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abtrehearn
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I click on Rbrass33's question about setting up an integral, and I get nothing but blank space. Something's terribly amiss.
 3 years ago
 3 years ago
abtrehearn Group Title
I click on Rbrass33's question about setting up an integral, and I get nothing but blank space. Something's terribly amiss.
 3 years ago
 3 years ago

This Question is Closed

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
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?
 3 years ago

Rbrass33 Group TitleBest ResponseYou've already chosen the best response.0
hey sorry i don't know what wrong
 3 years ago

Rbrass33 Group TitleBest ResponseYou've already chosen the best response.0
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
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
Looking for volume of region?
 3 years ago

Rbrass33 Group TitleBest ResponseYou've already chosen the best response.0
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
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
We can set up the integral using either the disk method or the cylindrical shell method.
 3 years ago

Rbrass33 Group TitleBest ResponseYou've already chosen the best response.0
okay which one is easier
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
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.
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
The disk has radius z, so its volume dV is pr*z^2 dz. Since z goes from 0 to 1, \[V =\pi \int\limits_{0}^{1} z ^{2}dz.\]
 3 years ago

Rbrass33 Group TitleBest ResponseYou've already chosen the best response.0
so u don't need to set up a triple integral to find the volume?
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
What for? Keep it simple.
 3 years ago

Rbrass33 Group TitleBest ResponseYou've already chosen the best response.0
lol i like that
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
Even the cylindrical shell would be easier than the triple integral method. We could do it if we had to, though.
 3 years ago

Rbrass33 Group TitleBest ResponseYou've already chosen the best response.0
thanks for the help :)
 3 years ago

Rbrass33 Group TitleBest ResponseYou've already chosen the best response.0
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
 3 years ago

Rbrass33 Group TitleBest ResponseYou've already chosen the best response.0
would that still be equal to the volume= integral pi z^2 dz?
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
Yhe integral from 0 to 1, that is.
 3 years ago
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