# Triple integrals solution check

1. May 11, 2015

### Sociomath

Are these correct?

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

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

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

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

2. May 11, 2015

### Staff: Mentor

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.

3. May 11, 2015

### mathman

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

4. May 13, 2015

### HallsofIvy

Staff Emeritus
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).