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

1. abtrehearn

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?

2. Rbrass33

hey sorry i don't know what wrong

3. Rbrass33

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

4. abtrehearn

Looking for volume of region?

5. Rbrass33

yupp

6. Rbrass33

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

7. abtrehearn

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

8. Rbrass33

okay which one is easier

9. abtrehearn

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.

10. abtrehearn

The disk has radius z, so its volume dV is pr*z^2 dz. Since z goes from 0 to 1, $V =\pi \int\limits_{0}^{1} z ^{2}dz.$

11. Rbrass33

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

12. abtrehearn

What for? Keep it simple.

13. Rbrass33

lol i like that

14. abtrehearn

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

15. Rbrass33

thanks for the help :)

16. Rbrass33

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

17. Rbrass33

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

18. abtrehearn

Yes.

19. abtrehearn

Yhe integral from 0 to 1, that is.