Requesting help for this simple triple integral..

Jan 2017
209
3
Toronto
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
Sep 2015
2,959
1,673
USA
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.
 
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