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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    we are to assume r is a constant correct?

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    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 years ago
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    http://www.youtube.com/watch?v=k7ND70gFTLU&feature=related

  5. anonymous
    • 5 years ago
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    http://www.youtube.com/watch?v=9FulyvBhdkw&feature=related

  6. anonymous
    • 5 years ago
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    that should help. You can use the polar coordinates too.

  7. anonymous
    • 5 years ago
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    the thing is for this section we haven't learned the polar coordinates, it would make this problem somewhat easier i feel

  8. anonymous
    • 5 years ago
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    well, then, just go with the first video. That should help you evaluate multi variable double integrals.

  9. anonymous
    • 5 years ago
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    okay thank you very much

  10. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    I would think that since you are calculating volume, you should have a triple integral.

  12. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    chaguanas, why are you assuming r = 1?

  18. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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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