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

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- 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

hey sorry i don't know what wrong

- 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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## More answers

- anonymous

Looking for volume of region?

- anonymous

yupp

- 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

We can set up the integral using either the disk method or the cylindrical shell method.

- anonymous

okay which one is easier

- 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

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

so u don't need to set up a triple integral to find the volume?

- anonymous

What for? Keep it simple.

- anonymous

lol i like that

- anonymous

Even the cylindrical shell would be easier than the triple integral method. We could do it if we had to, though.

- anonymous

thanks for the help :)

- 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

would that still be equal to the volume= integral pi z^2 dz?

- anonymous

Yes.

- anonymous

Yhe integral from 0 to 1, that is.

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