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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we are to assume r is a constant correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Find 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that should help. You can use the polar coordinates too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the thing is for this section we haven't learned the polar coordinates, it would make this problem somewhat easier i feel

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, then, just go with the first video. That should help you evaluate multi variable double integrals.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay thank you very much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would think that since you are calculating volume, you should have a triple integral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.youtube.com/watch?v=AQHcZklcltI That video and the part 2 of it explain how to calculate volumes bounded by two regions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Scratch that. I've got it. \[x ^{2}+y ^{2}=r ^{2} \] \[x ^{2}+y ^{2}=1\] That's one surface without r.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Other surface \[y ^{2}+z ^{2}=r ^{2}\] \[y ^{2}+z ^{2}=1\] \[z ^{2}=1y ^{2}\] \[z =\sqrt{1y ^{2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Other surface \[y ^{2}+z ^{2}=r ^{2}\] \[y ^{2}+z ^{2}=1\] \[z ^{2}=1y ^{2}\] \[z =\sqrt{1y ^{2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0chaguanas, why are you assuming r = 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your 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 (1y^2)
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