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math1234

  • 3 years ago

Find the volume between two intersecting cylinders x^2+y^2=r^2 and y^2+z^2=r^2 using polar coordinates and double integrals only.

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  1. TuringTest
    • 3 years ago
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    |dw:1352144686186:dw|

  2. TuringTest
    • 3 years ago
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    in the xy-plane|dw:1352144919394:dw|\[-\sqrt{r^2-y^2}\le x\le\sqrt{r^2-y^2}\]in the yz-plane|dw:1352145089843:dw|\[-\sqrt{r^2-y^2}\le z\le\sqrt{r^2-y^2}\]leaving\[-r\le y\le r\]hm...

  3. TuringTest
    • 3 years ago
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    in the xz-plane the intersection is square with sides=2r|dw:1352145471598:dw|

  4. TuringTest
    • 3 years ago
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    I think the integral is\[\iint\limits_Dr\sin\theta dA=\int_0^{2\pi}\int_0^r r^2\sin\theta drd\theta\]can you check somewhow?

  5. TuringTest
    • 3 years ago
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    actually it would have to be\[\iint\limits_Dr\sin\theta dA=2\int_0^\pi\int_0^r r^2\sin\theta drd\theta\]to avoid getting zero

  6. math1234
    • 3 years ago
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    But the integrand should be r^2 * cos(theta) ?

  7. math1234
    • 3 years ago
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    since i am integrating z = sqrt(z^2-y^2)

  8. math1234
    • 3 years ago
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    so in polar i thought it was sqrt(r^2-y^2) = x = rcos(theta)

  9. TuringTest
    • 3 years ago
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    yeah I just came to that conclusion as well

  10. TuringTest
    • 3 years ago
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    I guess I was right the first time

  11. math1234
    • 3 years ago
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    but if I integrate with rcos(theta), I end up with zero in the end

  12. math1234
    • 3 years ago
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    I will get sin(2*pi) - sin(0) = 0.

  13. TuringTest
    • 3 years ago
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    Hm.. now I am not sure again. No way to check I presume...

  14. TuringTest
    • 3 years ago
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    @mahmit2012 double integral help

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