Let sigma be the upper hemisphere with radius R. Use cartesian coordinates to write a triple integral that represents the volume V of sigma. Indicate both terminals on each integral?

- anonymous

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- anonymous

\[\int\limits_{-R}^{R}\int\limits_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\int\limits_{0}^{\sqrt{R^2-x^2-y^2}}1dzdydx\]
is this correct? because I'm getting 0 out of this.

- anonymous

hii nikola !!!

- anonymous

hey dan, whenever u integrate a constant, in this case 1, u get the variable here wrt x, u get an x similarly with respect to y & z.. are u familiar with basic integrals?? :)

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## More answers

- anonymous

u der??

- anonymous

The integrals looks right. You sure you're getting zero?

- anonymous

yes, I do familiar with the basic integral, but when I get 0 that means something wrong with the terminals

- anonymous

yeah I got 0 after I integrated it wrt to y

- anonymous

I have a question on how to do this as well. I'm learning this right now too.
How are you integrating with respect to y? You have y^2 in the sqrt and no y outside.

- anonymous

you got:
\[\int\limits_{-R}^{R}\int\limits_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} z dy dx\]
and you put in sqrt(R^2-x^2-y^2)
So:
\[\int\limits_{-R}^{R}\int\limits_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \sqrt{R^2-x^2-y^2} dy dx\]
then u integrate again wrt to y and when u put the terminals in, it becomes 0 :(

- anonymous

I don't know if this would help...but doing this in spherical coordinates would give you the answer of (R^2 * pi^2)/4
I'm pretty sure you can convert that to cartesian by changing the R^2, which, in spherical to Cartesian would be x^2 + y^2 + z^2 (sorry, i don't know how to do the formula thing to make it look pretty)

- anonymous

yeah, i know we can solve it via spherical coordinates, but the question ask me to do in cartesian and it's obvious the answer wouldn't be a nice and simple one

- anonymous

the general formula of spherical would be, \[x^{2}+y^{2}+z^{2} = r^{2}\]
use the equation box below the chat box :)

- anonymous

I'm going to try this. I'm pretty sure you're doing the integration wrt y wrong. I'm going to substitute (R^2 - x^2 - y^2)^.5 as u and integrate using substitution, it shouldn't be zero.
and lol, thanks, but I clicked it and I can't seem to make heads or tails out of the interface. so I'll stick to plain and simple text :D

- anonymous

correction: just R^2 - x^2 - y^2 as u

- anonymous

yup, i got
something that's not 0...
now let's see if i can't figure this equation thing out...

- anonymous

hahaha, ok i'll wait. let's see.

- anonymous

\[1/3 \int\limits\limits_{-R}^{R} (R ^{2} - x ^{2} - y ^{2})^{3/2} dz\]
where y = plus or minus (R^2 - x^2)
^-- i didn't plug it in yet and I dont know how to use the equations.... sighs

- anonymous

ahh..i see what you mean....

- anonymous

yeah? u got 0 after u plug the terminals in? that must be something wrong with the terminals, but I don't know which part :(

- anonymous

i think it's -1/3y instead of 1/3 because we use the chain rule of integral?

- anonymous

i don't believe so...but i'm not sure

- anonymous

what r u studying btw? vector calculus?

- anonymous

because if you substitute, you would just get \[\int\limits_{long equation}^{longequation} 1/2 \sqrt{u}du\]
which would be 1/2 * 2/3 * u ^ 3/2
which is 1/3 * u ^ 3/2
take the 1/3 out caz it's a constant and replace u with original equation.
I'm doing line integrals right now. and no one has helped me with my question T-T

- anonymous

ok, i have a solution
or rather, i found somewhere there's a solution
the question here is the same but it's a whole sphere, so the z value is different.
but the part where we are stuck, you need trig identities, which has always been one of my weakest areas
here's the link
http://www.physicsforums.com/showthread.php?t=386226

- anonymous

i still don't quite understand it....but I believe that the solution makes sense

- anonymous

yup but if you let
\[u = R^{2}-x{2}-y{2}\]
and
\[du/dy = -2y\]
so,
\[dy= du/-2y\]
then,
\[-1/2y\int\limits\limits_{}^{} u^{1/2} du\]

- anonymous

ok thanks i'll check it out

- anonymous

Yeah I think you're "more" right than I am.
I think i just broke every single rule of integration that I've ever learned.
But I don't think you can take out a variable like that...
because the y doesn't necessarily need to be the same inside and outside of the (u)^/5.
I looked that up as well and it says that you need trig identities again.
you have to substitute x, or y in our case, with 2 *cos(u) and onwards!
I'm completely lost now. :(
I have to go now, but I hope that site will help.
And also, if you have time, can you take a look at my problem? It's two questions below yours. The one that starts with a intimidating integral...and relating to line integrals
Good luck!

- anonymous

ok i'll try, that physics guy made me lost too haha.. thanks :)

- anonymous

got it, it's a very helpful link up there. :)

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