Question

double intergral, multi variable
\[\[\int\limits\limits_{-r}^{r}\int\limits\limits_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\ \sqrt{r^2-y^2} dx dy\]

Answer 1

we are to assume r is a constant correct?

Answer 2

yes

Answer 3

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

Answer 4

http://www.youtube.com/watch?v=k7ND70gFTLU&feature=related

Answer 5

http://www.youtube.com/watch?v=9FulyvBhdkw&feature=related

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

okay thank you very much

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

chaguanas, why are you assuming r = 1?

Answer 18

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

Answer 19

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)