Are These Triple Integrals Set Up Correctly?

[Sociomath]

Are these correct?
Thanks in advance!

1.) Set up the triple integral for $f(x,y,z) = xy + 2xz$ on the region $0 ≤ x ≤4, 0 ≤ y ≤ 2$ and $0 ≤ x ≤ 3xy + 1$.

$\displaystyle \int_0^4 \int_0^2 \int_0^{3xy+1} 2y +2xz\ dz\ dy\ dx$

\text{2.) Set up the triple integral in cylindrical coordinates to find the volume bounded by}\\ <br /> z = x^2 + y^2, z = 0, x^2 + y^2 = 1\ \text{and}\ x^2 + y^2 = 4.

$\displaystyle \int_1^2 \int_0^{2\pi} \int_0^{\sqrt{2}} r^2 r\ dr\ d\theta\ dz$

 

[mfb]

How did xy become 2y in the first integral?

I don't understand how you set up the second one, it does not look right.

 

[mathman]

The first one looks funny. Did you mean 0 \le z \leq 3xy+1 ?

 

[HallsofIvy]

First, yes, in problem 1, you mean "0\le z\le 3xy+ 1". But the integral is set up correctly.

In problem 2, there is no good reason to write "r^2r" rather than "r^3".
More importantly, the upper bound on the figure is z= x^2+ y^2 which is z= r^2 in cylindrical coordinates.

What you have, \int_1^2\int_0^{2\pi}\int_0^\sqrt{2} r^3 drd\theta dz, would be the integral of r^2 over the cylinder whose base is a circle with center at (0, 0), radius \sqrt{2}, and extending from z= 1 to z= 2. What you want is z to go from 0 go r^2 (so the z-integral will have to be inside the r-integral).