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anonymous
 5 years ago
I click on Rbrass33's question about setting up an integral, and I get nothing but blank space. Something's terribly amiss.
anonymous
 5 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is 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
 5 years ago
Best ResponseYou've already chosen the best response.0hey sorry i don't know what wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Looking for volume of region?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0We can set up the integral using either the disk method or the cylindrical shell method.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay which one is easier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Looks 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
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0so u don't need to set up a triple integral to find the volume?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What for? Keep it simple.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Even the cylindrical shell would be easier than the triple integral method. We could do it if we had to, though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for the help :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0would that still be equal to the volume= integral pi z^2 dz?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yhe integral from 0 to 1, that is.
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