anonymous
  • anonymous
I click on Rbrass33's question about setting up an integral, and I get nothing but blank space. Something's terribly amiss.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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?
anonymous
  • anonymous
hey sorry i don't know what wrong
anonymous
  • anonymous
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

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anonymous
  • anonymous
Looking for volume of region?
anonymous
  • anonymous
yupp
anonymous
  • anonymous
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
anonymous
  • anonymous
We can set up the integral using either the disk method or the cylindrical shell method.
anonymous
  • anonymous
okay which one is easier
anonymous
  • anonymous
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.
anonymous
  • anonymous
The disk has radius z, so its volume dV is pr*z^2 dz. Since z goes from 0 to 1, \[V =\pi \int\limits_{0}^{1} z ^{2}dz.\]
anonymous
  • anonymous
so u don't need to set up a triple integral to find the volume?
anonymous
  • anonymous
What for? Keep it simple.
anonymous
  • anonymous
lol i like that
anonymous
  • anonymous
Even the cylindrical shell would be easier than the triple integral method. We could do it if we had to, though.
anonymous
  • anonymous
thanks for the help :)
anonymous
  • anonymous
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
anonymous
  • anonymous
would that still be equal to the volume= integral pi z^2 dz?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Yhe integral from 0 to 1, that is.

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