Here's the question you clicked on:
abtrehearn
I click on Rbrass33's question about setting up an integral, and I get nothing but blank space. Something's terribly amiss.
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?
hey sorry i don't know what wrong
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
Looking for volume of region?
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
We can set up the integral using either the disk method or the cylindrical shell method.
okay which one is easier
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.
The disk has radius z, so its volume dV is pr*z^2 dz. Since z goes from 0 to 1, \[V =\pi \int\limits_{0}^{1} z ^{2}dz.\]
so u don't need to set up a triple integral to find the volume?
What for? Keep it simple.
Even the cylindrical shell would be easier than the triple integral method. We could do it if we had to, though.
thanks for the help :)
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
would that still be equal to the volume= integral pi z^2 dz?
Yhe integral from 0 to 1, that is.