox{div }F\ dV$ inside the cube.

3. Originally Posted by godiva
I need to find the surface integrals for these various questions. Some I don't know where to begin, others I don't know how to continue. Help would be appreciated

I get: $\int \int F \cdot n dS$
$F = (x,y,2z)$
$n = \frac {2x}{\sqrt {4x^2 + 4y^2 + 1}} + \frac {2y}{\sqrt {4x^2 + 4y^2 + 1}} + \frac {1}{\sqrt {4x^2 + 4y^2 + 1}}$
$\int \int F \cdot n dS = \int \int 2 dA$

But not sure what the limits are to proceed

I'm guessing similar manner to question above, but do I get z on its own i.e. $z = 3-3x-\frac {3}{2}y$ then proceed. But once again, not sure what the limits would be

Using divF = 3 then $\int^1_0 \int^1_0 \int^1_0 3 dzdxdy = 3$ Is that right?

Need to use divergence again, but not sure how to proceed.

I could do the same as below, but not sure how to proceed

I got the curl to be i + j + k and $\sqrt { ( \frac {\partial f}{\partial x})^2 + ( \frac {\partial f}{\partial y})^2 + 1 } = \sqrt {3}$
Not sure how to proceed

for the second problem change it into polar and use the limits

$\int^{1}_{0}\int^{2\pi}_{0}d\theta dr$

for the 3rd one the limits are

$\int^{\frac{3}{2}}_{0} \int^{-2x+3}_{0} dydx$

Correct on 4

For question 5 find the divergence and then use

$\int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} drd\theta dz$

4. Originally Posted by 11rdc11
For question 5 find the divergence and then use

$\int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} drd\theta dz$
Thanks. For the above would an r need to be added? I remember something like that when i did coordinates, i.e. should it be $\int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} r drd\theta dz$

Also, any ideas for the final 2? I've gone through my book twice and it doesn't have a proper example

5. Originally Posted by godiva
Thanks. For the above would an r need to be added? I remember something like that when i did coordinates, i.e. should it be $\int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} r drd\theta dz$

Also, any ideas for the final 2? I've gone through my book twice and it doesn't have a proper example
Yes for the 6th one remember what Stokes theorem says. It does not matter what surface you use as long as C is the same. In this case it be alot easier to use the surface of z=1 with the domain being projected to the xy plane and you get the integral

$\int^{1}_{0} \int^{2\pi}_{0} d\theta dr$

Another way to do 6 is let

$r(t)= \cos{t}i +\sin{t}j + k$

$f(r(t)) = \cos^2{t}i + \sin^2{t}j +k$

$r'(t) = -\sin{t}i +\cos{t}j + 0k$

Now just evaluate the line integral

6. Originally Posted by godiva
Thanks. For the above would an r need to be added? I remember something like that when i did coordinates, i.e. should it be $\int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} r drd\theta dz$

Also, any ideas for the final 2? I've gone through my book twice and it doesn't have a proper example

And for the last one I got a different curl than you did.

Mine was

$-i +j -k$

so

$\int \int [-Mf_x - Nf_y +P]dxdy$

f_x = -1

f_y = -1

$\int^{1}_{0} \int^{1-x}_{0} dydx$