Question

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

Answer 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?

Answer 2

hey sorry i don't know what wrong

Answer 3

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

Answer 4

Looking for volume of region?

Answer 5

yupp

Answer 6

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

Answer 7

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

Answer 8

okay which one is easier

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

What for? Keep it simple.

Answer 13

lol i like that

Answer 14

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

Answer 15

thanks for the help :)

Answer 16

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

Answer 17

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

Answer 18

Yes.

Answer 19

Yhe integral from 0 to 1, that is.