# Requesting help for this simple triple integral..

#### zollen

The region is bounded by x^2+y^2=3, z=-1, z=2
$$\displaystyle \int_{}^{} \int_{}^{} \int_{}^{} y dxdydz$$

I managed to setup the following boundary, but it yielded zero
$$\displaystyle \int_{-1}^{2} \int_{-sqrt(3)}^{sqrt(3)} \int_{-sqrt(3-y^2)}^{sqrt(3-y^2)} y dxdydz$$

so.. I reshape the boundaries to avoid getting zero...

$$\displaystyle \int_{-1}^{2} \int_{0}^{sqrt(3)} \int_{0}^{sqrt(3-y^2)} 4 * y dxdydz$$

This new boundaries yields 12 * sqrt(3).

But to my surprise the correct answer is 6 * sqrt(3).

Any idea where I got wrong??

#### romsek

Math Team
The integrand doesn't depend on $z$ at all, so this can be reduced to

$\displaystyle I = 3 \int \int_A ~y ~dx~dy$

where $A$ is the disk $x^2+y^2 \leq 3$

We can easily convert this polar coordinates to obtain

\begin{align*} \displaystyle I &= 3 \int_0^{2\pi} \int_0^3~r^2 \sin(\theta)~dr~d\theta \\ \\ &= 27 \int_0^{2\pi}~\sin(\theta)~d\theta = 0 \end{align*}

You've got a typo somewhere.

Last edited by a moderator:
1 person