CalcIII: Solving Double Integral with F(x,y,z)

[tdusffx]

Find
s\int\intF*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\intr^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.

 

[Defennder]

Uh, what is delG? What is G here?

 

[tdusffx]

Uh, what is delG? What is G here?

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

 

[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.

 

[tdusffx]

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.

how is it help me?

 

[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.

 

[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?

 

[tdusffx]

thanks euclid. Sorry for the several grammatical errors. I have 4 finals coming up tomorrow, tuesday, so I am sort of 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.