## anonymous 5 years ago double intergral, multi variable $\[\int\limits\limits_{-r}^{r}\int\limits\limits_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\ \sqrt{r^2-y^2} dx dy$

1. anonymous

we are to assume r is a constant correct?

2. anonymous

yes

3. anonymous

Find the volume of the solid bounded by the cylinders x^2+y^2=r^2 and y^2+z^2=r^2, and That is how i formed the intergral, if that helps anybody

4. anonymous
5. anonymous
6. anonymous

that should help. You can use the polar coordinates too.

7. anonymous

the thing is for this section we haven't learned the polar coordinates, it would make this problem somewhat easier i feel

8. anonymous

well, then, just go with the first video. That should help you evaluate multi variable double integrals.

9. anonymous

okay thank you very much

10. anonymous

also, he explains how to calculate double integrals for regions bounded by two curves. Keep that in mind and see if you have done it that way to evaluate your problem.

11. anonymous

I would think that since you are calculating volume, you should have a triple integral.

12. anonymous

http://www.youtube.com/watch?v=AQHcZklcltI That video and the part 2 of it explain how to calculate volumes bounded by two regions.

13. anonymous

OK, if you are not using polar coordinates you have to get rid of r. $r ^{2}=x ^{2}+y ^{2}$ Convert in your original eq see if you have something to work with.

14. anonymous

Scratch that. I've got it. $x ^{2}+y ^{2}=r ^{2}$ $x ^{2}+y ^{2}=1$ That's one surface without r.

15. anonymous

Other surface $y ^{2}+z ^{2}=r ^{2}$ $y ^{2}+z ^{2}=1$ $z ^{2}=1-y ^{2}$ $z =\sqrt{1-y ^{2}}$

16. anonymous

Other surface $y ^{2}+z ^{2}=r ^{2}$ $y ^{2}+z ^{2}=1$ $z ^{2}=1-y ^{2}$ $z =\sqrt{1-y ^{2}}$

17. anonymous

chaguanas, why are you assuming r = 1?

18. anonymous

It is equal to $x ^{2}+y ^{2}=1$And I can also derive it by converting one of the equations to $r ^{2}\cos ^{2}\theta + r ^{2}\sin ^{2}\theta = r ^{2}$ Factor the r out gives me equivalent r=1

19. anonymous

Your region is the top semi circle of 1 which makes your x go from -1 to +1; your y goes from 0 to square root of (1-y^2)