anonymous
  • anonymous
double intergral, multi variable \[\[\int\limits\limits_{-r}^{r}\int\limits\limits_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\ \sqrt{r^2-y^2} dx dy\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
we are to assume r is a constant correct?
anonymous
  • anonymous
yes
anonymous
  • 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

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anonymous
  • anonymous
http://www.youtube.com/watch?v=k7ND70gFTLU&feature=related
anonymous
  • anonymous
http://www.youtube.com/watch?v=9FulyvBhdkw&feature=related
anonymous
  • anonymous
that should help. You can use the polar coordinates too.
anonymous
  • anonymous
the thing is for this section we haven't learned the polar coordinates, it would make this problem somewhat easier i feel
anonymous
  • anonymous
well, then, just go with the first video. That should help you evaluate multi variable double integrals.
anonymous
  • anonymous
okay thank you very much
anonymous
  • 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.
anonymous
  • anonymous
I would think that since you are calculating volume, you should have a triple integral.
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
Scratch that. I've got it. \[x ^{2}+y ^{2}=r ^{2} \] \[x ^{2}+y ^{2}=1\] That's one surface without r.
anonymous
  • anonymous
Other surface \[y ^{2}+z ^{2}=r ^{2}\] \[y ^{2}+z ^{2}=1\] \[z ^{2}=1-y ^{2}\] \[z =\sqrt{1-y ^{2}}\]
anonymous
  • anonymous
Other surface \[y ^{2}+z ^{2}=r ^{2}\] \[y ^{2}+z ^{2}=1\] \[z ^{2}=1-y ^{2}\] \[z =\sqrt{1-y ^{2}}\]
anonymous
  • anonymous
chaguanas, why are you assuming r = 1?
anonymous
  • 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
anonymous
  • 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)

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