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abtrehearn

  • 4 years ago

I click on Rbrass33's question about setting up an integral, and I get nothing but blank space. Something's terribly amiss.

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  1. abtrehearn
    • 4 years ago
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    Is there supposed to be an integrand other than 1, or is the integral used to find the volume of the region? Is the integral supposed to be a single integral of is it supposed to be a multiple integral?

  2. Rbrass33
    • 4 years ago
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    hey sorry i don't know what wrong

  3. Rbrass33
    • 4 years ago
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    but i can post it here :) need help setting up integral. Let W be the region inside the cone z =sqrt(x^2 + y^2) and below the plane z = 1

  4. abtrehearn
    • 4 years ago
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    Looking for volume of region?

  5. Rbrass33
    • 4 years ago
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    yupp

  6. Rbrass33
    • 4 years ago
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    i was trying to use cylindrical coordinates and i had r is from 0 to 1, theta is from 0 to 2pi and z is from 0 to 1

  7. abtrehearn
    • 4 years ago
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    We can set up the integral using either the disk method or the cylindrical shell method.

  8. Rbrass33
    • 4 years ago
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    okay which one is easier

  9. abtrehearn
    • 4 years ago
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    Looks like disk method is easier. Dependent variable z goes from 0 1o 1, and we can partition it. In a particular subinterval of length dz is some particular z.

  10. abtrehearn
    • 4 years ago
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    The disk has radius z, so its volume dV is pr*z^2 dz. Since z goes from 0 to 1, \[V =\pi \int\limits_{0}^{1} z ^{2}dz.\]

  11. Rbrass33
    • 4 years ago
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    so u don't need to set up a triple integral to find the volume?

  12. abtrehearn
    • 4 years ago
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    What for? Keep it simple.

  13. Rbrass33
    • 4 years ago
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    lol i like that

  14. abtrehearn
    • 4 years ago
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    Even the cylindrical shell would be easier than the triple integral method. We could do it if we had to, though.

  15. Rbrass33
    • 4 years ago
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    thanks for the help :)

  16. Rbrass33
    • 4 years ago
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    i was just looking over the problem again and it says to evaluate the triple integral (1+sqrt(x^2+y^2) dV bound by W= the region inside the cone z =sqrt(x^2 + y^2) and below the plane z = 1

  17. Rbrass33
    • 4 years ago
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    would that still be equal to the volume= integral pi z^2 dz?

  18. abtrehearn
    • 4 years ago
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    Yes.

  19. abtrehearn
    • 4 years ago
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    Yhe integral from 0 to 1, that is.

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