## anonymous 5 years ago 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?

1. 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.

2. anonymous

hii nikola !!!

3. 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?? :)

4. anonymous

u der??

5. anonymous

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

6. anonymous

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

7. anonymous

yeah I got 0 after I integrated it wrt to y

8. 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.

9. 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 :(

10. 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)

11. 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

12. anonymous

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

13. 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

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. 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

18. anonymous

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

19. 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 :(

20. anonymous

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

21. anonymous

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

22. anonymous

what r u studying btw? vector calculus?

23. 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

24. 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

25. anonymous

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

26. anonymous

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

27. anonymous

ok thanks i'll check it out

28. 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!

29. anonymous

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

30. anonymous