## 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

1. abtrehearn Group Title

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 Group Title

hey sorry i don't know what wrong

3. Rbrass33 Group Title

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 Group Title

Looking for volume of region?

5. Rbrass33 Group Title

yupp

6. Rbrass33 Group Title

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 Group Title

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

8. Rbrass33 Group Title

okay which one is easier

9. abtrehearn Group Title

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 Group Title

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 Group Title

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

12. abtrehearn Group Title

What for? Keep it simple.

13. Rbrass33 Group Title

lol i like that

14. abtrehearn Group Title

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

15. Rbrass33 Group Title

thanks for the help :)

16. Rbrass33 Group Title

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 Group Title

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

18. abtrehearn Group Title

Yes.

19. abtrehearn Group Title

Yhe integral from 0 to 1, that is.