# Homework Help: CalcIII questions

1. Dec 8, 2007

### tdusffx

Find
$$s\int\int$$F*Nds

F(x,y,z) = x*i + y*j - 2z*k

S: Z = $$\sqrt{}a^2 - x^2 -y^2$$

I solved for delG and doted F to del G. then I converted it polar since I had x^2 + y^2

I got:

$$\int\int$$r^3/(a^2-r^2)^(1/2)drdo - 2$$\int\int$$(a^2-r^2)rdrdo

I evaluated the double integral and got 4/3pi(a^2+a^3)

sorry for the shortcut guys, its just so much stuff to type if I try to some the work. Plz, tell me if my answer is wrong.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 9, 2007

### Defennder

Uh, what is delG? What is G here?

3. Dec 9, 2007

### tdusffx

del G is fxi + fyj + fzk which is

del G = x/sqrt(a^2-x^2-y^2)*i + y/sqrt(a^2-x^2-y^2)*j -2k

then the formula says I have to dot del G to F(x,y,z) which gives me

(x/sqrt(a^2-x^2-y^2)*i + y/sqrt(a^2-x^2-y^2)*j -2k) * (x*i + y*j - 2z*k)

delG*F = (x^2+y^2)/(sqrt(a^2-x^2-y^2) -2sqrt(a^2-x^2-y^2)

then I converted it to polar to make the integral easier

$$\int\int$$delG*Fds = $$\int\int$$(r^2/(a^2-r^2) - 2sqrt(a^2-r^2))rdrd$$\Theta$$

4. Dec 9, 2007

### HallsofIvy

What you have failed to say is that your surface is given by G(x,y,z)= $\sqrt{a^2- x^2- y^2}- z$= 0.

5. Dec 9, 2007

### tdusffx

how is it help me?

6. Dec 9, 2007

### tdusffx

oh, ya. I remember my question now. Thanks for pointing out the surface. Actually I got up to that part...z-sqrt(a^2-x^2-y^2)..my question was the limits of integrating in this problem. Thanks for pointing out the surface euclide, however, I knew what the surface was its just the graphing, and finding the limits of integrating I am worrying about.

7. Dec 10, 2007

### HallsofIvy

It helps us by telling us what you are talking about! You initially said "I solved for delG and doted F to del G" without saying what you meant by "G"!

Defennnder asked "Uh, what is delG? What is G here?" and you still did not say what G was.

$z= \sqrt{a^2- x^2- y^2}$ is the upper half of the sphere $x^2+ y^2+ z^2= a^2$ Since you don't say anything about integrating over a portion of that, I must assume (hesitantly, given your reluctance to clearly state the entire problem) that you are to integrate over the entire hemisphere. You might want to $x= 2cos(\theta)sin(\phi)$, $y= 2sin(\theta)sin(\phi)$, $z= 2cos(\phi)$, spherical coordinates with $\rho$ set equal to 2, as parameters. To cover the entire sphere we take $\theta= 0$ to $2\pi$, $\phi= 0 to \pi$. What should the limits be to get only the upper hemisphere?

8. Dec 10, 2007

### tdusffx

thanks euclid. Sorry for the several grammatical errors. I have 4 finals coming up tomorrow, tuesday, so I am sorta stressng a bit. And not to mention that it is my first semester in college. ;D

anyways, I have to convert it to spherical? hmm, I usually convert things into polar coordinates, and my instructor rushed the discussion on the last 3 sections of our text, which are the surface integrals, divergence and stoke's. I'm really having troule in those sections (mostly algebraic problems). and thanks for stating the regions of integration.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook